Computing the joint CDF of $(X,X^3)$ with $X$ exponential

Question

Let $X\sim \text{Exponential}(\lambda)$, and let $Y=X^3$. Compute the joint CDF $F_{X,Y}$ of $(X,Y)$.

With my current understanding, I can only come up with using the one dimensional change in variable where I set $F_Y(y) = P(Y \leq y) = P((\lambda e^{-\lambda x})^3\leq y)$, then isolating for y, and then apply the transformation.

I am not sure if this is even a valid solution.

Answer 1

The PDF of $(X,Y)$ does not exist since the support of the distribution of $(X,Y)$ is a subset of the set $\{(x,y)\in\mathbb R^2\,;\,y=x^3\}$, whose measure is zero.

To compute the CDF $F$ of $(X,Y)$, note that $$\{X\leqslant x,Y\leqslant y\}=\{X\leqslant x,X^3\leqslant y\}=\{X\leqslant x,X\leqslant y^{1/3}\}=\{X\leqslant\min\{x,y^{1/3}\}\}$$ In terms of the CDF $F_X$ of $X$, one gets

$$F(x,y)=F_X(\min\{x,y^{1/3}\})$$

In the present case, if $x$ or $y$ is negative, then $F(x,y)=0$, and, if $x$ and $y$ are nonnegative, then $$ F(x,y)=1-e^{-\lambda\min\{x,y^{1/3}\}} $$