From meyer@imap.theory.csail.mit.edu  Mon Nov 28 14:37:57 2005
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From: Praveen Pamidimukkala <mukkala@MIT.EDU>
Subject: [Sayan] Week 12 Comments
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In section 4.3 of the Course Notes on Probability, disjointness is 
mentioned in comparison to independence.  I understand that mathematically, 
when two events are disjoint, Pr{AintB} = 0 as opposed to Pr{AintB} = Pr{A} 
x Pr{B}, which applies for independence.  However, in terms of events, how 
could this be described?

Thanks,
Praveen Pamidimukkala


From meyer@imap.theory.csail.mit.edu  Mon Nov 28 18:12:32 2005
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Subject: David-tp13 reading comment
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Hey,

Reading and lecture were clear.  I found the ability to get a percentage 
confidence for sampling data to be interesting.

- Steven Zhou


From meyer@imap.theory.csail.mit.edu  Mon Nov 28 19:19:41 2005
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Date: Mon, 28 Nov 2005 19:19:37 -0500
From: Isaac Charny <icharny@MIT.EDU>
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Subject: [Sayan] Week 13 Comments
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I don't understand binomial distribution (section 3.4, page 11). Please discuss
this in the next lecture.

~Isaac Charny

From meyer@imap.theory.csail.mit.edu  Mon Nov 28 19:51:40 2005
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The section on polling was very confusing, I haven't quite understood it 
yet.  I would like this to be explained better in class.  There are a 
lot of variables and equations and it's hard to follow.  I forgot what 
the goal of all this was by the fourth paragraph, and had to reread it a 
couple of times to understand why each equation brought us closer to 
what we wanted. 

Also, in the tutor problems, there was a question that included the 
concept of variance, but this was not mentioned in the notes (or at 
least I couldn't find it).  This made it a lot more difficult to solve 
the problem. 

Adriana

From meyer@imap.theory.csail.mit.edu  Mon Nov 28 22:02:53 2005
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Subject: [Sayan] Week 13 Comments
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X-Keywords: NotJunk                                                                                            

I would like the approximation methods of page 13 explained more
clearly.

From meyer@imap.theory.csail.mit.edu  Mon Nov 28 23:02:17 2005
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Will we be going over those horrible equations for confidence in class? 
They are still pretty confusing even in the notes.

Thanks,
Andrew
----------------------------
Illegitmitatum Non Carborundum Est
Andrew Shafer, MIT Blog
http://shaferandrew.blogspot.com
Si hoc legere scis numium eruditionis habes.
----------------------------


From meyer@imap.theory.csail.mit.edu  Tue Nov 29 01:01:10 2005
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From: "Chris Yang" <yangc@MIT.EDU>
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Subject: [David] Week 13 Comments
Date: Tue, 29 Nov 2005 01:03:06 -0500
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For the derivation of the confidence interval stuff, why can we assume p =
=3D =BD
to continue the calculations for delta?  Is it because delta is like the
upper bound on the error, and not the error itself?

=20

Thanks,

Chris Yang=20


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<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'>For the derivation of the confidence interval stuff, =
why can
we assume p =3D =BD to continue the calculations for delta?=A0 Is it =
because delta is
like the upper bound on the error, and not the error =
itself?<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
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<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
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<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
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font-family:Arial'>Chris Yang <o:p></o:p></span></font></p>

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From meyer@imap.theory.csail.mit.edu  Tue Nov 29 01:47:52 2005
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To: 6042-probs@theory.lcs.mit.edu
From: Hamidou Soumare <hsoumare@MIT.EDU>
Subject: {Sayan} Weekly Email Comment
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X-Keywords: NonJunk NotJunk                                                                                            

Hi,

I didnt quite understand the section on polling and would like to  
have it gone over in class in more detail, especially the part about  
the confidence interval and its derivation. Thanks.

Hamidou

From meyer@imap.theory.csail.mit.edu  Tue Nov 29 02:19:04 2005
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Variance crept in by accident; will be covered next week.  I've revised 
the tutor problem to indicate this.  Thanks for pointing this out.

regards, A.

Adriana Lopez wrote:

> The section on polling was very confusing, I haven't quite understood 
> it yet.  I would like this to be explained better in class.  There are 
> a lot of variables and equations and it's hard to follow.  I forgot 
> what the goal of all this was by the fourth paragraph, and had to 
> reread it a couple of times to understand why each equation brought us 
> closer to what we wanted.
> Also, in the tutor problems, there was a question that included the 
> concept of variance, but this was not mentioned in the notes (or at 
> least I couldn't find it).  This made it a lot more difficult to solve 
> the problem.
> Adriana



From meyer@imap.theory.csail.mit.edu  Tue Nov 29 02:28:27 2005
Message-ID: <438C031A.4040302@csail.mit.edu>
Date: Tue, 29 Nov 2005 02:28:26 -0500
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Subject: Re: [Sayan] Week 12 Comments
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Are you asking what independent events "look like"?  Best answer I can 
give to that is the picture in section 4.3.

By the way, happy to have you reviewing Notes 12, but this week's email 
is supposed to be about Notes 13.

regards, A.

Praveen Pamidimukkala wrote:

> In section 4.3 of the Course Notes on Probability, disjointness is 
> mentioned in comparison to independence.  I understand that 
> mathematically, when two events are disjoint, Pr{AintB} = 0 as opposed 
> to Pr{AintB} = Pr{A} x Pr{B}, which applies for independence.  
> However, in terms of events, how could this be described?
>
> Thanks,
> Praveen Pamidimukkala
>


From ksindi@MIT.EDU Tue Nov 29 02:36:40 2005
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I didn't like the binomial sampling part. Maybe we could list the assumptions on
it made in class. I also still don't understand what confidence levels are. I
found the random varaibles bit to be very clear.

From meyer@imap.theory.csail.mit.edu  Tue Nov 29 01:17:29 2005
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Indeed, you are correct.  Thanks for catching this.  The TP has been 
adjusted accordingly.  Let me know if you need to redo this problem.

