# Prove that $N$ has a Poisson distribution with parameter $\lambda$.

Let $\lambda > 0$, define the random variable $N$ as follows: where $U_1$, $U_2$, $\ldots$ is a sequence of iid $U(0,1)$ random variables.

$U_1 \geq e^{-\lambda}, U_1U_2 \geq e^{-\lambda}, U_1U_2 \ldots U_N \geq e^{-\lambda}$ but $U_1U_2 \ldots U_N U_{N+1} < e^{-\lambda}$.

Can we show that $N$ has a poission distribution with paramter $\lambda$? This is a homework problem. My thought is that we might possibly use $\log U_i/\lambda$, but I'm stuck so far...

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Your proposed method works. Let $U$ have uniform distribution on $(0,1)$, and let $$X=-\frac{\log U}{\lambda}.$$ Then $$P(X \le x)=P(-\log U \le \lambda x)=P(\log U \ge - \lambda x)=P(U \ge e^{- \lambda x})=1-e^{-\lambda x}.$$ Thus $X$ has cumulative distribution function $F_X(x)=1-e^{- \lambda x}$ (for $x>0$). It follows that $X$ has density function $\lambda e^{-\lambda x}$ (for $x>0$).

Thus $X$ has exponential distribution with parameter $\lambda$. Now use the following fact, that I assume has been proved in your course. Suppose that the waiting time between successive occurrences of an event has exponential distribution with parameter $\mu$. Then the number of occurrences of that event in the time interval $[0,T]$ has Poisson distribution with parameter $\mu T$. Take $\mu=\lambda$ and $T=1$.

To apply to your case, let $X_i=-\frac{\log U_i}{\lambda}$. Then $$U_1U_2\cdots U_k \ge e^{-\lambda} \quad\text{iff}\quad X_1+X_2 +\cdots +X_k \le 1.$$ So if $X_1$ is the waiting time until the first event, and $X_2$ is the waiting time between the first event and the second, and so on, then the random variable $N$ is the number of occurrences of the event in the time interval $[0,1]$. (Since the product of the $U_i$ up to $i=N+1$ is $<e^{-\lambda}$, the $(N+1)$-th event happens after time $1$.)

Remark: For no good reason, I would prefer to let $X=-\log U$. Then just as above, we conclude that $X$ has exponential distribution with parameter $1$. If waiting times are exponentially distributed with parameter $\mu$, then the number of occurrences of the event in a time interval of length $T$ has Poisson distribution with parameter $\mu T$. In our case, pick $\mu=1$ and $T=\lambda$.

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Awesome explanation! Thanks Andre! –  LBJ Mar 1 '12 at 0:19

One can proceed indirectly and, for some suitable $s$ and $x$, try to compute the double integral $$A(x,s)=\int_0^{+\infty}\mathrm e^{-x\lambda}\mathrm E(s^{N_\lambda})\mathrm d\lambda.$$ Since $[N=n]=[V_n\geqslant\mathrm e^{-\lambda}\geqslant V_{n+1}]$ where $V_0=1$ and $V_n=U_1U_2\cdots U_n$ for every $n\geqslant1$, $$\mathrm E(s^{N_\lambda})=\sum_{n=0}^{+\infty}s^n\mathrm P(V_n\geqslant\mathrm e^{-\lambda}\geqslant V_{n+1}).$$ Hence, $A(x,s)=\sum\limits_{n=0}^{+\infty}s^n\mathrm E(X_n(x))$ with $$X_n(x)=\int_0^{+\infty}\mathrm e^{-x\lambda}[V_n\geqslant\mathrm e^{-\lambda}\geqslant V_{n+1}]\mathrm d\lambda=\int_{-\log V_n}^{-\log V_{n+1}}\mathrm e^{-x\lambda}\mathrm d\lambda=\frac1x(V_n^x-V_{n+1}^x).$$ Since $X_n(x)=\frac1xV_n^x(1-U_{n+1}^x)=\frac1xU_1^x\cdots U_n^x(1-U_{n+1}^x)$, one gets $$A(x,s)=\sum\limits_{n=0}^{+\infty}s^n\frac1xu(x)^n(1-u(x))=\frac1x(1-u(x))\frac1{1-su(x)},$$ with $u(x)=\mathrm E(U_1^x)=\frac1{x+1}$. After some simplifications, this yields $$A(x,s)=\frac1{1+x-s}.$$ One recognizes $$\frac1{1+x-s}=\int_0^{+\infty}\mathrm e^{-x\lambda}\mathrm e^{-\lambda(1-s)}\mathrm d\lambda,$$ hence, for every $\lambda$, $$\mathrm E(s^{N_\lambda})=\mathrm e^{-\lambda(1-s)}.$$ One recognizes $$\mathrm e^{-\lambda(1-s)}=\sum_{n=0}^{+\infty}p_{\lambda}(n)s^n,\quad\text{with}\ p_\lambda(n)=\mathrm e^{-\lambda}\frac{\lambda^n}{n!},$$ hence $\mathrm P(N_\lambda=n)=p_\lambda(n)$ for every $n$ and every $\lambda$. In other words, for every $\lambda$, $N_\lambda$ is a Poisson random variable with parameter $\lambda$.

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Thanks for your answer! It's very detailed. :) –  LBJ Mar 1 '12 at 0:20

$V_n=U_1...U_n$ is in (0,1).

$P(V_n<v)=P(\sum_{i=1}^n - \log U_i > - \log v)=\int_{- \log v}^{\infty}\frac{e^{-x}}{\Gamma(n)}x^{n-1}dx$, this because $- \log U_i$ are exponential and the sum is a gamma. The density of $V_n$ is $\frac{(- \log v )^{n-1}}{\Gamma(n)}$.

$N=0$ with probability $e^{-\lambda}$; for $n \geq 1$, $N=n$ iff $V_n>e^{-\lambda},V_nU_{n+1}<e^{-\lambda}$ with probability:

$\int_{e^{-\lambda}}^1 \frac{(- \log v )^{n-1}}{\Gamma(n)} dv \int_0^{e^{- \lambda}/v}du= \frac{e^{- \lambda}}{\Gamma(n)} \int_{e^{- \lambda}}^1 \frac{(-\log v )^{n-1}}{v}dv=e^{-\lambda} \frac{\lambda^n}{\Gamma(n+1)}$.

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