# Product of three independent random variables

We are given three independent random variables $X, Y, Z$.

$X$ has the following Bernoulli distribution: $P(X=1)=\frac{3}{4}, \ P(X=0) = \frac{1}{4}$

$Y$ has a uniform distribution on the interval $[-1,1]$

$Z$ has a uniform distribution on the interval $[0,2]$

How to establish the distribution of random variable $W=XY - (1-X)Z$?

I know how to solve a similar problem, that is if we are given a random variable $X$ and its distribution and are asked to find the distriobution of $g(X)$, for example $X^2, \ aX+b,$ etc

But I don't know what to do if there are several random variables and of different types of distributions.

Could you help?

Using the tower rule and the independence of our random variables we write that

$F_W(x)=P(W(X,Y,Z)<x)=E[P(W(X,Y,Z)<x|X)]=$ $P(W(X,Y,Z)<x|X=1)\frac{3}{4}+P(W(X,Y,Z)<x|X=0)\frac{1}{4}=$ $P(W(1,Y,Z)<x)\frac{3}{4}+P(W(0,Y,Z)<x)\frac{1}{4}.$

Since $W(1,Y,Z)=Y,\text{ and }W(0,Y,Z)=-Z$ we have finally $$F_W(x)=P(Y<x)\frac{3}{4}+P(-Z<x)\frac{1}{4}.$$

Given that $Y$ is uniformly distributed over $[-1,1]$ and $-Z$ is uniformly distributed over $[-2,0]$, $$P(Y<x)=\begin{cases}0 &\text{ if } x<-1\\ \frac{1}{2}(x+1) &\text{ if } -1\le x \le 1\\ 1&\text{ if } x>1 \end{cases},$$ and $$P(-Z<x)=\begin{cases}0 &\text{ if } x<-2\\\frac{1}{2}x+1& \text{ if } -2\le x \le 0\\ 1&\text{ if } x>0 \end{cases}.$$

The rest is easier to be drawn than told. Here is $F_W(x)$ the distribution function of $W$:

Edited to better explain the first steps of the calculation

Rather than referring to the tower rule we can do the first steps directly:

$P(W(X,Y,Z)<x)=$ $P(\{W(X,Y,Z)<x\}\cap \{X=1\}\cup \{W(X,Y,Z)<x\}\cap \{X=0\})=$ $P(\{W(X,Y,Z)<x\}\cap \{X=1\})+P(\{W(X,Y,Z)<x\}\cap \{X=0\})=$ $P(\{W(1,Y,Z)<x\}\cap \{X=1\})+P(\{W(0,Y,Z)<x\}\cap \{X=0\}).$

And now, here comes independence:

$P(\{W(1,Y,Z)<x\}\cap \{X=1\})=P(W(1,Y,Z)<x)P(X=1)=P(W(1,Y,Z)<x)\frac{3}{4}$ $P(\{W(0,Y,Z)<x\}\cap \{X=0\})=P(W(0,Y,Z)<x)P(X=0)=P(W(0,Y,Z)<x)\frac{1}{4}.$

• Thank you very much. Could you tell me why this equation holds? $$P(W(X,Y,Z)<x)=E[P(W(X,Y,Z)<x|X)]$$ Feb 28 '15 at 7:45
• The rest of calculations is clear to me. I suppose it somehow follows from en.wikipedia.org/wiki/Law_of_total_expectation , but I don't see how, at the moment. Feb 28 '15 at 8:15
• @Hagrid: Yes, the first step of my solution is based on en.wikipedia.org/wiki/Law_of_total_expectation. I will add some explanatory notes to my answer so you get full satisfaction.
– zoli
Feb 28 '15 at 9:08
• @Hagrid: I did the promised edits.
– zoli
Feb 28 '15 at 9:31
• Thank you. Everything is clear to me now. Feb 28 '15 at 12:01