# How to find a sequence of random variables of infinite expectation that converges to zero?

I want to find an example such that $X_n\rightarrow0$ as $n\rightarrow \infty$ while $E(X_n)=\infty$.

I am thinking of making use the divergence of $\sum\frac{1}{n}$, but fail to find suitable $X_n$ and $\Pr(X_n=k)$.

Can you show me some examples or give me some hints?

• Presumably $X_n$ and $x_n$ are sequences of random variables and you mean the non convergence of the Harmonic series. But it needn't be that complicated. For instance, consider a discrete distribution $F_n$ with $\Pr_{F_n}(0)=1-1/n$ and $\Pr_{F_n}(n^2)=1/n$. If $X_n \sim F_n$ are iid, does $X_n$ converge to $0$ in the sense you intend? What happens to their expectations? Now this does not answer your question, because $E(X_n)$ are all finite, but it might suggest some simple modifications that do work.
– whuber
Commented Apr 9, 2013 at 17:48
• Thanks, but the I just can't work it out. Commented Apr 10, 2013 at 3:18

To make a series of random variables $$X_n$$ approach $$0$$ in probability, we want the chance that $$X_n$$ is not $$0$$ to become vanishingly small. At the same time, to ensure the expectations of each of the $$X_n$$ diverge, we need to leave some probability out in the tails (that is, where the values of the $$X_n$$ are large enough to overwhelm the small probabilities that $$X_n$$ is nonzero).

To do this, let's consider discrete distributions supported on the natural numbers. Their probability mass functions $$f_n(k)$$, $$k=0, 1, 2, \ldots,$$ need to have these properties:

1. The limiting value of $$f_n(0)$$ must be $$1$$ as $$n\to\infty$$.

2. Every expectation, equal to $$\sum_{k=0} k f_n(k)$$, must diverge to $$\infty$$.

The question suggests we make use of the fact that the Harmonic series $$\sum_{k=1} 1/k$$ diverges. Comparing that to (2) would have us try

$$k f_n(k) = 1/k,$$

at least for sufficiently large $$k$$ ($$k=0$$ would be a problem, of course). So, solving these equations for $$k$$, let's see what happens when $$f(k)$$ is made proportional to $$1/k^2$$, $$k \gt 0$$. For this to define a probability distribution, we would need the masses to be nonnegative and sum to unity. But $$\sum_{k=1} 1/k^2 = \pi^2/6$$. That's exactly what's needed to normalize $$f$$. Now as $$n$$ increases, take a little mass out of this positive part of $$f$$ and move it to $$0$$. An easy way to accomplish this would be to cut down the positive mass by $$1/n$$. Whence, defining

$$f_n(k) = \frac{1}{n}\frac{6}{\pi^2}\frac{1}{k^2} = \frac{6}{n\pi^2 k^2}, \ k \gt 0;\quad f_n(0) = 1 - \frac{1}{n}$$

will do it.