# The Poisson Distribution

Customers arrive at a casino as a Poisson process of rate 100 customers per hour. Upon arriving, each customer must flip a coin, and only those customers who flips heads actually enter the casino.

So how can I find

• the PMF of N? Let's say for the number of customers who arrive between 5PM to 7PM.
• the pdf of T? The time between successive customers arriving.
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The key observation is that the new process, denote it by $N = \lbrace N(t) : t \geq 0 \rbrace$, has rate $\lambda$ which is half the original, that is $\lambda=50$. Then the answers to your questions follow immediately. Denote the original process by $N_0$, and set $\lambda_0 = 100$. Then, the distribution of $N(1)$ (that is, the number of customers who actually enter the casino in one hour) is obtained (using the law of total probability) as follows. $${\rm P}(N(1) = k) = \sum\limits_{j = k}^\infty {{\rm P}(N(1) = k|N_0 (1) = j) {\rm P}(N_0 (1) = j)}.$$ Given $N_0 (1) = j$, $N(1)$ is binomial$(j,1/2)$. Hence, $${\rm P}(N (1) = k) = \sum\limits_{j = k}^\infty {\frac{{e^{ - \lambda _0 } \lambda _0^j }}{{j!}}{j \choose k}\frac{1}{{2^j }}} = \cdots = \frac{{e^{ - \lambda _0 /2} (\lambda _0 /2)^k }}{{k!}}.$$
EDIT: For a proof that the new process $N$ indeed has stationary independent increments, see, for example, p. 211 here (where $\lambda$ and $p$ correspond to $100$ and $1/2$, respectively).
Of course, for your first question, you need to consider $N(2)$ (but the answer can be adapted immediately to any $t>0$). –  Shai Covo Mar 7 '11 at 10:04
Note that the pdf of $T$ is an exponential distribution.