DS

Kevin Wang wrote:

> Hi,
>
> I think 8. has a typo: "The probability that the second voter chosen 
> will favor Gore, given that the *first *voter chosen is from the same 
> state as the first, may not equal /p/."
>
> It should be *second* right?
>
> Thanks,
> Kevin

From lye@MIT.EDU Tue Nov 29 11:25:25 2005
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3.3 The Numbers Game

This was especially counterintuitive, and thus difficult. It seems
strange that by guessing (assuming information you do not know), you
would improve your chances.



From pgroudas@MIT.EDU Tue Nov 29 16:11:34 2005
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I found it difficult to really absorb everything past page 14, it just seemed
that once it got to polling it started throwing so much math around that it was
too much to get all at once.  Hopefully this will all get cleared up in class.

-Paul Groudas

From nedzel@MIT.EDU Tue Nov 29 18:44:29 2005
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I'm fairly comfortable with this material, I've seen most of it in 
6.041. It might be nice to have a refresher during lecture on the 
expectation & variance of different distributions.

- David

From brevzin@MIT.EDU Tue Nov 29 20:13:57 2005
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You guys go over a lot of definitions for expectation, but what about the 
more general one ?

Let f: X --> R+ be a random variable, the expection is integral over X of f 
du, where u (mu) is the measure.

Also, this entire section will be extremely weird since in 18.103 (Fourier 
Analysis, Measure Theory) we learn the measure theory side which has its 
own explanations and descriptions... and symbols. The overlap will be 
confusing.

Barry


From mwangi@MIT.EDU Tue Nov 29 20:26:12 2005
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Passage:  2  The Birthday Principle
Page:       4
I found this passage most surprising because of the large probability of 
two people in a room with n people having the same birthday given that 
there are d days in a year. It seemed very counter-intuitive.

Sincerely,
Timothy M. Mwangi 


From bens@MIT.EDU Tue Nov 29 21:02:44 2005
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This number-picking game is very cool.
I'm surprised that the best strategy isn't just guess that the numbers 
span 50.
I guess it has to do with the envelope-giver guessing your strategy
and working around it.

From jehan@MIT.EDU Tue Nov 29 21:11:01 2005
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I thought the mean time to failure thing was pretty cool.
no questions

jehan


From aston@MIT.EDU Tue Nov 29 22:02:29 2005
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From: "Aston Motes" <aston@MIT.EDU>
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Pages 12 and 13 mention a closed form formula for the binomial density
function approximation, but I was extremely confused by the lack of
rationalization for elements in the formula. I suppose we can take it on
faith, but I feel kind of uncomfortable about that.

 

            - Aston


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<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'>Pages 12 and 13 mention a closed form formula for the
binomial density function approximation, but I was extremely confused by =
the
lack of rationalization for elements in the formula. I suppose we can =
take it
on faith, but I feel kind of uncomfortable about =
that.<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'><o:p>&nbsp;</o:p></span></font></p>

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style=3D'font-size:10.0pt;
font-family:Arial'>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=
&nbsp;&nbsp; -
Aston<o:p></o:p></span></font></p>

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From hectorb@MIT.EDU Tue Nov 29 22:05:01 2005
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5.1 Equivalent Definitions of Expectation
"There are some other ways of writing the definition of expectation. 
Sometimes using one
of these other formulations can make computing an expectation a lot 
easier. One option
is to group together all outcomes on which the random variable takes on 
the same value." p. 18-19

This section was not so clear and I would like to see it explained in 
lecture.

From meyer@imap.theory.csail.mit.edu  Tue Nov 29 22:17:32 2005
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18.103 is not a prereq for 6.042.  I expect few 6.042 students go on to 
take it later, and the consideration of using consistent notation has 
never come up.

Since all sample spaces in 6.042 are discrete (countable), we do fine 
using summations and avoiding the complexities of measure theory, most 
notably the existence of non-measurable sets. On the other hand, 
everything we do is consistent with the measure-theoretic perspective.

I hope the conflicting notation does not burden you unduly.

regards, A.

Barry Revzin wrote:
> You guys go over a lot of definitions for expectation, but what about 
> the more general one ?
> 
> Let f: X --> R+ be a random variable, the expection is integral over X 
> of f du, where u (mu) is the measure.
> 
> Also, this entire section will be extremely weird since in 18.103 
> (Fourier Analysis, Measure Theory) we learn the measure theory side 
> which has its own explanations and descriptions... and symbols. The 
> overlap will be confusing.
> 
> Barry
> 


From avalys@MIT.EDU Tue Nov 29 22:34:12 2005
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From: Alex Valys <avalys@MIT.EDU>
Subject: 6.042: [Hanson] Required Reading Comments
Date: Tue, 29 Nov 2005 22:34:07 -0500
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I am a little bit unclear about how to actually apply the section on  
expected value to actual problems.  Some examples would be nice.

Alex Valys


From ridell@MIT.EDU Tue Nov 29 22:46:43 2005
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Can you explain more about the probability density function on page 6?

Rebecca Idell

-- 
Rebecca Idell
Massachusetts Institute of Technology
Electrical Engineering and Computer Science, Class of 2007

479 Commonwealth Avenue
Boston, MA 02215
(617) 875-0889

From fluff@MIT.EDU Tue Nov 29 22:51:21 2005
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Subject: [Jelani] week 13 reading
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Everything starting from the binomial density approximation onwards  
went completely over my head. I think I might just have a 10-page-per- 
day math comprehension limit. The expected value stuff seemed  
understandable with the six-sided die example, but I didn't get  
anything after that (starting from 5.1).

~Crystal 

From sil_03@MIT.EDU Tue Nov 29 23:14:58 2005
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It is interesting that the population of the country has no effect on poll size!
 I don't see how.  I couldn't really follow the Binomial Sampling theorem so it
would be nice if we went over it again in class.

From dowgun@MIT.EDU Wed Nov 30 00:00:52 2005
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Right, so this week's lecture notes introduced a lot of math that I can sort of
follow, but I doubt if I'll be able to remember. Thus, my question for the
week: do we need to remember how to derive the following formulas?
e^(-n(n-1)/2d)
The upper bound for PDF (#1 on page 13)
The upper bound for CDF (bottom of 13)
The formula for delta (#3 on page 15) and
The binomial sampling equation.

