Probability of a Poisson random variable taking a specific value

Question

Let $X$ be a random variable with Poisson distribution with mean $\lambda=9$.

Knowing that $P(X=8)=0.131756$, compute $P(X=9)$.

How do I approach this problem?

Answer 1

$$P(X=k\ |\ \lambda) = \frac{\lambda^k}{k!}e^{-\lambda}$$ We are interested when $\lambda = 9$, and $k = 9$, so just plug them in... $$P(X=9\ |\ \lambda = 9) = \frac{9^9}{9!}e^{-9} = \frac{9^8}{8!}e^{-9}$$ I believe the info they gave you about $P(X=8) = \dots$ is a red herring, but if you feel it must be used... notice that $$P(X=k\ |\ \lambda) = \frac{\lambda^k}{k!}e^{-\lambda} = \frac{\lambda}{k}\left(\frac{\lambda^{k-1}}{(k-1)!}e^{-\lambda}\right) = \frac{\lambda}{k}P(X=k-1\ |\ \lambda)$$ So $$P(X = 9\ |\ \lambda = 9) = P(X = 8\ |\ \lambda = 9)$$

Answer 2

The knowledge of $\mathbb P(X=8)$ is not actually needed. Since $X\sim\operatorname{Poisson}(9)$, we have

$$\mathbb P(X=9) = \frac{e^{-9}9^9}{9!} = \frac{4782969}{4480}e^{-9}\approx 0.131756. $$

However, as @Nitin pointed out, it just so happens that $$\frac{e^{-9}9^9}{9!} = \frac{e^{-9}9^8}{8!},$$ so $\mathbb P(X=9)=\mathbb P(X=8)$. So I would assume that is the intention of this exercise.

Answer 3

In general, if a Poisson random variable $X$ has an integer mean $\lambda \ge 1$, then $$P(X = \lambda) = P(X = \lambda - 1).$$ Because

$$P(X = \lambda) = e^{-\lambda}\frac{ \lambda^\lambda}{\lambda !} = e^{-\lambda} \frac{\lambda^{\lambda-1}}{(\lambda -1)!} = P(X = \lambda -1).$$

The problem is for the case $\lambda = 9.$ Also, these two adjacent values are the 'double' mode of the distribution. (That is, they are the two most likely values of $X$.)