Let $X_1$ have p.d.f $$p_1(x_1)=x_1 \cdot \text{exp} \left( \frac{-x_1^2}{2} \right),$$ and $X_2$ have p.d.f $$p_2(x_2) = \frac{1}{\sqrt{2 \pi}} \text{exp} \left( \frac{-x_2^2}{2} \right). $$ Calculate the distribution of $$X=X_1X_2.$$
In the solution it is claimed that $$p(x) = \iint_{\mathbb{R}^2} p_1(x_1)p_2(x_2) \delta(x-x_1x_2) \, dx_1dx_2=...,$$ where $\delta$ is the dirac-delta function. I have two questions:
- Can someone give me the intuition behind this formula?
- Does this hold independent of $X_1$ and $X_2$ beeing independent?
My personal intuition: the Dirac delta is $=0$ for all values of $x$ that are not equal to $x_1 x_2$. So the probability of $X$ taking a value different from $X_1 X_2$ is zero. When $X=X_1 X_2$, the probability of $X=x$ is loosely speaking given by the sum of all possible combination of $X_1=x_1$ and $X_2=x_2$ (weighted with the probability of their occurrence) such that their product is $=x$.If $X_1$ and $X_2$ were not independent, the formula would work as $$ p(x) = \iint_{\mathbb{R}^2} p(x_1,x_2) \delta(x-x_1x_2) \, dx_1dx_2\ ,$$ where $p(x_1,x_2) $ is the joint probability density function of $X_1$ and $X_2$. If you have a more general function $X=F(X_1,X_2)$, the formula would be $$ p(x) = \iint_{\mathbb{R}^2} p(x_1,x_2) \delta(x-F(x_1,x_2)) \, dx_1dx_2\ .$$