Thanks, Neil

From ozcan@MIT.EDU Wed Nov 30 00:17:55 2005
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Subject: [hanson]reading comments
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The section on confidence levels, and how we might interpret the confidence 
levels wrong, is quite interesting.

Section 4: Confidence Levels

Suppose a pollster uses a sample of 662 random voters to estimate the 
fraction of voters
who prefer Clinton, and the pollster finds that 364 of them prefer Clinton. 
It’s tempting,
but sloppy, to say that this means:
False Claim. With probability 0.95, the fraction, p, of voters who prefer 
Clinton is 364/662±0.04.
Since 364/662-0.04 > 0.50, there is a 95% chance that more than half the 
voters prefer Clinton.
What’s objectionable about this statement is that it talks about the 
probability or “chance”
that a real world fact is true, namely that the actual fraction, p, of 
voters favoring Clinton
is more than 0.50. But p is what it is, and it simply makes no sense to 
talk about the
probability that it is something else. For example, suppose p is actually 
0.49; then it’s
nonsense to ask about the probability that it is within 0.04 of 364/662 —it 
simply isn’t.
A more careful summary of what we have accomplished goes this way:
We have described a probabilistic procedure for estimating the value of the
actual fraction, p. The probability that our estimation procedure will 
yield a value
within 0.04 of p is 0.95.
This is a bit of a mouthful, so special phrasing closer to the sloppy 
language is commonly
used. The pollster would describe his conclusion by saying that
At the 95% confidence level, the fraction of voters who prefer Clinton is 
364/662± 0.04.
It’s important to remember that confidence levels refer to the results of 
estimation procedures
for real-world quantities. The real-world quantity being estimated is typically
unknown, but fixed; it is not a random variable, so it makes no sense to 
talk about the
probability that it has some property.



From jjmonzon@MIT.EDU Wed Nov 30 00:34:15 2005
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Subject: [David] - reading comments
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This subject matter is a bit complex but I like the examples given - 
some of them are a bit counterintuitive and confusing like the 
birthday principle. I liked the "making baby girls" example though - was funny.

Josh


Joshua Jen C. Monzon
Massachusetts Institute of Technology
Electrical Engineering with Computer Science
jjmonzon@mit.edu   617-803-7497


From zacharyozer@gmail.com Wed Nov 30 00:46:24 2005
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From: Zachary Adam Ozer <zozer@mit.edu>
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Subject: Tutor Comments: Mean time to failure LN 13 Pg 20
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I was just wondering how mean time to failure is affected by redundant
systems, ie, what if I have 2 systems where one kicks in when the
other fails. It seems like the probability of failure the same, since
I'm still relying on an identical system, yet that doesn't make sense
from a practical standpoint, since then there'd be little advantage to
using redundant systems. Are they mulitiplicative? IE if the mean
proabaility of failure of one system is 1/100, does the redundancy
cause a mean probability of failure of (1/100)(1/100)? What about as
the redundancy tends towards infinity?

Just some thoughts,

-zozer


From bakster@MIT.EDU Wed Nov 30 00:54:10 2005
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This wasn't implicit in the reading I thought: if x and y are two
variables, is E[xy] = E[x] * E[y]?

Alexander Bakst

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I don't understand the difference between events and random variables.

From meyer@imap.theory.csail.mit.edu  Wed Nov 30 01:06:08 2005
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only if X,y are independent

counterexample when not independent:  Let X=Y = +1 or -1 with equal 
probability (1/2).  So E[X] = E[Y] = 0, but E[XY] = E[X^2] = 1.

Alexander G Bakst wrote:
> This wasn't implicit in the reading I thought: if x and y are two
> variables, is E[xy] = E[x] * E[y]?
> 
> Alexander Bakst


From rshearer@MIT.EDU Wed Nov 30 01:07:49 2005
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Subject: [Jelani] TP13 Comments
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The reading didn't explain Expected Values very well at all, and I 
found myself confused when doing the tutor problems.

- Rachel Shearer


From meyer@imap.theory.csail.mit.edu  Wed Nov 30 01:09:39 2005
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give me a hint where your understanding fails.  for a start, an event is 
a set (of outcomes, that is, points in the sample space), while an RV is 
a function (from outcomes to whatever you want).
regards, A.

Amanda Seybold wrote:
> I don't understand the difference between events and random variables.


From rnjacobs@MIT.EDU Wed Nov 30 01:11:32 2005
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 Nope, no comments. I still remember most of my 041. (A review of
continuous probability would be nice but apparently will not be
covered in 042...)

 - Robert Jacobs

From sergiob@MIT.EDU Wed Nov 30 01:17:28 2005
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Date: Tue, 29 Nov 2005 23:17:21 -0700
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From: Sergio Bacallado <sergiob@MIT.EDU>
Subject: [David] Week 13 Comments
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I liked the part about confidence. It's a very useful concept in every-day 
experimental calculations. Probability doesn't cease to amaze me; intuition 
can be so misleading.

Sergio Bacallado
Group 1


From yaser@MIT.EDU Wed Nov 30 01:21:31 2005
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From: "Yaser Khan" <yaser@MIT.EDU>
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Hi David,
 
As a whole, the reading was a bit dense with formulas, to the extent that I
keep getting similar definitions mixed up (for instance the distinctions
between the distributions). However, my question is: will we be concerned
with the t, poisson, or chi distributions in terms of random variables?

Thanks!
 
_Yaser

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<p class=3DMsoNormal><font size=3D2 color=3Dnavy face=3DVerdana><span =
style=3D'font-size:
10.0pt;font-family:Verdana;color:navy'>Hi =
David,<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 color=3Dnavy face=3DVerdana><span =
style=3D'font-size:
10.0pt;font-family:Verdana;color:navy'><o:p>&nbsp;</o:p></span></font></p=
>

<p class=3DMsoNormal><font size=3D2 color=3Dnavy face=3DVerdana><span =
style=3D'font-size:
10.0pt;font-family:Verdana;color:navy'>As a whole, the reading was a bit =
dense
with formulas, to the extent that I keep getting similar definitions =
mixed up
(for instance the distinctions between the distributions). However, my =
question
is: will we be concerned with the t, <span =
class=3DSpellE>poisson</span>, or chi
distributions in terms of random variables?<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 color=3Dnavy face=3DVerdana><span =
style=3D'font-size:
10.0pt;font-family:Verdana;color:navy'><br>
Thanks!<o:p></o:p></span></font></p>

