[Math] The product of integrable random variables need not be integrable

probability theory

The product of integrable random variables need not be integrable

@Did gave a great example showing that in general the product
of two Lebesgue integrable functions need not be integrable. I thought
I could just tweak that example and prove this. However, I found the
“tweaking'' is not easy. Suppose random variable $X,Y\in L^{1}$.
It means that $E(X),E(Y)<+\infty$ , that is, $\int XdP,\int YdP<+\infty$.
I want to find an example where $\int XYdP$ is infinite.

The problem with the definition of expectation is that it is not convenient
to calculate. The convenient way of computing the expectation is to
use the density function : $E(X)=\int_{-\infty}^{\infty}xf(x)dx$
. However, if I use the density function, then unless $X$ and $Y$
are independent, contructing the density function for $XY$ and making
$E(XY)$ infinite will be tricky.

Any suggestion? Thank you.

Best Answer

Explaining the ingredients of the counterexample by Did:

  1. Taking $X=Y$ does not limit our chances of finding a counterexample, because $XY\le X^2+Y^2$. (So, if $XY$ has infinite expected value, then one of $X^2$ and $Y^2$ does too.)
  2. The focus is on the large values of $X$, because this is where squaring hurts.
  3. The integral that gives the expected value of $X$ should converge, but not too rapidly (like the Gaussian), more like $\int_1^\infty x^{-p}\,dx$ with $p>1$ to be determined.
  4. According to 3, the pdf of $X$ is $c_px^{-p-1}\cdot [x>1]$ where $c_p$ is normalization constant and $[\ \ ]$ the Iverson bracket.
  5. Instead of computing the pdf of $X^2$, we can directly calculate $E(X^2)$ using the pdf of $X$: it is $c_p\int_1^\infty x^2x^{-p-1}\,dx=c_p\int_1^\infty x^{-p+1}\,dx$
  6. The integral in 5 diverges when $p\le 2$. Thus, any $p$ in the range $1< p\le 2$ could be used. We can just as well pick the integer value $p=2$.
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