# how to apply change of variable formula to find joint distribution of 2 random variables given the joint distribution of 3 random variables

I have trouble with find the joint distribution of $X_1,X_2$.

I know I need to use the change of variable formula and the first them is to express $Y_1,Y_2,Y_3$ in terms of $X_1,X_2,X_3$, where extra is just another random variable I needed to add and $X_3=Y_1$.

So, solving simultaneously, I get: $$y_1=x_3$$ $$y_2=\frac{x_3x_2-x_3x_1}{x_1}$$ $$y_3 = \frac{x_3-x_3x_2}{x_1}$$

But when I compare this with the given solution:

Looks like I am close to the provided solution but I don't know what else I can do to get the exact solution I doubt that I made any mistakes in my workings. Please help.

Also, what was the argument(s) used to determine the support of $f_{x_1,x_2,x_3}$? I know from my workings, where $x_3=y_1$, I can let $x_3>0$ because $y_1$ follows an exponential distribution. But I am not sure how to argue for $x_1,x_2$ because they are functions of 3 variables and this is the first time I am seeing this kind of question.

EDIT:

The rest of the solution is as follows:

EDIT 2: If I start with assuming that $\lambda>0$:

The integral I need to evaluate will be:

$\int\limits_0^\infty \lambda^3 e^{-\lambda s}s^2 \,ds = \lambda^3 \int\limits_0^\infty e^{-\lambda s}s^2 \,ds \tag{1}$

So,

\begin{align} \int e^{-\lambda s}s^2 \,ds & = \frac{-s^2e^{-\lambda s}}{\lambda}+\frac{2}{\lambda}\int 2s \frac{e^{-\lambda s}}{\lambda} \,ds \text{ ,integration by parts.} \\ & = \frac{-s^2e^{-\lambda s}}{\lambda} -\frac{2se^{-\lambda s}}{\lambda^2} - \frac{2e^{-\lambda s}}{\lambda^3} \text{ ,integration by parts again and then simplify} \tag{2} \end{align}

Evaluating (2) from $0$ to $\infty$ gives: \begin{align} \int\limits_0^\infty e^{-\lambda s}s^2 \,ds & = [0-0] - [0-0] - \left[0-\frac{2}{\lambda^3} \right] \\ &= \frac{2}{\lambda^3}, \tag{3} \end{align}

Putting (3) and (1) together gives: \begin{align} \int\limits_0^\infty \lambda^3 e^{-\lambda s}s^2 \,ds & = \lambda^3 \int\limits_0^\infty e^{-\lambda s}s^2 \,ds \\ & = \lambda^3 \frac{2}{\lambda^3} \\ & = 2 \end{align}

Which gives the desired result independent of the the value of $\lambda$.

Suppose that $Y_i\sim Exp(\lambda=1)$

We denote by $S$ the sum $Y_1+Y_2+Y_3$.

The joint pdf of the random variables $Y_1, Y_2, S$ is: $$\begin{array}{l}f_{Y_1, Y_2, S}(y_1, y_2, s)=f_{Y_1, Y_2}(y_1,y_2)f_{S|Y_1=y_1, Y_2=y_2}(s)=\\ =e^{-y_1}e^{-y_2}f_{Y_3}(s-y_1-y_2)=e^{-y_1-y_2}e^{-s+y_1+y_2}=e^{-s}\end{array}$$ for $y_1\geq 0, y_2\geq 0, s\geq y_1+y_2$. Here $f_{S|Y_1=y_1, Y_2=y_2}$ stands for the conditional pdf of the rv S, under conditions $Y_1=y_1$, $Y_2=y_2$, which amounts to the pdf of $Y_3$ evaluated at $s-y_1-y_2$.

In order to get the joint distribution of the random variables $X_1=\displaystyle\frac{Y_1}{S}, X_2=\displaystyle\frac{Y_1+Y_2}{S}, S$ we consider the transformation $\psi: (y_1, y_2, s)\mapsto (x_1, x_2, s)$, with: $$x_1=y_1/s,\quad x_2=(y_1+y_2)/s, \quad s=s$$

Hence $\psi^{-1}(x_1, x_2, s)=(y_1, y_2, s)$ is defined by:

$$y_1=sx_1, \quad y_2=-sx_1+sx_2, \quad s=s$$ and its Jacobian is:

$$\left|\begin{array}{rrr}s&0&x_1\\ -s&s &x_2-x_1\\0&0&1\end{array}\right|=s^2$$

Thus the joint distribution of the variables $X_1, X_2, S$ is: $g(x_1, x_2, s)=e^{-s}s^2$, for $sx_1, sx_2-sx_1\geq 0$ and $s\geq sx_1+sx_2-sx_1$. But these conditions amount to: $x_1, x_2\geq 0$, $x_1\leq x_2\leq 1$.

To get the joint density distribution, $h_{X_1, X_2}(x_1, x_2)$, of the random variables $X_1,X_2$ we integrate with respect to $s$ the density $g(x_1, x_2, s)$: $$h_{X_1, X_2}(x_1, x_2)=\int_0^\infty s^2 e^{-s}ds=2! \quad x_1, x_2\in[0,1], \quad x_1\leq x_2$$
But this is just the joint distribution of the order statistics, as you stated in your question.

If you start with $Y_k\sim Exp(\lambda)$, $k=1,2,3$, with $\lambda>0$ arbitrary you get that the joint distribution of the variables $X_1, X_2$ coincides with that of the order statistics $U_{(1)}, U_{(2)}$, where $U_1, U_2\sim Unif[0,1]$, iff $\lambda=1$.

• Thank you for your detailed solution. There a some parts that I don't follow, but before I get to them, it looks like it does not agree with the provided solution (i've updated my question with this). Specifically, the provided solution does not have an iif condition on $\lambda$ in order for the joint distributions to coincide. Is this because you used a different transformation that than the solution? – mauna Mar 22 '15 at 20:11
• No, my solution is not related to the chosen transformation. It is well known in the theory of random variate generation that in order to simulate the order statistics $U_{(1)}, U_{(2)}, \ldots, U_{(n)}$, one generates $n+1$ independent values from $Exp(\lambda=1)$ (see for example the Devroye's book Non-Uniform Random Variate Generation). – xecafe Mar 22 '15 at 21:17
• So the solution provided in my text is incorrect? – mauna Mar 22 '15 at 23:17
• @mauna I suppose that the author of your problem forgot to tell you that $\lambda=1$. – xecafe Mar 23 '15 at 5:21
• I've done your solution assuming arbitrary $\lambda>0$ and was able to get a $2$ from the integration (see my edit 2). Did I do the integration wrong. I don't have access to the Devroye's book. Do you know if there are online materials that discuss this point? – mauna Mar 23 '15 at 11:22