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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?

Farhood ET
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1 Answers1

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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

Another way to prove independence is to use Basu's theorem.

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

tommik
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  • Does uniformity of $f_Z(z)$ means that it is independent of $W$? – Farhood ET Nov 26 '21 at 17:12
  • @Farhood ET: yes because it does not depends on $\lambda$ anymore. I added another way to solve your problem – tommik Nov 26 '21 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? – Farhood ET Nov 26 '21 at 17:17
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    @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

     – tommik Nov 26 '21 at 17:28
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    Writing $f_Z(z)=\mathbb{1}_{[0,1]}(z)$ is more advisable than $f_z(z)=1$. – Jean Marie Nov 26 '21 at 18:15