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Suppose that $X$ and $Y$ are i.i.d. r.v.'s with an exponential distribution of parameter $1$.Then it is known that the ratio $$Z = \frac{X}{X+Y}$$ has a uniform distribution on $(0,1)$. See for instance: X,Y are independent exponentially distributed then what is the distribution of X/(X+Y)

More generally, assume that $X$ and $Y$ are i.i.d. continuous r.v.'s with convex support in $\mathbb{R}^+$. What is a set of sufficient conditions such that $$Z = \frac{X}{X+Y}$$ has a uniform distribution on $(0,1)$?

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Let us assume that $f$, the PDF of $X$ and $Y$, is smooth and supported on a subset of $\mathbb{R}^+$.
For any $\alpha\in(0,1)$ we want that $$\mathbb{P}\left[\frac{X}{X+Y}\leq \alpha\right]=\mathbb{P}\left[X \leq \frac{a}{1-\alpha} Y\right]=\alpha $$ or $$\alpha= \int_{0}^{+\infty}\int_{0}^{\frac{\alpha}{1-\alpha}y}f(x)\,f(y)\,dx\,dy = \int_{0}^{+\infty}\int_{0}^{\frac{\alpha}{1-\alpha}}f(y t)\,yf(y)\,dt\,dy$$ or $$1-\alpha= \int_{0}^{+\infty}\int_{0}^{1}f\left(y\frac{\alpha}{1-\alpha} t\right)\,yf(y)\,dt\,dy$$ or by setting $\frac{\alpha}{1-\alpha}=\beta$ $$\forall \beta>0,\qquad\frac{1}{\beta+1}=\int_{0}^{+\infty}\int_{0}^{1}f(y\beta t)y\,f(y)\,dt\,dy = \iint_{0\leq t\leq y\leq+\infty}f(\beta t)\,f(y)\,dt\,dy.$$ This certainly is a peculiar differential equation. This identity at $\beta=1$ is fulfilled by any PDF, but by differentiating both sides with respect to $\beta$ we get $$ -\frac{1}{(\beta+1)^2}=\iint_T t\, f'(\beta t)\,f(y)\,dt\,dy $$ then by evaluating at $\beta=1$ and by applying integration by parts $$ -\frac{1}{4}=\int_{0}^{+\infty}f(y)\int_{0}^{y}t f'(t)\,dt\,dy=\int_{0}^{+\infty}f(y)\left[y f(y)-\int_{0}^{y}f(t)\,dt\right]\,dy$$ we get that the value of $\int_{0}^{\infty}y\,f(y)^2\,dy $ is fixed and it equals $\frac{1}{4}$. By reiterating the process (multiple differentiation with respect to $\beta$, evaluation at $\beta=1$, integration by parts) we get that all the moments of $f(y)^2$ are fixed. In particular, the smoothness assumption implies that $f(y)=\lambda e^{-\lambda y}$, i.e. the trivial solution, is the only solution.

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  • $\begingroup$ Incredible! A pleasure to read as always. $\endgroup$ – nbubis Nov 19 '17 at 6:48

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