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?

  • $\begingroup$ $X$ and $Y$ are independent? If not, the problem is false (consider $X=Y$) $\endgroup$ Nov 26, 2021 at 16:49
  • $\begingroup$ Yes they are independent. I will edit my description. $\endgroup$
    – Farhood ET
    Nov 26, 2021 at 16:54
  • $\begingroup$ @MartínVacasVignolo I've edited my question. $\endgroup$
    – Farhood ET
    Nov 26, 2021 at 16:55
  • 1
    $\begingroup$ 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}$ $\endgroup$
    – Math Lover
    Nov 26, 2021 at 17:53
  • $\begingroup$ How to prove these two random variables are independent? $\endgroup$ Nov 26, 2021 at 19:15

1 Answer 1


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

  • $\begingroup$ Does uniformity of $f_Z(z)$ means that it is independent of $W$? $\endgroup$
    – Farhood ET
    Nov 26, 2021 at 17:12
  • $\begingroup$ @Farhood ET: yes because it does not depends on $\lambda$ anymore. I added another way to solve your problem $\endgroup$
    – tommik
    Nov 26, 2021 at 17:15
  • $\begingroup$ 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? $\endgroup$
    – Farhood ET
    Nov 26, 2021 at 17:17
  • 1
    $\begingroup$ @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 $\endgroup$
    – tommik
    Nov 26, 2021 at 17:28
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    $\begingroup$ Writing $f_Z(z)=\mathbb{1}_{[0,1]}(z)$ is more advisable than $f_z(z)=1$. $\endgroup$
    – Jean Marie
    Nov 26, 2021 at 18:15

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