# Show that the random variables $Y_1$ and $Y_2$ are independent

Let $X_1,X_2$ be i.i.d with pdf $$f_X(x)=\begin{cases} e^{-x} & \text{for } 0< x<\infty{} \\0 & \text{elsewhere } \end{cases}$$ Show that the random variables $Y_1$ and $Y_2$ with $Y_1=X_1+X_2$ and $Y_2=\frac{X_1}{X_1+X_2}$ are independent.

I know that for $Y_1$ and $Y_2$ to be independent. $P(Y_1\cap Y_2)=P(Y_1)P(Y_2)$.

• What you wrote down is indepence of events. For independent random variables, see here: probabilitycourse.com/chapter3/3_1_4_independent_random_var.php – ToucanNapoleon Oct 3 '16 at 8:32
• More information is needed. I suspect that $X_1$ and $X_2$ are iid, but that is not mentioned in your question. – drhab Oct 3 '16 at 8:40
• @HarrySmit. Thanks for the clarification. – angelo086 Oct 3 '16 at 8:43
• @drhab. The question doesn't mention it. But all the problems in class had variables that were iid so that is a fair assumption. – angelo086 Oct 3 '16 at 8:44
• A direct route is to compute the PDF of $(Y_1,Y_2)$ using the arch classical change of variables formula and to check that this PDF factorizes as a product. Did you try that? – Did Oct 3 '16 at 8:46

It seems that you already have found that $X_1=Y_1Y_2$ and $X_2=Y_1(1-Y_2)$. But you are not done. What is the domain of $Y_1$ and $Y_2$? Since $X_1,X_2>0$ you have that $Y_1>0$ and $0<Y_2<1$. Hence \begin{align*}f_{Y_1,Y_2}(y_1,y_2)&=f_{X_1,X_2}(y_1y_2,y_1(1-y_2))\begin{vmatrix}y_2&y_1\\1-y_2&-y_1\end{vmatrix}\mathbf{1}_{\{0<y_1, 0<y_2<1\}}\\[0.3cm]&=e^{-y_1y_2}e^{-y_1+y_1y_2}|-y_1|\mathbf{1}_{\{0<y_1, 0<y_2<1\}}\\[0.3cm]&=y_1e^{-y_1}\mathbf{1}_{\{0<y_1\}}\mathbf{1}_{\{0<y_2<1\}}\\[0.3cm]&=\underbrace{y_1e^{-y_1}\mathbf{1}_{\{0<y_1\}}}_{f_{Y_1(y_1)}}\underbrace{\mathbf{1}_{\{0<y_2<1\}}}_{f_{Y_2(y_2)}}\end{align*} So, never forget the domain! The result is that $Y_2 \sim U(0,1)$ and $Y_1\sim f_{Y_1}(y_1)=y_1e^{-y_1}$ for $0<y_1$.
Here is a simulation of 100,000 $(Y_1, Y_2)$-pairs from R statistical software. The $X_i$ are iid $Exp(rate=1),$ $Y_1 \sim Gamma(shape=2, rate=1)$ and $Y_2 \sim Unif(0, 1).$ Also, $Y_1$ and $Y_2$ are uncorrelated. (If these distributions are not covered in your text, you can see Wikipedia articles on 'exponential distribution' and 'gamma distribution'.)
In the figure below, the first panel shows no pattern of association between $Y_1$ and $Y_2$. Of course, this is no formal proof of independence, but if you do the bivariate transformation to get the joint density function of $Y_1$ and $Y_2,$ you should be able to see that it factors into the PDFs of $Y_1$ and $Y_2$. These PDFs are plotted along with the histograms of the simulated distributions of $Y_1$ and $Y_2$ in the second and third panels, respectively.