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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.

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marked as duplicate by Did probability-distributions Mar 1 '17 at 8:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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 $(0,1)$.


*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 $X$.

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  • $\begingroup$ Thanks for your answer. I don't understand what you put inside your first integral. I would have simply said Fz(z) = 1 - df of Y with value x(1/z - 1). Alright, then after it won't help because we have two variables. I need to find the formula for your integral $\endgroup$ – endlessend2525 Mar 1 '17 at 7:29
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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, $$ f_U(u) = \int f_{U,V}(u,v)dv = \mathcal{B}eta(1,1) = \mathcal{U}(0,1). $$

See here for more details and examples.

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