%from the spider problem, cp13W.tex

\problempart Ben argues that we should model the spider's situation by
assuming that the captures of large and small flies are independent
Poisson processes with respective rates of 100 small flies captured per 10
hours and 300 large ones per 10 hours.  Under these assumptions, what are
the Markov, Chebyshev, and Chernoff bounds on the spider's probability of
meeting her quota?

\solution{ Let $S$ be the random variable equal to the number of small
flies caught in 10 hours, and $L$ the number of large flies.  We are given
that $S$ has a Poisson distribution with rate $\lambda_S = 100$ and
likewise for $L$ with rate $\lambda_L = 300$.  The total number of flies
caught in 10 hours is $R \eqdef S+L$.  We know that the arrival rate and
mean of a Poisson distribution are the same, so $\mu_S = 100$ and $\mu_L =
300$.  So $\mu_R = 100 + 300 = 400$ by linearity of expectation.

The Markov bound depends only on the expected number of flies caught.
Since this is still 400, we have the same probability bound of 0.8 as in
the previous case.

But since $S$ and $L$ are independent, we know from Notes 14 that $R$ has
a Poisson distribution with rate $\lambda_R = \lambda_S + \lambda_L =
400$.  So not only is $\mu_R = 400$, but $\variance{R} = 400$, so the
one-sided Chebyshev bound becomes $400/(400+100^2) = 1/26$, a little
larger than the previous case.

The standard Chernoff Bound \emph{Theorem} does not apply because $R$ is
not a sum of Bernoulli variables.  However, we know from Notes 14 that the
standard Chernoff \emph{bound} applies to the Poisson distribution even
though it is not such a sum.  Since the value of the Chernoff bound
depends only on the expectation, the bound in this case remains the same
$1/100,000$ as before.}
\fi
