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The question is about a particular exercise.

Let X,Y be independent random variables with exponential distribution where both have the same parameter $\lambda$. Define $Z_1:=X_1+X_2$. Calculate the density of $Z_1$. Also define $Z_2:=\frac{X_1 X_2}{X_1+X_2}$.

$Z_3:=\frac{X_1^2}{X_1+X_2}$

$Z_4:=\frac{X_2}{X_1+X_2}$

Are $Z_1,Z_2$ independent?

Are $Z_1,Z_3 $independent?

Are $Z_1,Z_4$ independent?

My Approach: I calculated the densitiy of Z$_1$by using the convolution formula, but now I'm struggling to show that $Z_1$ and $Z_4$ are independent. (I got a hint that these two are independent.)

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Note $f_{X_1X_2}(x_1,x_2)=\lambda^2 e^{-(x_1+x_2)} \cdot 1_{[0,\infty)^2}$. Also note $Z_4$ is supported on $(0,1)$. If $z_4\in (0,1)$ is fixed then $$\begin{eqnarray*} F_{Z_4}(z_4)&=&P(Z_4 \leq z_4) \\&=& P\Bigg(X_2\leq\frac{z_4X_1}{1-z_4}\Bigg) \\ &=& \int_0^{\infty} \int_0^{\frac{z_4x}{1-z_4}}\lambda^2e^{-\lambda(x+y)}\mathrm{d}y\mathrm{d}x \\ &=& z_4\end{eqnarray*}$$ This means $Z_4\sim \mathcal{U}(0,1)$. You should have gotten with convolution that $f_{Z_1}(z_1)=\lambda^2 z_1e^{-\lambda z_1}\cdot 1_{(0,\infty)}$. Using multivariate change of variables we get for $(z_1,z_4)\in (0,\infty)\times (0,1)$ that $$\begin{eqnarray*} f_{Z_1Z_4}(z_1,z_4)&=&f_{X_1X_2}\Big(z_1-z_1z_4,z_1z_4\Big)\Bigg|\frac{\partial(z_1-z_1z_4,z_1z_4)}{\partial(z_1,z_4)}\Bigg| \\ &=& \lambda^2 z_1e^{-\lambda z_1} \\ &=& f_{Z_1}(z_1)f_{Z_4}(z_4)\end{eqnarray*}$$ So you get independence.

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