How to compute this marginal density when the integral does not converge?

Question

I have this mass density function: $$f_{X,Y}(x,y) = \frac{2\ {e}^{-2\ x}}{x}\ I_{[0,\infty)}(x)\ I_{[0,x]}(y)$$ and I need to find $Cov[X,Y]$. So far, so good, I found $E[XY] = \frac{1}{4}$ and found $f_X(x)$ by integrating $f_{X,Y}(x,y)$ in $Y$, then found $E[X] = \frac{1}{2}$. But when I try to do the same for $Y$ the integral does not converge: $$f_Y(y) = \int_{0}^{\infty} \frac{2\ {e}^{-2\ x}}{x} dx$$ How to pass that?

Answer 1

The marginal density of $Y$ can only be expressed in terms of the upper incomplete gamma function, but the map $$(x,y)\mapsto \frac{2y e^{-2x}}x\mathsf 1_{[0,\infty)\times [0,x]}(x,y)$$ is measurable and nonnegative, so we may use Tonelli's theorem to compute \begin{align} \mathbb E[Y]&=\int_0^\infty y\int_y^\infty \frac{2y e^{-2x}}x\,\mathsf dx\,\mathsf dy\\ &= \int_{[0,\infty)\times[0,x]}\frac{2y e^{-2x}}x\,\mathsf d(x\times y)\\ &= \int_0^\infty \int_0^x \frac{2y e^{-2x}}x\,\mathsf dy\,\mathsf dx\\ &= \int_0^\infty xe^{-2x}\,\mathsf dx\\ &= \frac14. \end{align}

It follows that \begin{align} \operatorname{Cov}(X,Y) &= \mathbb E[XY] - \mathbb E[X]\mathbb E[Y] \\ &= \frac14 - \frac12\cdot\frac14\\ &= \frac18. \end{align}