\problemdata     
{S01-PS10-5}           
{expectation}              
{} 		
{S01-PS10}
{S01 Karger}





\begin{problem}


 Suppose that I choose a listing of the
numbers $1, 2, \ldots, n$ uniformly at random.  What is the expected
number of entries that are greater than all preceding entries?  For
example, in the listing 4, 2, 1, 5, 3, the numbers 4 and 5 are greater
than all preceding entries\footnote{This problem can be equivalently phrased as follows: There are n people whose heights are $1, 2, \ldots, n$. I choose a random permutation, and line them up accordingly. What is the expected number of people who can see what's going on in front of the line? We are assuming that a person can see if and only if all the people in front of him/her are shorter.}

\hint What's the probability that the first entry is greater than all the preceding entries? What about the second one? 

\solution{
\textbf{Solution:}\\
Let $i_k$ be an indicator for the event that the $k$-th entry is
greater than the preceding $k-1$ entries.  By linearity of expectation

\begin{eqnarray*}
\Ex(\text{\# new maxima})
    & = &   \sum_{k=1}^n \Ex(i_k) \\
    & = &   \sum_{k=1}^n 1 \cdot \Pr(i_k = 1) + 0 \cdot \Pr(i_k = 0) \\
    & = &   \sum_{k=1}^n \Pr(i_k = 1)
\end{eqnarray*}

All that remains is to compute $\Pr(i_k = 1)$, which is the
probability that the $k$-th entry is greater than all preceding
entries.  Note that each of the first $k$ entries is equally likely to
be the largest.  Therefore, $\Pr(i_k = 1) = \frac{1}{k}$.
Substituting into the equation above gives:

\begin{eqnarray*}
\Ex(\text{\# new maxima})
    & = &   \sum_{k=1}^n \Pr(i_k = 1) \\
    & = &   \sum_{k=1}^n \frac{1}{k} \\
    & = &   H_n
\end{eqnarray*}
}

\end{problem}
