Let $X_1$ and $X_2$ be independent and identically distributed continuous random variables, with probability density function

$$p(x)=\begin{cases} \exp(-x), & \text{if}\ x>0 \\ 0, & \text{otherwise}. \end{cases}$$


$$Y_1=\frac{X_1}{X_1+X_2}, \ Y_2=2X_2$$



Derive the probably density function of $Y_1$, $Y_2$ and $Z$. For $Z$ it is sufficient to give the required pdf in the form of an integral of a joint pdf.

Firstly, note that $y_1,y_2>0$ with the given transformations.

Now, I found the transformation of $Y_2=2X_2$ by doing the following.

The cdf of $X_2$ is $C(x_2)=-\exp(-x_2)$ and so the cdf for $Y_2$ is

\begin{align} G(y_2)&=\Pr(Y_2 \leq y_2) \\ &=\Pr(2X_2 \leq y_2) \\ &=\Pr\left(X_2 \leq \frac{y_2}{2}\right) \\ &=C\left(\frac{y_2}{2}\right) \\ &=-\exp\left(-\frac{y_2}{2}\right) \\ &\therefore \frac{dG(y_2)}{dy_2}=g(y_2)=\frac12\exp\left( -\frac{y_2}{2} \right) \end{align}

Hence, the pdf for $Y_2$ is $g(y_2)=\begin{cases} \frac12\exp\left( -\frac{y_2}{2} \right), & \text{if}\ y_2>0 \\ 0, & \text{otherwise}. \end{cases}$

I'm pretty sure I've done this correctly but now I am rather lost for the transformation of $Y_2$.

The joint pdf of $X_1$ and $X_2$ is $f(x_1,x_2)=\begin{cases} \exp(-x_1-x_2), & \text{if}\ x_1,x_2>0 \\ 0, & \text{otherwise}. \end{cases}$

as $X_1$ and $X_2$ are independent. And so for $Y_1$ and $Y_2$, the inverse functions are $x_2= v_2(y_1,y_2) =\frac12y_2$ and $x_1=v_1(y_1,y_2)=\frac{y_1y_2}{2(1-y_1)}$. Hence, the Jacobian matrix is

$$J=\begin{bmatrix} \frac{y_2}{2(y_1-1)^2} & \frac{y_1}{2(1-y_1)} \\ 0 & \frac12 \\ \end{bmatrix} \implies |J|=\frac{y_2}{4(y_1-1)^2}$$

But this doesn't seem nice at all, have I done something wrong here?


2 Answers 2


Instead of finding the joint density of $Y_1$ and $Y_2$, find that of $Y_1$ and $Y_3=X_1+X_2$. Then the Jacobian works out very neatly. That gives you the distribution of $Y_1$ and you don't need that of $Y_3$.

(You'll find out that $Y_1$ is uniformly distributed on the interval $[0,1]$.)

\begin{align} y_1 & = \frac{x_1}{x_1+x_2} \\[8pt] y_3 & = x_1+x_2 \\[20pt] x_1 & = y_1 y_3 \\[8pt] x_2 & = (1-y_1)y_3 \end{align}

  • $\begingroup$ This is what I got. So $x_1=v_1(y_1,y_3)=y_1y_3$ and $x_2=v_2(y_1,y_3)=y_3(1-y_1)$ and so $|J|=y_3$. Hence, the joint pdf of $y_1$ and $y_3$ is $g(y_1,y_3)=f(v_1(y_1,y_3),v_2(y_1,y_3))|J|=y_3\exp(-y_1y_3-y_3(1-y_1))=y_3\exp(-y_3)$. Now, since $0<x_1<\infty$ and $0<x_2<\infty$, $0<y_1y_3<\infty$ and $0<y_3(1-y_1)<\infty \implies 0<y_1<1$ and $0<y_3<\infty$. So the marginal pdf for $y_1$ is $g(y_1)=\int_0^{\infty}y_3\exp(-y_3) \ dy_3=1$. Is this correct? $\endgroup$ Dec 9, 2015 at 10:13
  • $\begingroup$ Also, for $Z=Y_1+Y_2$, how would I approach this? As $Y_1$ and $Y_2$ do not have a joint distribution and they are not necessarily independent. $\endgroup$ Dec 9, 2015 at 10:39

For distribution of $Y_1$ you have (for $t \in (0,1)$): $$P(Y_1<t)=P\left( \frac{X_1}{X_1+X_2}<t \right)=P\left( X_2>\frac{1-t}{t}X_1 \right).$$ All you need to do is find the integral.

As for the joint distribution everything is fine so far.


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