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. $P(Y_1\cap Y_2)=P(Y_1)P(Y_2)$.
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'.)
x1 = rexp(10^5); x2 = rexp(10^5)
y1 = x1 + x2; y2 = x1/y1
cor(y1,y2)
## 0.002440974 # consistent with 0 population correlation
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.
