3
$\begingroup$

Consider a population of people. Each person's height is IID (you're not given the distribution). You randomly choose a person and observe that their height is $X$. Let $N$ be the number of additional samples needed to randomly select a person who's height is greater than $X$. What is $E[N]$?

Apparently, $E[N] = \infty$ and the logic is that $X$ is a random variable.

My first concern is that I am not sure why $X$ is a random variable. Once you obtain a sample with a value $X$, isn't $X$ now fixed/observed (the instructions state the height is observed) and no longer random? In that case, the expectation should be bounded?

If we assume $X$ is indeed a random number. Then I believe you do the following

$$ E[N] = \sum_i E[N|X = X_i] P(X_i) = $$

We apply the law of total expectation and condition on all possible values of $X$. Is the expectation infinity because $X$ can take on the value $\max(heights)$, in which case you will never get a person with a height greater than $> \max(heights)$ no matter how long you sample for? If there was a constraint that you know $X$ isn't $\max(heights)$, then in this case, would $E[N]$ be bounded?

$\endgroup$

2 Answers 2

1
$\begingroup$

Suppose you select the tallest person in the population; there is an assumption that the population is finite. There is a nonzero probability that this happens. If you select the tallest person, you'll never find a higher one, so $N=\infty$. Since $\infty×p=\infty$ if $p>0$, when $X$ is treated as a random variable on which $N$ depends, we have $E[N]=\infty$.

If you do not select the tallest person, then the expectation would indeed be bounded.

$\endgroup$
1
  • 1
    $\begingroup$ What if the population was not finite? $\endgroup$ Mar 2, 2021 at 5:09
0
$\begingroup$

Let's say you need $n$ people to find the one who is taller than the first. You sort these $n+1$ people by height and it happens that the first one got into the place number $n$ and the last one got into the place number $n+1$. The probability of this is $$P\{N=n\}=\frac{(n-1)!}{(n+1)!}=\frac1{n(n+1)}$$ Therefore the expectation is indeed infinity $$E[N]=\frac12+\frac13+\frac14+...=\infty$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .