Suppose $X_1 $ and $X_2$ are i.i.d standard normal r.v.s and $Y=X_1^2+X_2^2$, then we know $Y \sim \chi_2^2 $ and $f_Y(y)= \frac{1}{2}e^{\frac{-y}{2}}$. Using the identity $f_{X,Y}=f_{X\mid Y} \cdot f_Y $, we can calculate the joint density of $X=(X_1, X_2)$ and $Y$.

$f_{X\mid Y}(x\mid y)=f_X(x)=f_{X_1}(x_1)f_{X_2}(x_2)=\frac{1}{2\pi}e^{\frac{-y}{2}} $ if $x_1^2 +x_2^2 =y$ and zero otherwise. Finally we get $f_{X\mid Y}f_Y=\frac{1}{4\pi}e^{-y}$ if $x_1^2+x_2^2=y$ and zero otherwise. However this must not be the joint density because it does not integrate to one. Demonstration:

$$\int \int f_{X,Y} \,dx\,dy=\int_0^\infty\frac{e^{-y}}{4\pi}\int_{x_1^2+x_2^2=y} 1\,dx\, dy= \int_0^\infty\frac{e^{-y}}{4\pi}2\pi y^{\frac{1}{2}}\,dy=\frac{1}{2}\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{4} \neq 1$$

What did I do wrong?


There are (at least) two issues with your calculation.

  1. The equality $$\int_{x_1^2+x_2^2 =y}1 \, dx = 2\pi \sqrt{y}$$ is not correct. Note that you integrate with respect to the two-dimensional Lebesgue measure and that the (two-dimensional) Lebesgue measure of the set $\{x \in \mathbb{R}^2; x_1^2+x_2^2=y\}$ is $0$. Hence, $$\int_{x_1^2+x_2^2 =y}1 \, dx=0.$$ I just want to mention this, obviously, this doesn't answer your question.
  2. You didn't calculate the conditional density $f_{X \mid Y}$ correctly and therefore you do not obtain the correct joint density. But since $Y=X_1^2+X_2^2$ is fully determined by $X=(X_1,X_2)$ we can calculate the joint distribution $\mu_{X,Y}$ of $(X,Y)$ directly: Recall that $\mu_{X,Y}$ is the distribution of $(X,Y)$ if, and only if, $$\mathbb{E}g(X,Y) = \int g(x,y) \, d\mu_{X,Y}(x,y)$$ for all $g$ such that $g(X,Y) \in L^1$. It follows from the definition and the image measure theorem that $$\mathbb{E}g(X,Y) = \mathbb{E}(X,X_1^2+X_2^2) = \int g((x_1,x_2),x_1^2+x_2^2) \, d\mu_{X_1,X_2}(x_1,x_2).$$ Now we can rewrite the last expression using the Dirac measure $\delta_{x_1^2+x_2^2}$ concentrated at $x_1^2+x_2^2$: $$\mathbb{E}g(X,Y) = \int \int g((x_1,x_2),y) \delta_{x_1^2+x_2^2}(y) \, d\mu_{X_1,X_2}(x_1,x_2).$$ Finally, we can use that the joint distribution $\mu_{X_1,X_2}$ equals -by the independence- the product of the marginals, i.e. $$\mathbb{E}g(X,Y) = \int \int \int g((x_1,x_2),y) \delta_{x_1^2+x_2^2}(y) \, f_{X_1}(x_1) f_{X_2}(x_2) \, dx_1 \, dx_2.$$ Consequently, $$d\mu_{X,Y}(x,y) = \delta_{x_1^2+x_2^2}(y) \, f_{X_1}(x_1) f_{X_2}(x_2) \, dx_1 \, dx_2.$$
  • $\begingroup$ Can do this in a non-measure-theoretic argument? $\endgroup$ – Patrick Sep 27 '14 at 14:37
  • $\begingroup$ I still have no idea how to do it correctly. Can you give me some tricks? $\endgroup$ – Patrick Sep 27 '14 at 14:51
  • $\begingroup$ @Patrick No, I can't do it without measure theory. $\endgroup$ – saz Sep 27 '14 at 16:26
  • $\begingroup$ Fair enough, but I still need to calculate the joint density in question. How should I go about doing it? $\endgroup$ – Patrick Sep 27 '14 at 16:37
  • $\begingroup$ @Patrick In exactly the same way as in the example above. Show that $$\mathbb{E}g(X,Y) = \int g((x_1,x_2),y) \underbrace{f_{X_1,X_2}(x_1,x_2)}_{f_{X_1}(x_1) f_{X_2}(x_2)} \, \delta_{x_1^2+x_2^2}(dy) \, dx_1 \, dx_2$$ holds for any integrable function $g$. $\endgroup$ – saz Sep 27 '14 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.