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10.0pt;font-family:Verdana;color:navy'><o:p>&nbsp;</o:p></span></font></p=
>

<p class=3DMsoNormal><font size=3D2 color=3Dnavy face=3DVerdana><span =
style=3D'font-size:
10.0pt;font-family:Verdana;color:navy'>_Yaser<o:p></o:p></span></font></p=
>

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From ereid@MIT.EDU Wed Nov 30 01:28:15 2005
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To: 6042-probs@theory.lcs.mit.edu
From: Elizabeth Reid <ereid@MIT.EDU>
Subject: [Hanson] Reading comments
Date: Wed, 30 Nov 2005 01:28:09 -0500
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I was initially surprised (and skeptical) about the numbers game  
strategy, but its proof makes sense. I guess my only question would  
be how to pick the "x" you hope is between L and H, but that's just a  
matter of personal opinion. :)

Oh, and did you get my reading comments last week? You never  
responded, and the 6042-probs list was acting oddly when I tried to  
email it...

Elizabeth

From mpapi@MIT.EDU Wed Nov 30 01:28:44 2005
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Subject: [Jelani] Week 13 comments
From: Matt Papi <mpapi@MIT.EDU>
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Section 3, particularly the calculations performed on pages 15 and 16 -
I could definitely use a clearer explanation of polling and confidence
levels in lecture. I also feel that the notes weren't very helpful with
the last tutorial problem, but reading the answer to the problem after
submitting helped a lot.

- Matt


From kevin08@MIT.EDU Wed Nov 30 01:46:13 2005
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From: Kevin Wang <kevin08@MIT.EDU>
Subject: [Hanson] Reading Comments
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Hi,

I was curious about where the entropy function H(a) came originated 
from (Section 3.4.2, Pg. 12). Thanks.

Kevin


From meyer@imap.theory.csail.mit.edu  Wed Nov 30 01:54:55 2005
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Subject: Re: LN13
References: <p05230124bfb2ef9d5ba5@[18.243.2.26]> <438D4229.6020807@csail.mit.edu> <p05230125bfb2f25d007a@[18.243.2.26]>
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Sure they are related: events correspond (via indicator variables 
defined in Notes 13) to zero-one valued RV's, so they are essentially a 
special case of RV's.  Conversely, with an RV, R, and value v, there is 
an event [R=v] consisting of those outcomes w such that R(w)=v.  So you 
could specify an RV by giving all the events [R=v].

Is it clearer now?

regards, A.

Amanda Seybold wrote:
> It seems like they are expressing the same idea: creating a set of 
> outcomes with something in common.
> 
>> give me a hint where your understanding fails.  for a start, an event 
>> is a set (of outcomes, that is, points in the sample space), while an 
>> RV is a function (from outcomes to whatever you want).
>> regards, A.
>>
>> Amanda Seybold wrote:
>>
>>> I don't understand the difference between events and random variables.


From meyer@imap.theory.csail.mit.edu  Wed Nov 30 01:11:50 2005
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Date: Wed, 30 Nov 2005 01:09:34 -0500
To: "Prof. Albert R. Meyer" <meyer@csail.mit.edu>
From: Amanda Seybold <vixen@MIT.EDU>
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X-Keywords: NotJunk                                                                                            

It seems like they are expressing the same idea: creating a set of 
outcomes with something in common.

>give me a hint where your understanding fails.  for a start, an 
>event is a set (of outcomes, that is, points in the sample space), 
>while an RV is a function (from outcomes to whatever you want).
>regards, A.
>
>Amanda Seybold wrote:
>>I don't understand the difference between events and random variables.

From meyer@imap.theory.csail.mit.edu  Wed Nov 30 02:00:22 2005
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most recent email comment we have for you is 11/17.  if you have a copy 
of last weeks, send it now.  (It would help if you include the TP# in 
the subject line.)

regards, A.

Elizabeth Reid wrote:
> I was initially surprised (and skeptical) about the numbers game  
> strategy, but its proof makes sense. I guess my only question would  be 
> how to pick the "x" you hope is between L and H, but that's just a  
> matter of personal opinion. :)
> 
> Oh, and did you get my reading comments last week? You never  responded, 
> and the 6042-probs list was acting oddly when I tried to  email it...
> 
> Elizabeth


From hkhall@MIT.EDU Wed Nov 30 02:10:20 2005
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From: Harrison King Hall <hkhall@MIT.EDU>
Subject: [David] Week 13 Comments
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David-
Well syntactically I am doing better this week.  I am having some major 
problems though, primarily with approximating the binomial density 
function and the applications that follow.  I understand the formula as 
it is just plug and chug, it is its roots that I do not understand, 
however.  I would like to see a formal definition of where it comes 
from if it is not beyond the scope of the class.  From there I really 
didn't understand any of the equivalencies used in determining the 
polling but that is probably because it is my first time through the 
material.  I hope we go over some of this stuff in class tomorrow.
-Harrison


From rshroff@MIT.EDU Wed Nov 30 02:13:50 2005
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I was wondering if we could go over the approximation to the binomial
distribution in some detail.

Thanks,

Rahul Shroff

From lkini@MIT.EDU Wed Nov 30 02:17:36 2005
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Hi Hanson,

I was a little confused about approximating cumulative binomial 
distribution functions, especially the caveat (Section 3.5, page 13). 
How does thinking it in complementary terms help?

Lohith

From jstritar@MIT.EDU Wed Nov 30 02:21:09 2005
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Subject: [Sayan] Reading comments
From: Jon Stritar <jstritar@MIT.EDU>
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X-Keywords: NonJunk NotJunk                                                                                   

3.3 The Numbers game

I thought the winning strategy for this was interesting. It is not very
intuitive... but it still makes sense. 

Jon Stritar


From dnreshef@MIT.EDU Wed Nov 30 02:46:10 2005
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I found the binomial distribution concepts and their applications (especially to
polling) particularly interesting.  Also, could we go over confidence levels in
class?
Thanks,
-Dave

From ctsims@MIT.EDU Wed Nov 30 03:34:55 2005
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Subject: [6042] Weekly reading problem - Clayton Sims
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I found that the section on page 18 about expected values could use more 
examples about the use of expected values with variables. I would like to 
see that presented in class.

