I'm dealing with this problem

Suppose that $X, Y$ are iid $N(0,1)$ random variables and let $R= \sqrt{X^2+Y^2}$. Find the joint pdf of $R$ and $R^2$.

I'm not sure how to solve the joint pdf of dependent random variables. I just tried to find cumulative distribution and differentiate it.

By change of variables, I got $f_R(r) = \frac{1}{2\pi}r\exp{-\frac12r}$. $$f_{R,R^2}(x,y) = \frac{\partial^2}{\partial x \partial y} P(R \le x, R^2 \le y)= \frac{\partial^2}{\partial x \partial y} P(R \le x , -\sqrt{y} \le R \le \sqrt{y} ) $$ so this is nonzero only if $-\sqrt{y} \le x \le \sqrt{y}$, and this is equal to $\frac{\partial^2}{\partial x \partial y} \int^x_{-\sqrt{y}} f_R(r) dr =0$, but this doesn't make sense. How should I deal with this?


1 Answer 1


You are right that there is an issue here. In general if $X$ is an RV, $X$ and $X^2$ will not have a joint density (though they will certainly have a joint distribution). The issue is that the 2D distribution will be singular. This is because $X^2$ can only take one value conditional on $X,$ so the density will be concentrated along a $1$-dimensional curve in $X-X^2$ space.

This isn't true for all dependent variables. For instance if $X$ and $Y$ are jointly Gaussian with correlation $.5,$ they are dependent but they have a joint density that is a function. However if they have correlation $1,$ they will be singular since one is just a multiple of the other.

However, we can still write down a density in terms of delta functions. Let $Y=X^2$ to clean up the notation a little bit. Then if $f_X(x)$ is the PDF for $X$ we will have $$ f_{X,Y}(x,y) = f_X(x)\delta(y-x^2)$$ where $\delta$ is the Dirac delta function.

Alternatively, just work with CDFs as you were above. To see where you went wrong in deriving that the PDF is zero, consider that there is an implicit step function in the CDF that you forgot to write in before differentiating.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .