I just wanted to confirm my understanding related to the distributions of functions of random variables. Can someone please tell me if all of my points are correct and make sense? I also have an example that I can't grasp yet. As a convention, I use $v(y)$ as an inverse function of $x$

  1. For the discrete r.v $X$, we can find the pdf of $Y=U(X)$ by: $$P(Y=y)=P(U(X)=y)=P(X=v(y)), y\in S_Y$$. where $\forall y \in S_y \ \forall x \in S_x \ y=u(x)$(my notation here might be slightly incorrect, but I imply that we can simply find the Support of $Y$ by applying the $u(x)$ on every $x$)

  2. For the continuous r.v there are 2 ways to do that: either through the distribution function technique or through the change of variable technique.

For the distribution function technique, we make use of the fact that $F'_Y(y)=f_y(y)$ ,so we simply express the probability $P(Y\le y)=P(X\le v(y))$, solve the integral, and differentiate the result to obtain $f_y(y)$

For the change of variable technique, we take into account that $f_Y(y)=f(v(y))|v'(y)|$ for one-to-one function and $f_Y(y)=f(v(y1))|v'(y1)|+f(v(y2))|v'(y2)|$ for a two-to-one function and do the same steps as for one-to-one function.

3)For the joint probability distribution of functions of r.v, $Y1=g1(X1,X2)$ and $Y2=g2(X1,X2)$ we use the following formula: $$f_{Y_1Y_2}(y1,y2)=f_{X_1,X_2}(x_1,x_2)|J(x_1,x_2)|^{-1}$$, where $x_1=h_1(y_1,y_2)$ and $x_2=h_2(y_1,y_2)$ , $J$ is a Jacobian determinant.

If everything so far is correct and I haven't made a single mistake, there's an example related to this topic I don't seem to understand.


Let $X_1\sim \chi^2(2),X_2\sim \chi^2(2)$ , $X_1,X_2$ are independent. Find the pdf of $Y=X_1+X_2$. My main issue with this question is that it doesn't seem to be one of the cases I discussed above. How would you approach this problem? Also, how would your approach change if $X_1, X_2$ were actually not independent?

P.S. Sorry for the lengthy question but I felt that the problem I am having with the given example is related to my understanding, so I included my explanation as well.


1 Answer 1


You are on the right lines. You have assumed that the inverse function $v$ exists and that $u$ is increasing.

If $u$ is not increasing then your distribution function equation $P(Y\le y)=P(X\le v(y))$ won't hold. Instead, $P(Y\le y)=\int_{u^{-1}((-\infty,y])}f_X(x)dx$, where $u^{-1}(S)$ is the pre-image of the set $S$ under $u$.

And if $u$ is not invertible, then your change of variable equation doesn't hold because $v$ doesn't exist. Instead you would need: $f_Y(y) = \sum_{x \in u^{-1}(\{y\})}\frac{f_X(x)}{|u'(x)|}$.

Your last question is about finding the pdf of the sum of two random variables. Google "convolution of probability distributions" to get started.

  • $\begingroup$ Thanks for the clarification! I checked the convolutions and it's definitely what I was looking for. $\endgroup$
    – Alex.Kh
    Nov 26, 2020 at 21:53

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