-Clayton 


From arup@MIT.EDU Wed Nov 30 03:44:46 2005
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Section 3.4.2: I'm kindof confused as to where the approximation for the 
binomial density function came from, so it would be nice if it were 
explained in a little more detail in lecture.

|Arup|

From bilodeau@MIT.EDU Wed Nov 30 04:03:09 2005
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From: "Peter Bilodeau" <bilodeau@MIT.EDU>
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Subject: [jelani] reading comments
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For this week's reading the most confusing part was probably the part on
confidence, or possibly the approximations for the binomial distributions.
Beyond that a lot of this seems understandable.

 

Peter Bilodeau


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<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
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probably the part on confidence, or possibly the approximations for the
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From tonyng@MIT.EDU Wed Nov 30 04:46:54 2005
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From: Tony Ng <tonyng@MIT.EDU>
Subject: [Jelani] Reading Comments Week 13
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I found "The Numbers Game" (Section 3.3, starting on pg8) interesting. I 
doesn't seem intuitive at all that selecting a number at random can help 
make me more likely to win the game. It almost seems like it is the same as 
choosing one of the two numbers at random, but it's surprising that the 
math works out to show that it is in fact a winning strategy.

- Tony Ng


From kjhollen@MIT.EDU Wed Nov 30 04:48:22 2005
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the numbers game is a neat example (p. 8-9).

Kate

From benlu@MIT.EDU Wed Nov 30 05:02:42 2005
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My eyes definitely started to glaze over during the section on polling 
and confidence intervals. Also, the online tutor problem took me 
forever. It would be great to go over those topics in lecture.

~Ben

From lana@MIT.EDU Wed Nov 30 06:50:19 2005
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The part I found the most intriguing was the halting problem, and the intuitive
feeling that it must have some kind of an answer. Can it be modeled by Markov
chains?
Lana

From kromer@MIT.EDU Wed Nov 30 09:51:22 2005
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p.21 "If every couple follows the strategy of having children until they
get a girl, what will eventually happen to the fraction of girls born in
this world?"
Will it remain the same (because each couple expects to have 2
children)?

From veracarr@MIT.EDU Wed Nov 30 10:12:41 2005
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If we could cover when we would use each of the distributions and there 
importance, that would be very useful.

From rian@MIT.EDU Wed Nov 30 10:20:12 2005
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Date: Wed, 30 Nov 2005 10:19:55 -0500
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In the part that is the "Analysis of the Winning Strategy" 3.3.2 what  
if you have different probabilities for each event, for instance not  
50-50? I don't exactly understand the variable assignments for not  
50-50 tree probabilities. Thanks

rian hunter

From cbossard@MIT.EDU Wed Nov 30 10:26:01 2005
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I'm really not sure what to ask.  The distribution stuff seems a
little tricky though.
Cynthia

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From: Hanson Zhou <hmzhou@theory.csail.mit.edu>
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Subject: Re: [Hanson] Week 13 Lecture Notes
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Hi...I think it's saying that in the case a > p, the formula does not hold
(since the geometric sum does not hold).  But if we look at tails instead
of heads, then we can calculate 1-Pr[>=(1-a)n tails] to get the CDF we
want(note that now the fraction is 1-a).  The probability of a tail is 1-p
and we have 1-a < 1-p by assumption.  Then, we can do a similar
analysis/formula to approximate Pr[>=(1-a)n tails].  You might want to
check it or ask Prof. Meyer since lots of details are omitted here, but I
think that's the right idea.

-Hanson

On Wed, 30 Nov 2005, Lohith Kini wrote:

> Hi Hanson,
>
> I was a little confused about approximating cumulative binomial
> distribution functions, especially the caveat (Section 3.5, page 13).
> How does thinking it in complementary terms help?
>
> Lohith
>

From petek@MIT.EDU Wed Nov 30 10:29:12 2005
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This is a multi-part message in MIME format.
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I'm up for hearing more on probability distributions and also about 
something that Ronnitt talked about last lecture to our breakout group.  
There is some complication in assigning a probability to whether a 
certain person prefers a candidate....something about it being either 1 
or 0.  Want to have that expanded a bit. 

-P

-- 
Pete Kruskall
28 The Fenway
Boston, MA 02215

:::::::::::::::::::::::::::::
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508.843.5861 ::::Cell Phone::
::::http://tege.mit.edu::::::



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<font size="-1">I'm up for hearing more on probability distributions
and also about something that Ronnitt talked about last lecture to our
breakout group.&nbsp; There is some complication in assigning a probability
to whether a certain person prefers a candidate....something about it
being either 1 or 0.&nbsp; Want to have that expanded a bit.&nbsp; <br>
<br>
-P<br>
</font>
<pre class="moz-signature" cols="72">-- 
Pete Kruskall
28 The Fenway
Boston, MA 02215

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From lmccart@MIT.EDU Wed Nov 30 10:38:00 2005
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I did not find anything particularly confusing about this reading.  I found the
Birthday Principle most interesting.
-lauren

From meyer@imap.theory.csail.mit.edu  Wed Nov 30 10:40:25 2005
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This week it's Notes and TP 13.

The Halting Problem is covered in 6.045 and 16.840 and, best of all :-) 
my course 6.044.

regards, A.

Svetlana Goldenberg wrote:

>The part I found the most intriguing was the halting problem, and the intuitive
>feeling that it must have some kind of an answer. Can it be modeled by Markov
>chains?
>Lana
>  
>


From cvnguyen@MIT.EDU Wed Nov 30 10:40:33 2005
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From: "Chieu Nguyen" <cvnguyen@MIT.EDU>
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Subject: [Jelani] Week 13 comments
Date: Wed, 30 Nov 2005 10:40:26 -0500
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(page 21) If every couple has children until getting a girl, the fraction of 
girls in the world does not change (if we ignore factors that may affect 
children of different genders differently... disease, abortion, infanticide, 
etc.). This is interesting. Though the fraction remains the same, the fact 
that each nuclear family has one and only one girl may introduce some 
interesting social issues....

--Chieu Nguyen 


From meyer@imap.theory.csail.mit.edu  Wed Nov 30 10:43:52 2005
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Subject: Re: [Hanson] Week 13 Lecture Notes
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yeah, that's right.  the condition a>p is necessary technically, but 
because of the complement trick (look at tails), it is not a limitation.
regards, A.

Hanson Zhou wrote:

>Hi...I think it's saying that in the case a > p, the formula does not hold
>(since the geometric sum does not hold).  But if we look at tails instead
>of heads, then we can calculate 1-Pr[>=(1-a)n tails] to get the CDF we
>want(note that now the fraction is 1-a).  The probability of a tail is 1-p
>and we have 1-a < 1-p by assumption.  Then, we can do a similar
>analysis/formula to approximate Pr[>=(1-a)n tails].  You might want to
>check it or ask Prof. Meyer since lots of details are omitted here, but I
>think that's the right idea.
>
>-Hanson
>
>On Wed, 30 Nov 2005, Lohith Kini wrote:
>
>  
>
>>Hi Hanson,
>>
>>I was a little confused about approximating cumulative binomial
>>distribution functions, especially the caveat (Section 3.5, page 13).
>>How does thinking it in complementary terms help?
>>
>>Lohith
>>
>>    
>>


From meyer@imap.theory.csail.mit.edu  Wed Nov 30 05:52:46 2005
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Subject: [sayan] probability
From: Jesus Medrano <medrano@MIT.EDU>
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I'm still uncertain about the theorem in the notes about the binomial
sampling.  What does this mean?  Is it true that only 600 people are
needed to be sampled out of any size n population to get an accurate
view of the entire population.  This does not seem very reasonable?  Is
this done in practice?

Jesus Medrano


From mukkala@MIT.EDU Wed Nov 30 10:48:49 2005
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To: 6042-probs@theory.csail.mit.edu
From: Praveen Pamidimukkala <mukkala@MIT.EDU>
Subject: [Sayan] Week 13 Comments
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In section 3.4 of the Notes, the first sentence reads, "Of the more complex 
distributions, the binomial distribution is surely the most important in 
computer science."  I understand that "an enormous number  of analyses in 
computer science come down to proving that the tails of the binomial and 
similar distributions are very small," but what is an example of this?

Thanks,
Praveen Pamidimukkala


From mracich@MIT.EDU Wed Nov 30 10:49:55 2005
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Subject: [Sayan] Comments for Course Notes, Week 13 (Random Variables,
	Distributions, and Expectation)
From: Moira Racich <mracich@MIT.EDU>
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I found section 3, "Probability Distributions", (starting on page 6)
confusing.  I would appreciate it if this was explained further.  

Moira Racich


From xiaoranz@MIT.EDU Wed Nov 30 10:54:07 2005
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Subject: [Hanson] Week 13 Comments
From: Xiaoran Zhang <xiaoranz@MIT.EDU>
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The Birthday principle on page 4-5 is very surprising. The confidence
levels is interesting in that it provides a good way to determine the
accuracy of the estimations made. 

Xiaoran Zhang

From vbrobbey@MIT.EDU Wed Nov 30 10:54:39 2005
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Statement 8 on the last tutor problem tricked me. In addition to that I =
didn't quite get the winning stategy for the numbers game. Aren't the =
numbers that the other guy puts in the envelope completely independent =
of the number x that I guess?
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<DIV><FONT face=3DArial size=3D2>Statement&nbsp;8 on the last tutor =
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From antonk@MIT.EDU Wed Nov 30 10:55:58 2005
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From: "Anton Katz" <antonk@MIT.EDU>
To: <6042-probs@theory.lcs.mit.edu>
Subject: [david] discussed more fully
Date: Wed, 30 Nov 2005 10:55:46 -0500
Organization: Massachusetts Institute of Technology
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Hi,

Is it possible to discuss the differences between 2 approximations?  :

Binominal density function

Cumulative Binominal distribution function.

 

Thank you,

 

Anton.

 


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<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
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font-family:Arial'>Hi,<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'>Is it possible to discuss the differences between 2 =
approximations?
&nbsp;:<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'>Binominal density =
function<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'>Cumulative Binominal distribution =
function.<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'><o:p>&nbsp;</o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'>Thank you,<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'><o:p>&nbsp;</o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'>Anton.<o:p></o:p></span></font></p>

<p class=3DMsoNormal><font size=3D2 face=3DArial><span =
style=3D'font-size:10.0pt;
font-family:Arial'><o:p>&nbsp;</o:p></span></font></p>

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From jeffhoff@MIT.EDU Wed Nov 30 11:19:10 2005
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Section 3.4

I found the binomial distribution a little confusing in 
general.  could there be an extension done on this?


From crowell@MIT.EDU Wed Nov 30 11:21:35 2005
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From: Robert Crowell <crowell@MIT.EDU>
Subject: [Hanson] Reading Comments
Date: Wed, 30 Nov 2005 11:21:26 -0500
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I would like to see more about Expected Value in lecture (this is  
section 5.2 on page 20).

I had some trouble on the tutor problem dealing with this subject.

--Rob Crowell



From hmzhou@blackbird.csail.mit.edu Wed Nov 30 11:37:42 2005
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From: Hanson Zhou <hmzhou@theory.csail.MIT.EDU>
To: Rebecca Idell <ridell@MIT.EDU>
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Subject: Re: reading 13, ta: hanson
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Well, most simply: PDF(x) = Pr[R=x].

R is the random variable and PDF(x) asks for the probability that the
value of R turns out to be x.

-Hanson

On Tue, 29 Nov 2005, Rebecca Idell wrote:

>
> Can you explain more about the probability density function on page 6?
>
> Rebecca Idell
>
> --
> Rebecca Idell
> Massachusetts Institute of Technology
> Electrical Engineering and Computer Science, Class of 2007
>
> 479 Commonwealth Avenue
> Boston, MA 02215
> (617) 875-0889
>

From kktyan@MIT.EDU Wed Nov 30 11:42:56 2005
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From: Karena Tyan <kktyan@MIT.EDU>
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Subject: [Hanson] (Belated) Reading Comments 13
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I apologize for the late reading comments - I nearly forgot to do them, and only
just realized that I hadn't done them last night as I'd meant to.

