# Given Independent $X, Y$, Prove $X+Y=W$ and $\frac{X}{X+Y}=Z$ are independent if $X$ and $Y$ are identical exponential distributions [duplicate]

Given independent, $$X$$ and $$Y$$, I have to prove that if $$X + Y = W$$ and $$\frac{X}{X+Y} = Z$$ are independent random variables given that $$X \sim \text{exponential}(x; \lambda), \space Y \sim \text{exponential}(y; \lambda)$$.

I have derived $$F_W(w)$$ and $$F_Z(z)$$ from $$F_{XY}(x,y)$$ using integration, but I have no idea how to prove these two integrations are independent.

$$F_Z(z) \sim \text{uniform}(0, 1),\\ F_W(w) = u(w)[1 - (1 + \lambda w)e^{-\lambda w}]$$

I know that I have to use $$\text{Pr}\{W \le w, Z\le z\} = \text{Pr}\{W \le w\}\cdot \text{Pr}\{Z \le z\}$$, but I cannot integrate on the joint distribution $$f_{XY}(x,y)$$ with those conditions.

Any idea how to prove these two variables are independent?

• $X$ and $Y$ are independent? If not, the problem is false (consider $X=Y$) Nov 26, 2021 at 16:49
• Yes they are independent. I will edit my description. Nov 26, 2021 at 16:54
• @MartínVacasVignolo I've edited my question. Nov 26, 2021 at 16:55
• There is a mistake in $F_W(w)$. $\displaystyle F_W(w) = \int_0^{w} \int_0^{w-y} e^{- \lambda (x+y)} ~ dx ~ dy = 1 - (1 + \lambda w) e^{-\lambda w}$ Nov 26, 2021 at 17:53
• How to prove these two random variables are independent? Nov 26, 2021 at 19:15

You ca use jacobian method. You already know that $$W\sim\text{Gamma}[2;\lambda]$$ thus when you get $$f(w;z)$$ you realize that $$Z\sim U(0;1)$$ independent from W
First observe that exponential distribution is a scale family. $$W$$ is Complete and Sufficient while $$Z$$, scale invariant, is Ancillary. Thus invoking Basu's theorem, $$Z;W$$ are independent
• Does uniformity of $f_Z(z)$ means that it is independent of $W$? Nov 26, 2021 at 17:12
• @Farhood ET: yes because it does not depends on $\lambda$ anymore. I added another way to solve your problem Nov 26, 2021 at 17:15
• Thanks! I have another question, I have found out that $f_Z(z)$ and $f_W(w)$ only depend on $z$ and $w$ respectively, is this sufficient to say that these two variables are independent? Or is it not? Nov 26, 2021 at 17:17
• @Farhood ET :You found that $$f_{WZ}(w,z)=\lambda^2 w e^{-\lambda w}$$ $$f_W(w)=\lambda^2 w e^{-\lambda w}$$ $$f_Z(z)=1$$ Thus $f_W(w)\cdot f_Z(z)=f_{WZ}(w,z)$ which is exactly the definition of independence Nov 26, 2021 at 17:28
• Writing $f_Z(z)=\mathbb{1}_{[0,1]}(z)$ is more advisable than $f_z(z)=1$. Nov 26, 2021 at 18:15