Expected Value: Calculate $E(1/X)$ where random variable $X$ denotes the number showing on the chip selected

Question

The question:

Urn contains $n$ chips numbered 1 through n. assume that the probability of choosing $i$ is equal to $ki$, $i=1,2,...,n$. If one chip is drawn, calculate $E(1/X)$, where the random variable $X$ denotes the number showing on the chip selected.

[HINT: recall that the sum of the first n integers is $n(n+1)/2$]

My solution:

Given: $p_x(i) = ki$ where $i=1,...,n$. Let $g(x)=1/x$. Then

$E(g(i)) = \sum^n_{i=1}g(i)p_x(i) = \sum^n_{i=1} (i)^{-1} (ki) = k\sum^n_{i=1}1 = k(n+1)$

I am struggling to figure out this question. The hint they gave have no use (which i doubt that the textbook will give the hint if its not used). I read on other online solutions that they have $kn(n+1)/2$ but I do not understand how that would be correct. Your feedback is appreciated.

Answer 1

Your answer needs a slight modification and at first hand should be: $$\mathsf EX^{-1}=kn\tag1$$ and not $k(n+1)$.

This simply because $\sum_{i=1}^n1=n$ (and not $n+1$).

But actually you are not ready here.

We also have the equality $\sum_{i=1}^nki=1$ which allows you to find and expression for $k$ in $n$ and substitute in $(1)$. This declares the hint in your textbook.

I leave that to you.

Answer 2

You have a minor error. $k\sum_{i=1}^{n}1 = nk$.

Aside from that, given the problem statement, as you've currently written it, your answer seems correct to my eyes.

Answer 3

Unless I'm making a fool of myself, your answer is certainly right (with the exception of one silly little mistake: $\sum_{i=1}^n = n$, not $n+1$)! Note, first, that $k = (\tfrac12n(n+1))^{-1}$; it might be worth writing your answer as $$ kn = 2/(n+1). $$

Note that the answer you're referencing is exactly equal to $1$. Let's consider some basic heuristics as to why this cannot be the case. Firstly, suppose $n$ is very large, $10^{100000}$ say. You're overwhelmingly likely to pick a number in the range $[n/2,n]$, (because squares increase much faster than linear). As such, we can say, informally, that the expectation is at most $$1/2 + \langle \text{what it would be if we restrict to $[n/2,n]$} \rangle.$$ In this latter range, $1/X$ can take the value at most $2/n \le 1/10$, and so the expectation is at most $1/2 + 1/10$.

Now, this is just heuristics. Perhaps as an exercise you could come up with a way to make such an upper bound rigorous? Use the equality $$ E(X^{-1}) = E(X^{-1} 1(X^{-1} > n/2)) + E(X^{-1} 1(X^{-1} \le n/2)), $$ where $1(A)$ is the indicator of an event $A$.