In this reading, I was confused by the "Expected Value" sections (starting on
page 18).  I wasn't sure exactly how one could compute/calculate the expected
value, or what the expected value really was, but then I reread the definition
of the expected value and figured it out.  For the most part, I understood the
rest of the reading, though; only the section on expected values was confusing.

- Karena

-- 
410 Memorial Drive
Cambridge, MA 02139
(585)957-5923

From hmzhou@blackbird.csail.mit.edu Wed Nov 30 11:45:29 2005
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From: Hanson Zhou <hmzhou@theory.csail.mit.edu>
To: Elizabeth Reid <ereid@MIT.EDU>
cc: 6042-probs@theory.lcs.mit.edu
Subject: Re: [Hanson] Reading comments
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Hi...I'm pretty sure I forwarded your email last week to Prof. Meyer,
though you might want to check.

-Hanson

On Wed, 30 Nov 2005, Elizabeth Reid wrote:

> I was initially surprised (and skeptical) about the numbers game
> strategy, but its proof makes sense. I guess my only question would
> be how to pick the "x" you hope is between L and H, but that's just a
> matter of personal opinion. :)
>
> Oh, and did you get my reading comments last week? You never
> responded, and the 6042-probs list was acting oddly when I tried to
> email it...
>
> Elizabeth
>

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Date: Wed, 30 Nov 2005 11:50:56 -0500 (EST)
From: Hanson Zhou <hmzhou@theory.csail.mit.edu>
To: Zachary Adam Ozer <zozer@mit.edu>
cc: 6042-probs@theory.lcs.mit.edu
Subject: Re: Tutor Comments: Mean time to failure LN 13 Pg 20
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X-Keywords: NotJunk                                                                                           

I'm not sure I understand your question.  If you have an identical backup
system that turns on when the first one fails then the mean time to
failure would seem to be 1/f + 1/f, where f is the failure probability of
each system.

-Hanson

On Wed, 30 Nov 2005, Zachary Adam Ozer wrote:

> I was just wondering how mean time to failure is affected by redundant
> systems, ie, what if I have 2 systems where one kicks in when the
> other fails. It seems like the probability of failure the same, since
> I'm still relying on an identical system, yet that doesn't make sense
> from a practical standpoint, since then there'd be little advantage to
> using redundant systems. Are they mulitiplicative? IE if the mean
> proabaility of failure of one system is 1/100, does the redundancy
> cause a mean probability of failure of (1/100)(1/100)? What about as
> the redundancy tends towards infinity?
>
> Just some thoughts,
>
> -zozer
>
>

From hmzhou@blackbird.csail.mit.edu Wed Nov 30 11:57:55 2005
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Date: Wed, 30 Nov 2005 11:57:55 -0500 (EST)
From: Hanson Zhou <hmzhou@theory.csail.mit.edu>
To: Valery Kwasi Brobbey <vbrobbey@MIT.EDU>
cc: 6042-probs@theory.csail.mit.edu
Subject: Re: [TA-name] Week 13 Comments
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X-Keywords: NotJunk                                                                                           

Hi Valery...by TA-name you mean Hanson :).  As for the winning strategy,
it is the very fact that your selection of x is random that thwarts
anything your opponent might do.  So yes, your selection of x is
independent, and this actually ends up helping you to do better than 50%.
See me if this remains confusing.

-Hanson

On Wed, 30 Nov 2005, Valery Kwasi Brobbey wrote:

> Statement 8 on the last tutor problem tricked me. In addition to that I
> didn't quite get the winning stategy for the numbers game. Aren't the
> numbers that the other guy puts in the envelope completely independent
> of the number x that I guess?

From ryan786@MIT.EDU Wed Nov 30 12:44:11 2005
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Apologies again for the late email, I have just now been able to read the pdf. 
The most difficult parts were the Probability Distributions (Section 3).  I do
think class today will clear it up though, as that is the topic of lecture (I
think).

-Ryan

From mitras@theory.csail.mit.edu Wed Nov 30 13:06:42 2005
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Expectation will be covered on Friday, and variance on next Wednesday.
Best,
sayan

David A. Nedzel wrote:

> I'm fairly comfortable with this material, I've seen most of it in 
> 6.041. It might be nice to have a refresher during lecture on the 
> expectation & variance of different distributions.
>
> - David



From mitras@theory.csail.mit.edu Wed Nov 30 13:08:26 2005
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Events are sets of points in the sample space. Events are disjoint when 
these sets are disjoint.
E.g, consider a 6-sided dice roll experiment. Define Event 1, A = even 
number comes up, B = odd number, C = prime number.
Then, A and B are disjoint, A and C are not.

-sayan

Praveen Pamidimukkala wrote:

> In section 4.3 of the Course Notes on Probability, disjointness is 
> mentioned in comparison to independence.  I understand that 
> mathematically, when two events are disjoint, Pr{AintB} = 0 as opposed 
> to Pr{AintB} = Pr{A} x Pr{B}, which applies for independence.  
> However, in terms of events, how could this be described?
>
> Thanks,
> Praveen Pamidimukkala
>


From mitras@theory.csail.mit.edu Wed Nov 30 13:20:15 2005
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 Distributions, and Expectation)
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Hope today's lecture is going to help.
Best,
Sayan

Moira Racich wrote:

>I found section 3, "Probability Distributions", (starting on page 6)
>confusing.  I would appreciate it if this was explained further.  
>
>Moira Racich
>
>  
>


From meyer@imap.theory.csail.mit.edu  Wed Nov 30 11:45:30 2005
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Hi,
You have to be careful about phrasing these polling related 
probabilistic claims. The sample size of 600 is derived for a particular 
accuracy (0.04) and for a particular probability of error (5%). The 
surprising thing is that, the sample size required is independent of the 
actual size of the population. We'll do a similar problem in class 
today, hopefully it will clear up these ideas.

Best,
Sayan


Jesus Medrano wrote:

>I'm still uncertain about the theorem in the notes about the binomial
>sampling.  What does this mean?  Is it true that only 600 people are
>needed to be sampled out of any size n population to get an accurate
>view of the entire population.  This does not seem very reasonable?  Is
>this done in practice?
>
>Jesus Medrano
>  
>

From meyer@imap.theory.csail.mit.edu  Wed Nov 30 02:01:47 2005
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Date: Wed, 30 Nov 2005 02:01:53 -0500
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poisson might get mentioned VERY briefly in the last week.  no t or chi.
regards, A.

