From SOA Sample #307:
The number of workplace accidents occurring in a factory on any given day is Poisson
distributed with mean $λ$ . The parameter $λ$ is a random variable that is determined by
the level of activity in the factory and is uniformly distributed on the interval $[0,3]$.
Calculate the probability of one accident on a given day.
Their solution starts:
$$P[X=k|λ]={{λ^ke^{-λ}}\over k!}$$
$$P[X=1]=\int_0^3 P[X=1|λ]f(λ)dλ=\int_0^3 P[X=k|1] {1 \over 3}dλ$$
I have two questions that may be the result of the same confusion:
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How does $P[X=k|λ]={{λ^ke^{-λ}}\over k!}$? What is the numerator of ${P[X=k] \cap P[λ=y]}\over P[λ=y]$, and what is the denominator to get the result they have?
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What is happening in the second step? I understand that they are trying to find the total probability of the uniform distribution of $λ$. I don't understand why they are multiplying by $f(λ)= {1\over 3}$. Does this mean the denominator of $P[X=k|λ]={{P[X=k] \cap P[λ=y]}\over P[λ=y]}$ is $f(λ) = {1\over 3}$? How does that show up in the first step?
Best Answer
The formula $P(A|B)=P(AB)/P(B)$ does not work when $P(B)=0$. Here, the conditional probability has to be understood as a limit. For example, $$ P(X=k|\lambda=2)=\lim_{\epsilon\to 0} P(X=k|2-\epsilon<\lambda<2+\epsilon)=\lim_{\epsilon\to0} \frac{P(X=k\text{ and }2-\epsilon<\lambda<2+\epsilon)}{P(2-\epsilon<\lambda<2+\epsilon)} $$ The reason for the $1/3$ is because the pdf of a uniform random variable between $a$ and $b$ is $1/(b-a)$.