A man goes fishing in a river (with a lot of fishes!) and only will stop when the last fish caught has less weight than the one before. Show that the fishes caught expected value is e(2,71828...).
The expected value of next fish is: $\frac{1}{n}\sum\limits_{i=1}^{n}w_i$, but how do i find the fish number expected value? Anyone help please?
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The probability that the man catches exactly $n$ fishes is:
$$P(n) = \frac{n-1}{n!}\;\;\; \text{for } n\ge2\;.$$
In order for the man to fish exactly $n$ fishes, you must have that the first $n-1$ are all in increasing order. If you assign values 1 to $n$ to each fish according to their weights, it's easy to realize that you have $n-1$ arrangements for the first $n-1$ fishes. For instance, if $n=4$, you have $[1,2,3]$, $[1,2,4]$ and $[2,3,4]$. You also have $n!$ possible arrangements for $n$ fishes.
So you have:
$$E[n] = \sum_{n=2}^\infty n \frac{n-1}{n!} = \sum_{n=0}^\infty \frac{1}{n!}$$
which leads to $E[n] = e$.
We can use the tail formula $\mathbb{E}[X]=\sum_{n=0}^\infty \mathbb{P}(X>n)$.
Fix any $n$. By symmetry, the probability that the first $n$ fish are increasing in weight is $\frac{1}{n!}$. (because there are exactly $n!$ possible ways to order their weights). Hence $\mathbb{P}(X>n)=\frac{1}{n!}$. And now use the well known equality $e=\sum_{n=0}^\infty \frac{1}{n!}$.