If $(X,Y)$ are i.i.d. exponential then $X/(X+Y)$ follows a Beta distribution [duplicate]

Question

I'm asked to show that if $X$ and $Y$ are independent exponential random variables with parameter , then has a Beta distribution.

Up to know I had to find the new pdf or df when $Y$ was of the kind $Y=aB + c$ which I understand now. Here a hint is given to use "Law of total probability" which I've only seen in measure theory.

My guess would be to compute using the fact that the df of exponential distribution is . I'm stuck at that step.

Answer 1

The naive approach: we have $$f_X(x) = e^{-x}, \quad f_Y(y) = e^{-y}, \quad x, y > 0.$$ Define $Z = X/(X+Y)$, which implies $Y = X(1/Z - 1)$, and note that we must have $0 < Z < 1$. Then $$\begin{align*} F_Z(z) &= \Pr[X/(X+Y) \le z] \\ &= \Pr[Y > X(1/z - 1)] \\ &= \int_{x=0}^\infty \Pr[Y > x(1/z - 1) \mid X = x]f_X(x) \, dx, \end{align*}$$ by the law of total probability*. Continuing, $$F_Z(z) = \int_{x=0}^\infty e^{-x(1/z-1)} e^{-x} \, dx = \int_{x=0}^\infty e^{-x/z} \, dx = \left[-ze^{-x/z}\right]_{x=0}^\infty = z,$$ hence $f_Z(z) = F_Z'(z) = 1$, and $Z$ is uniform on $(0,1)$.

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*Note. The law of total probability for a continuous random variable $X$ and some event of interest $A$ is $$\Pr[A] = \int_{x \in \Omega} \Pr[A \mid X = x] f_X(x) \, dx,$$ where $\Omega$ is the support of $X$.

Answer 2

Define $ U = X/(X+Y) = g_u(X,Y) $ and $V = X+Y = g_v(X,Y)$, thus $$ f_{U,V}(u,v) = f_X (g_u^{-1} (u,v))f_{Y} (g_v^{-1} (u,v))| \det\frac{\partial(X,Y)}{\partial(u,v)} | $$ where $X =UV$ and $Y = V (1-U)$, so $$ | \det\frac{\partial(X,Y)}{\partial(u,v)} |=v. $$ As such, $$ f_{U,V}(u,v)=e^{-uv}e^{-v(1-u)}v=ve^{-v}= \mathcal{G}amma(2,1)\mathcal{B}eta(1,1). $$ Finally, $$ f_U(u) = \int f_{U,V}(u,v)dv = \mathcal{B}eta(1,1) = \mathcal{U}(0,1). $$

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