Yaser Khan wrote:
> Hi David,
> 
>  
> 
> As a whole, the reading was a bit dense with formulas, to the extent 
> that I keep getting similar definitions mixed up (for instance the 
> distinctions between the distributions). However, my question is: will 
> we be concerned with the t, poisson, or chi distributions in terms of 
> random variables?
> 
> 
> Thanks!
> 
>  
> 
> _Yaser
> 


From dshin@MIT.EDU Wed Nov 30 14:09:03 2005
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Yes, that is correct.  Another way to see it is to define an indicator 
variable G(x) for each baby x that equals 1 if x is a girl and 0 if x is 
a boy.  Every baby x has E[G(x)] = 0.5, so by linearity of expectation 
the expected value of SUM(G(x)) is 0.5*population.

DS

Katherine Romer wrote:

>p.21 "If every couple follows the strategy of having children until they
>get a girl, what will eventually happen to the fraction of girls born in
>this world?"
>Will it remain the same (because each couple expects to have 2
>children)?
>  
>

From mdmurray@MIT.EDU Wed Nov 30 14:19:06 2005
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From: Michael D Murray <mdmurray@MIT.EDU>
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Hey,

Sorry for the delay, the reading was quite long to be done in two days. I 
am curious about the Bernoulli Distribution, do we need to know it in 
detail? There isn't alot of information in 3.1 and I don't really 
understand what the point of it is.

Thanks,
Michael Murray

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From: Shreyes Seshasai <shreyes@mit.edu>
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Subject: [Sayan] Week 13 Reading Comments
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Hi,

Here are my comments on week 13's reading comments:

My question deals with the the definition of Independence presented in the
notes.  The notes state on page 3:
Random variables R1 and R2 are independent if for all x1 in the codomain of
R1, and x2 in the
codomain of R2, we have:
Pr {R1 =3D x1 | R2 =3D x2} =3D Pr {R1 =3D x1} =B7 Pr {R2 =3D x2} .
This definition makes sense to me, but my questions is whether knowing this
is true implies Independence.  From other classes (6.041), we are able to
prove Independence by showing how the above equation holds, but there are
some questions where the equation holds but intuitively R1 and R2 are not
independent.  I can't think of an example off the top of my head, but i kno=
w
for other formulas describing Independence (like E[XY] =3D E[X]E[Y]), provi=
ng
the equation does not necessarily prove Independence.

Thanks,
Shreyes
group 7

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Hi,<br>
<br>
Here are my comments on week 13's reading comments:<br>
<br>
My question deals with the the definition of Independence presented in the =
notes.&nbsp; The notes state on page 3: <br>
Random variables R1 and R2 are independent if for all x1 in the codomain of=
 R1, and x2 in the<br>
codomain of R2, we have:<br>
Pr {R1 =3D x1 | R2 =3D x2} =3D Pr {R1 =3D x1} =B7 Pr {R2 =3D x2} .<br>
This definition makes sense to me, but my questions is whether knowing
this is true implies Independence.&nbsp; From other classes (6.041), we
are able to prove Independence by showing how the above equation holds,
but there are some questions where the equation holds but intuitively
R1 and R2 are not independent.&nbsp; I can't think of an example off
the top of my head, but i know for other formulas describing
Independence (like E[XY] =3D E[X]E[Y]), proving the equation does not
necessarily prove Independence.<br>
<br>
Thanks,<br>
Shreyes<br>
group 7<br>

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From mitras@theory.csail.mit.edu Thu Dec  1 17:28:19 2005
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Good question.
A. X and Y are independent if and only if (<=>) P(X =x, Y = y) = for all 
x, y,  P(X=x)P(Y=y).
This is the definition of independence.

B. X and Y are independent => E[XY] = E[X] E[Y], this is a one way 
implication.
Here's a quick proof:
E[XY] = \sum_{i,j} a_i b_j P[X = a_i, Y = b_j]
           = \sum_{i,j} a_i P[ X = a_i] b_j P[Y = b_j], since X and Y 
are independent.
          = \sum_i a_i P[X =a_i] \sum_j b_j P[Y = b_j]
          = E[X]E[Y]

But the converse is not necessarily true, i.e., E[XY] = E[X]E[Y] does 
not imply independence of X Y.
Here's an example:
Let (X, Y) assume values (1,0), (0,1), (-1,0), and (0,-1) with equal 
probabilities 1/4.
Then E[X] = E[Y] = 1 * 1/4 + (-1) * 1/4 = 0.
E[XY] = 0.
So, we do have E[XY] =E[X]E[Y].
Now lets see if they are independent. Choose x = 0, y = 0.
P(X=0) = P(Y =0) = 1/2 whereas P(X=0, Y=0) = 0.
So, P(X = 0, Y=0) is not = P(X=0)P(Y=0), i.e., they are not independent.

But remember, E[X+Y] = E[X] + E[Y] regardless of independence.

Best,
Sayan


Shreyes Seshasai wrote:

> Hi,
>
> Here are my comments on week 13's reading comments:
>
> My question deals with the the definition of Independence presented in 
> the notes.  The notes state on page 3:
> Random variables R1 and R2 are independent if for all x1 in the 
> codomain of R1, and x2 in the
> codomain of R2, we have:
> Pr {R1 = x1 | R2 = x2} = Pr {R1 = x1} · Pr {R2 = x2} .
> This definition makes sense to me, but my questions is whether knowing 
> this is true implies Independence.  From other classes (6.041), we are 
> able to prove Independence by showing how the above equation holds, 
> but there are some questions where the equation holds but intuitively 
> R1 and R2 are not independent.  I can't think of an example off the 
> top of my head, but i know for other formulas describing Independence 
> (like E[XY] = E[X]E[Y]), proving the equation does not necessarily 
> prove Independence.
>
> Thanks,
> Shreyes
> group 7



From dangut@MIT.EDU Thu Dec  1 19:39:06 2005
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I'm still comfortable with the material, I think the birthday example was pretty
cool.

