Sampling distribution of $Y = \frac{\ln U_1}{\ln U_1 + \ln (1 - U_2)}$, where $U_i \sim U(0,1), \forall i$

Question

For this problem I have used the fact, $-2 \ln U \sim \chi^2_{(2)}$. But I have doubt on the independence of numerator and the denominator which are $\ln U_1$ and $\ln U_1 + \ln (1 - U_2)$. If they are independent, then resultant statistic boils down to $F_{2,4}$ statistic. Please help.

Answer 1

Perhaps you could try looking at the distribution of $1/Y$ instead. Then, $1/Y=-2\ln(1-U_{2})/(-2\ln(U_{1}))+1$. Then the numerator and denominator would be independent chi-square distributed random variables.

Answer 2

Set $X_1 \equiv - \log(U_1)$ and $X_2 \equiv - \log(1 - U_2)$. Since $X_1$ and $X_2$ are independent exponential$(\lambda = 1)$ random variables then for $t \in (0, 1)$ \begin{align} P(X_1 / (X_1 + X_2) \leq t) &= P(X_1 + X_2 \geq X_1 / t) \\ &= P(X_2 \geq X_1 (1 / t - 1)) \\ &= \text{E} \left [ P(X_2 \geq X_1 (1 / t - 1) \mid X_1) \right ] \\ &= \text{E} \left ( e^{-X_1 (1 / t - 1)} \right ) \\ &= \int_{0}^{\infty} e^{-s (1 / t - 1)} e^{-s} ds \\ &= \int_{0}^{\infty} e^{-s / t} ds \\ &= t \end{align} which is the uniform$(0, 1)$ distribution function.