Random variable with infinite expectation but finite conditional expectation

Question

I've been very stuck on a question from Probability and Random Processes by Grimmett and Stirzaker for ages - so stuck that I flicked to the back to have a look at the answers. But, I can't seem to even understand the reasoning behind that!

The question is: Construct an example of two random variables X and Y for which $\mathbb{E}(Y)=\infty$ but $\mathbb{E}(Y|X)<\infty$.

The answer given is: Take Y to be a random variable with mean $\infty$, say $f_{Y}(y)=y^{-2}$ for $1 \leq y<\infty$, and let X=Y. Then $\mathbb{E}(Y|X)=X$ which is (almost surely) finite.

Firstly, how can a random variable have infinite mean? Secondly, I don't really understand the subtlety of letting X=Y and finally what's with the "(almost surely)"?

Thanks in advance for your help!

Answer 1

1. A random variable $Y$ has infinite mean if $\mathbb{E}[Y] = \infty$; in your case, this happens since $$ \int_1^{\infty} y f_Y(y) dy = \int_1^\infty \frac{1}{y} dy = \infty. $$

2. $X$ and $Y$ are just functions. As long as they are measurable (which you don't really need to worry about) you have valid random variables. I wouldn't call it sublte. Many other choices would have worked ($X = Y^3$ for instance). Here, once you "know" the value of $X$, you also "know" the value of $Y$, and it will always be finite. Each value of the function is finite, even though the integral above is infinite.

3. Any statement that has "almost surely" at the end of it means that the set on which the claim is true has measure $1$. In this case, the reason we write "a.s." in $$ \mathbb{E}[Y \mid X] = Y \ \ \mbox{a.s.} $$ is that conditional expectations are really choices of Radon-Nikodym derivatives (functions which satisfy 2 properties) and for some choices the equality holds everywhere, and for others it holds only "almost everywhere".

Answer 2

Take Y to be a random variable with mean ∞, say $f_Y(y)=\frac y2$ for $1≤y<∞$, and let $X=Y$. Then $(Y|X)=X$ which is (almost surely) finite.

Firstly, how can a random variable have infinite mean?

Consider a positive variable with a density $f$. An infinite mean means that $$ \int x f(x)dx = \infty $$ In your example:

$$ \int x f(x)dx = \int_1^\infty \frac x{x^2}dx = \int_1^\infty \frac {dx}{x}= \infty $$

Secondly, I don't really understand the subtlety of letting $X=Y$ and finally what's with the "almost surely"?

You have $$ (Y|X)=X\\ \Bbb P((Y|X) <\infty) = \Bbb P(X <\infty) =\int_1^\infty \frac {dx}x = 1 $$

So $(Y|X) <\infty$ a.s.

Answer 3

The "amost surely" is not necessary, $X$ is never infinite on the given domain. A random variable can have finite distribution but infinite mean if $\int f_Y(y) dy$ is a convergent integral but $\int y f_Y(y)$ is divergent. This is the case for example with $f_Y(y) = 1/y^2$, because $\int_1^\infty 1/y^2 dy$ is a convergent integral but $\int_1^\infty y(1/y^2) dy = \int_1^\infty 1/y$ is a divergent integral (equal to $\infty$). Finally, if you condition the expectation of a variable on itself then you just get the variable back, and in this case the variable is always finite which is why the example works.