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Let us model the number of vehicles crossing a bridge in a day as a Poisson random variable with parameter $\lambda$. Suppose that probability that a given vehicle is a truck is $p$ and is independent of other vehicles. I need to show that the number of trucks crossing the bridge is a Poisson random variable.

I know that the number of vehicles is a Poisson RV with a probability distribution of $\frac{e^{-\lambda}\lambda^k}{k!}$. Since mean $\lambda$ vehicles cross the bridge, then the mean number of trucks (expected) would be $\lambda p$ which would lead me to conclude that the Poisson RV distribution would be $\frac{e^{-\lambda p}(\lambda p)^k}{k!}$ but I am a bit lost as to how to derive this.

William
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1 Answers1

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Let $X$ be the number of vehicles, and let $Y$ be the number of trucks.

$$P(Y=k) = \sum_{n=k}^\infty P(X=n, Y=k) = \sum_{n=k}^\infty P(X=n) P(Y=k \mid X=n).$$

You know how to compute $P(X=n)$ because $X \sim \text{Pois}(\lambda)$.

What is $P(Y=k \mid X=n)$? (Hint: do you see a binomial distribution anywhere? You can think of each vehicle as a coin flip, with probability $p$ of being a truck.)

Solution sketch:

$$\sum_{n=k}^\infty e^{-\lambda} \frac{\lambda^n}{n!} \binom{n}{k} p^k(1-p)^{n-k} = \frac{e^{-\lambda} (\lambda p)^k}{k!} \underbrace{\sum_{n=k}^\infty \frac{[\lambda(1-p)]^{n-k}}{(n-k)!}}_{e^{\lambda(1-p)}}$$

angryavian
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