Let $X_1 ,X_2 ,X_3 ,…$ be a sequence of i.i.d. uniform $(0,1)$ random variables. Then, calculate the value of $$\lim_{n\to \infty}P(-\ln(1-X_1)-\ln(1-X_2)-\cdots-\ln(1-X_n)\geq n)?$$

My work:

Since $X_i$ are uniform $(0,1)$ r.v.s so $E(X_1)=\frac{1}{2}$ and $\operatorname{Var}(X_1)=\frac{1}{12}$. Now I tried to used Chebyshev's inequality but didn't work. These logarithms are confusing me. How should I approach the problem? Any help would be great. Thanks.

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    $\begingroup$ Hint: What is the distribution of $-\log(1-X_1)$? $\endgroup$ – Math1000 Jul 24 '16 at 14:44
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    $\begingroup$ @Math1000 I think exponential distribution with mean $1$ $\endgroup$ – Harry Potter Jul 24 '16 at 14:48
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    $\begingroup$ Hint: Use the CLT on the random variables $Y_k=-\log(1-X_k)-1$, independent, square integrable and centered. (Hence the desired limit is $\frac12$.) $\endgroup$ – Did Jul 24 '16 at 16:17
  • $\begingroup$ @Did Thanks for the hint. This is definitely more elegant than the revised argument I had been working on... $\endgroup$ – Math1000 Jul 24 '16 at 17:19

Recall that if $U\sim\mathsf U(0,1)$ and the distribution function $F_Y$ of $Y$ is absolutely continuous, then the random variable $F_Y^{-1}(U)$ has the same distribution as $Y$. In particular, if $F_Y(y) = (1-e^{-y})\mathsf 1_{(0,\infty)}(y)$ then $$F(Y) = U \iff Y = -\log(1-U). $$ Set $Y_k = -\log(1-X_k)-1$, then $Y_k\stackrel d= Y-1$ where $Y\sim\operatorname{Exp}(1)$, so $\mathbb E[Y_k]=0$ and $\mathbb E[Y_k^2]=1$ (and the $Y_k$ are independent since the $X_k$ are). Let $\xi_n = \frac1n\sum_{k=1}^n Y_k$, then $$\mathbb E[\xi_n] = \mathbb E\left[\frac1n\sum_{k=1}^n Y_k \right]=\frac1n\sum_{k=1}^n \mathbb E[Y_k] = 0$$ and $$\mathbb E[\xi_n^2] = \mathbb E\left[\left(\frac1n\sum_{k=1}^n Y_k\right)^2 \right] = \frac1{n^2}\mathbb E\left[\sum_{k=1}^n Y_k^2 \right]=\frac1{n^2}.$$ The central limit theorem then implies that $Z_n$ converges in distribution to the standard normal distribution, where $$Z_n = \frac{\xi_n-\mathbb E[Y_1]}{\operatorname{Var}(Y_1)/\sqrt n} = \sqrt n \xi_n. $$ It follows that \begin{align} \mathbb P\left(\sum_{k=1}^n-\log(1-X_k)\geqslant n \right) &= \mathbb P\left(\sum_{k=1}^n Y_k\geqslant 0\right)\\ &=\mathbb P\left(\frac1{\sqrt n}\sum_{k=1}^n Y_k\geqslant 0 \right)\\ &= \mathbb P\left(\sqrt n\xi_n\geqslant 0\right)\\ &= \mathbb P(Z_n\geqslant 0)\\ &\stackrel{n\to\infty}\longrightarrow 1-\Phi(0)=\frac12, \end{align} where $$\Phi(t) = \int_{-\infty}^t \frac1{\sqrt{2\pi}} e^{-\frac12 x^2}\,\mathsf dx;\ t\in\mathbb R $$ is the CDF of the standard normal distribution.

  • $\begingroup$ but the answer given is $1/2$ $\endgroup$ – Harry Potter Jul 24 '16 at 15:15

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