# Conditioning on a Discrete Random Variable

Could anyone help me out with this problem? I am preparing for an exam and this problem was given to us as practice, but I am not sure how to start.

Suppose we observe a Poisson process with arrival $\lambda$ per minute, but that we are not able to detect all the arrivals; rather each arrival has probability $p$ of being observed, independently of the others. Let $Y$ be the number of arrivals observed in the first minute. Find the mass function of $Y$ and identify the distribution name. [Hint: Condition on the total number of arrivals, observed and unobserved, in the first minute.]

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The title and tags mention conditioning, but I don't see any conditoning in the body. The distribution is simply that of a Poisson process with rate $\lambda p$ per minute. By the way, $\lambda$ is usually used to refer to the average rate of arrival, not the average number of arrivals in some time span. – joriki May 4 '13 at 17:14
Fixed it. Thanks. – Daniel May 4 '13 at 17:20
IMHO, the hint given in the problem and the two answers given to your question distract from a very straightforward intuition, which would be more valuable to develop than skills at manipulating conditional probabilities: If events arrive in Poisson fashion at an average rate of $\lambda$ per minute and you observe each event independently with probability $p$, then the observed events also occur in Poisson fashion, at an average rate of $\lambda p$ per minute. – joriki May 4 '13 at 18:28

Let $X$ be the mother Poisson. We have $\Pr(X=n)=e^{-\lambda}\frac{\lambda^n}{n!}$.
We want the probability that $Y=k$. In order for $Y$ have value $k$, we must have $n$ arrivals for some $n\ge k$, of which $k$ are observed.
For any fixed $n$, the number of particles observed has binomial distribution. In symbols, $$\Pr(Y=k|X=n)=\binom{n}{k}p^k(1-p)^{n-k}.$$ Thus $$\Pr(Y=k)=\sum_{n=k}^\infty \binom{n}{k}p^k (1-p)^{n-k}e^{-\lambda}\frac{\lambda^n}{n!}.\tag{1}$$ Now it is manipulation. Note that the expression $\frac{\binom{n}{k}}{n!}$ in $(1)$ simplifies to $\frac{1}{k!(n-k)!}$. Replace $n$ by $k+j$, and bring all the terms that do not involve $j$ outside the sum. Note that $\lambda^n=\lambda^k\lambda^j$. Then $(1)$ becomes $$\Pr(Y=k)=\frac{1}{k!}p^k e^{-\lambda}\lambda^k \sum_{j=0}^\infty \frac{(1-p)^j \lambda^j}{j!}.$$ We recognize the inner sum as $e^t$ where $t=(1-p)\lambda$, and minor algebra shows that $$\Pr(Y=k)=e^{-\lambda p}\frac{(\lambda p)^k}{k!}.$$
Remark: We have given an unattractive manipulational solution. Usually, when the relationship between Poisson and exponential is discussed, there is an "axiomatic" treatment of the Poisson. If one becomes comfortable with the nature of the Poisson, the fact that in our case $Y$ is Poisson becomes clear without computation. The value of the parameter is easier, since the description of $Y$ tells us immediately that $Y$ has expectation $\lambda p$.