1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ These two links may be helpful math.stackexchange.com/questions/659254/… and math.stackexchange.com/questions/375967/… $\endgroup$ Commented Jan 30, 2016 at 6:47
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jan 30, 2016 at 8:19

1 Answer 1

Reset to default
2
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Work out the distribution of $- \log(1+X_k)$. If $a=-1$, the result is the same, but with $2$ replaced with $b-a$. $\endgroup$ Commented Jan 30, 2016 at 22:43
  • $\begingroup$ 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.,$$\begin{array}{l} P\left( { - \log \left( {\frac{{1 + X\left( k \right)}}{{b - a}}} \right) \le z} \right) = P\left( {X\left( k \right) \ge \left( {b - a} \right){e^{ - z}} - 1} \right)\\ = 1 - P\left( {X\left( k \right) \le \left( {b - a} \right){e^{ - z}} - 1} \right)\\ = 1 - \left( {\frac{{\left( {b - a} \right){e^{ - z}} - 1 - a}}{{b - a}}} \right)\\ = 1 - {e^{ - z}} + \frac{{1 + a}}{{b - a}} \end{array}$$ $\endgroup$ Commented Jan 30, 2016 at 23:00
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jan 30, 2016 at 23:13

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.