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Let $X(k)$ be i.i.d random variable governed by uniform distribution $[-1,1]$ for $k=0,1,2,...N$. I would like to compute the following CDF $$ P\left( {\prod\limits_{k = 0}^{N - 1} {(1 + X(} k)) \le 1 + x} \right) $$

My goal is trying to make above CDF formula look like $ P \left( X(k) \le \text{Stuff} \right) $ but I failed. Here is my first try: taking logarithm on both side and I get $$P\left( {\sum\limits_{k = 0}^{N - 1} {\log (1 + X(k))} \le \log \left( {1 + x} \right)} \right)$$ Then I stuck to pursue further. By the way, I also get confused about the following two events: By i.i.d. of $X$, am I allowed to say $$\left\{ {\prod\limits_{k = 0}^{N - 1} {(1 + X(} k)) \le 1+x} \right\} = ? = \left\{ {{{(1 + X)}^N} \le 1 + x} \right\}$$

Any suggestion/hint is appreciated.

Fianra
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    These two links may be helpful http://math.stackexchange.com/questions/659254/product-distribution-of-two-uniform-distribution-what-about-3-or-more and http://math.stackexchange.com/questions/375967/probability-density-function-of-a-product-of-uniform-random-variables – robit Jan 30 '16 at 06:47
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    The last thought is incorrect, as each $X(k)$ is allowed to have different values. I guess the general form is quite tedious and maybe you can only use CLT to get an approximate answer. – BGM Jan 30 '16 at 08:19

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Note that $- \log( (1+X_k)/2)$ is an Exponential random variable with mean 1 (this follows from inverse transform sampling, and the fact that $(1+X_k)/2$ is U(0,1)).

Thus, $\sum_{k=1}^n - \log \left( \frac{1+X_k}{2} \right) = - \sum_{k=1}^n \log(1+X_k) + n \log 2 $ is Erlang(n,1) since $\{- \log( (1+X_k)/2) \}$ is a collection of i.i.d. Exp(1) r.v.'s.

From this, you can calculate the desired probability via the Erlang CDF.

Batman
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    Work out the distribution of $- \log(1+X_k)$. If $a=-1$, the result is the same, but with $2$ replaced with $b-a$. – Batman Jan 30 '16 at 22:43
  • Thanks, @Batman. I agree with you $a=-1$ same result hold. But if I replace 2 by $b-a$ with $a \neq -1$, then I get an extra constant $(1+a)/(b-a)$; i.e.,$$\begin{array}{l} P\left( { - \log \left( {\frac{{1 + X\left( k \right)}}{{b - a}}} \right) \le z} \right) = P\left( {X\left( k \right) \ge \left( {b - a} \right){e^{ - z}} - 1} \right)\ = 1 - P\left( {X\left( k \right) \le \left( {b - a} \right){e^{ - z}} - 1} \right)\ = 1 - \left( {\frac{{\left( {b - a} \right){e^{ - z}} - 1 - a}}{{b - a}}} \right)\ = 1 - {e^{ - z}} + \frac{{1 + a}}{{b - a}} \end{array}$$ – Fianra Jan 30 '16 at 23:00
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    You've done something wrong. Note that if $z \to \infty$, the RHS of that statement is $1+ \frac{1+a}{b-a} > 1$ so it can't be a probability. – Batman Jan 30 '16 at 23:13