0

I'm trying to understand what the expectation of $1/X$ would be when $X$ can take on values of $0$. I've looked this up, and I understand that for continuous distributions you can take limits. However, I couldn't find any information on how to do this for discrete distributions (e.g. Binomial(n, p)). It seems to me that you can't take a limit since you're not ever approaching that value since it's discrete - so would that expectation be undefined? Or infinite?

Let's just take that example - what would $E[1/X]$ where $X~Bin(n, p)$ be?

lennon310
  • 2,582
  • 2
  • 21
  • 30
Poisyn
  • 1
  • When I apply the expectation formula for the simplest example of a Bin(1,1/2) variable, I get the formula $$E[1/X]=1/2\left(\frac{1}{0}\right)+ 1/2\left(\frac{1}{1}\right).$$ The rules of arithmetic you are using will tell what that value is. – whuber Oct 16 '20 at 17:51

1 Answers1

2

That expectation would be $\infty$, as we can interpret, in this context, $1/0$ as infinity. One reason is that the post mentioned, as an example, the binomial distribution, which is nonnegative. So this solution can only be defended, if at all, for nonnegative random variables. The principled solution is in the next paragraph. See also I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?

Alternatively, and maybe better, the expectation is just undefined as it involves division by zero which is undefined.

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
  • 1
    I am sceptical, because an equally strong argument can be made to treat $1/0$ as $-\infty.$ Consider, for instance, a trinomial variable with equal chances of being $-1,0,1.$ What basis is there for choosing $+\infty$ in this case?? – whuber Oct 16 '20 at 17:50
  • Because of his binomial example, implying a non-negative random variable. Will clarify. – kjetil b halvorsen Oct 16 '20 at 17:52
  • I thought as much, but that strikes me as artificial for two reasons. First, it suggests you would compute an expression like "1/0" differently depending on context -- and that seems to fly in the face of basic mathematical principles. Second, adopting the more general measure-theoretic approach where $X:\Omega\to\mathbb{R}$ is a measureable function, we have to entertain the possibility that even though $X$ has a binomial distribution, *it might take on negative values on a set of measure zero.* Once again, then, we have to confront the question "why not $-\infty$?" – whuber Oct 16 '20 at 17:54
  • @whuber So it is undefined. Whereas for continuous distributions we can speak about the mean in terms of an integral which can be infinite (As far as you accept infinite anyway. I personally don't like it, infinities. But in a constructionist sense you can see it as a limit that diverges. This divergence and limit doesn't work for discrete distributions). – Sextus Empiricus Oct 16 '20 at 17:57
  • 1
    Yeah, it seems to me that 1/0 is undefined, not that it's some infinity. – Poisyn Oct 16 '20 at 18:21
  • 1
    @Sextus To a limited extent, $\pm\infty$ are definite members of the *extended reals* in measure theory. The concept of divergence is necessary for discrete distributions, too. Consider computing the expectation of $1/X$ where $\Pr(X=1/n)=6/(\pi^2n^2),$ $n=1,2,\ldots$ for instance. – whuber Oct 16 '20 at 18:52
  • 1
    @Whuber That is a nice example. So we can define a discrete distribution $P(X = n) = 6/\pi^2/n^2$ where the domain is all *finite* positive integers. And the variable $Y = 1/X$ will not need to include 0, but the sum that computes the expectation of $1/Y$ will diverge (due to the increasing density of points in the domain of $Y$ near 0). – Sextus Empiricus Oct 16 '20 at 19:35
  • @Sextus Right. (And thank you for correcting my omission of the power $n^2,$ which I have now fixed.) But I really did mean that these are the probabilities attached to $X=1/n,$ not $X=n.$ Thus the support of this distribution is the set $\{0,1,1/2,1/3,\ldots\}$ (because $0$ is in its closure) and the expectation of $1/X$ is proportional to the sum of $n/n^2,$ which diverges. – whuber Oct 16 '20 at 21:11
  • 0 is inside the support, but has zero probability that seems a bit artificial to me – Sextus Empiricus Oct 16 '20 at 22:15
  • 1
    @Sextus Empiricus: The support is defined as the smallest *closed* set with probability 1. – kjetil b halvorsen Oct 16 '20 at 22:28