Expected value of a variable related to uniform r.v.

Question

Let $X_i$ be i.i.d. uniform random variable on $[0,1]$.

Let $N=\inf(n \colon \prod_{k=1}^n X_k < 10^{-3})$.

What is $\mathbb{E}(N)$?

I am trying to solve this but I am totally confused. How do I find an expected value for an infimum?

Answer 1

For every $a$ in $(0,1]$, let $N_a=\inf\{n\geqslant1\mid X_1X_2\cdots X_{n}\lt a\}$.

Assume that $X_1=x$ with $x$ in $(0,1)$. If $x\lt a$, then $N_a=1$. If $x\geqslant a$, then $N_a=1+N'$ where $N'=\inf\{n\geqslant1\mid X_2\cdots X_{n+1}\lt a/x\}$ is independent of $X_1$ and distributed like $N_{a/x}$.

Thus, $m(a)=\mathbb E(N_a)$ solves the equation $$ m(a)=1+\int_a^1m\left(\frac{a}x\right)\mathrm dx\ \stackrel{(s=a/x)}{=}\ 1+a\int_a^1m(s)\frac{\mathrm ds}{s^2}. $$ Differentiating this, one sees that $$ m'(a)=a^{-1}(m(a)-1)-m(a)a^{-2}=-a^{-1}. $$ Since $N_1=1$ almost surely, this yields, for every $a$ in $(0,1]$, $$ \mathbb E(N_a)=1-\log a. $$ Edit: Another approach is to consider $Y_n=-\log X_n$ and to note that $N_a$ is $1$ plus the number of points in the interval $(0,-\log a)$ for the process with increments $(Y_n)_{n\geqslant1}$. This is a homogeneous Poisson process with intensity $1$, thus the mean number of points in an interval $I$ is the Lebesgue measure of $I$. For $I=(0,-\log a)$, this is the result above.

Answer 2

Substitute $Y_i=-\ln (X_i)\ \forall i$. You will find the $Y$s are iid distributed exponentially with parameter 1. Then $N=\inf_n \{\sum_{i=1}^n Y_i>3 \ln(10)\}$. You may proceed by noting the sum of iid exponentials is an Erlang distribution.

$$\begin{align}E(N)&=\sum_{n=1}^\infty P(N\ge n)\\ &=1+\sum_{n=1}^\infty P\left(\sum_{i=1}^{n} Y_i \le 3 \ln(10)\right) \\ &=1+\sum_{n=1}^\infty \frac{\gamma(n,3\ln(10))}{(n-1)!}\\ &=1+3\ln(10) \end{align}$$

Alternatively you could view this as a different question:

• Having waited for a total time of $3\ln(10)$, what is the expected number of arrivals I could have seen in a Poisson process where the interarrival time has a mean 1?

The memoryless property of the exponential distribution will ensure this is 1+3$\ln(10)$.

Answer 3

This can be solved much along the lines of another problem involving first passage time with sum of uniforms.

Observe that the event $\{N=n\}$ can be thought of as follows: $$ \{N=n\} = \{ \prod_{k=1}^{n-1} X_k > \epsilon, \prod_{k=1}^{n} X_k < \epsilon \} $$ Thus $$ \Pr\left(N=n\right) = \mathbb{P}\left( \prod_{k=1}^{n-1} X_k > \epsilon, \prod_{k=1}^{n} X_k < \epsilon \right) = \Pr\left( \sum_{k=1}^{n-1} \log X_k > \log \epsilon, \sum_{k=1}^{n} \log X_k < \log\epsilon \right) $$ Since $-\log X$ equals in distribution to an exponential distribution with zero mean, and that $Z_{n-1} = -\sum_{k=1}^{n-1} \log X_k$ is equal in distribution to Erlang distribution $\mathcal{E}(n-1,1)$, and denoting $\delta = -\log\epsilon$ we get $$ \Pr(N=n) = \Pr\left(Z_{n-1} < \delta, Z_{n-1} + Y > \delta\right) = \frac{\delta^{n-1}}{(n-1)!}\mathrm{e}^{-\delta} $$ Therefore $$ \mathbb{E}(N) = \sum_{n=1}^\infty n \Pr(N=n) = 1 + \delta = 1-\log\epsilon $$