Let $N$ a random variable such that $N \sim \operatorname{Pois}(\lambda)$. Furthermore, let $X$ a random variable such that $$\mathbb{P}( X = k\ | \ N=n)=\binom{n}{k}p^kq^{n-k},\ \ 0\leq k \leq n, \ \ p+q=1$$ Show that $X \sim \operatorname{Pois}(\lambda p)$

Honestly im hard stuck here. So i have to show that $\mathbb{P}(X=k)=\dfrac{e^{-p\lambda}(p\lambda)^k}{k!}$ and $\operatorname{Supp}(X)=\mathbb{N}$. Using the definition of conditional probability plus Bayes law i have $$\mathbb{P}( X = k\ | \ N=n)=\frac{\mathbb{P}( N = n\ | \ X=k)\mathbb{P}(X=k)}{\mathbb{P}(N=n)} \iff\binom{n}{k}p^kq^{n-k}\frac{e^{-\lambda}\lambda^n}{n!}=\mathbb{P}( N = n\ | \ X=k)\mathbb{P}(X=k) $$ So i need a expression for $\mathbb{P}( N = n\ | \ X=k)$ and i done. But i dont know how to get it.

  • 3
    $\begingroup$ This is not a good approach as you are switching the conditioning in a scenario where the given conditioning is what you need. So Baye's rule is not useful here. Use the law of total probability (LoTP) to compute $P[X=k]$ by conditioning on $N=n$ for all $n\in\{0,1,2,...\}$. Everything in the resulting LoTP expression is directly given to you and waiting for you to use. $\endgroup$
    – Michael
    Commented May 13 at 16:01
  • $\begingroup$ It does not help here since you are looking for a proof, but $\mathbb{P}( N = n \mid X=k) = \mathbb{P}( N -X= n-k)$ as $X$ and $N-X$ are independent. $N-X \sim \operatorname{Pois}(\lambda q)$, matching $X \sim \operatorname{Pois}(\lambda p)$. $\endgroup$
    – Henry
    Commented May 13 at 19:00

1 Answer 1


The law of total probability (LoTP) helps you compute the probability of an event by summing over simpler pieces, where the conditional probability of each piece is either given or is easy to compute.

Baye's rule is useful when it is conveninent to switch the given conditionings. However, in this problem, the given conditionings are exactly what is needed for application of LoTP. So Baye's rule is not useful in your problem.

Recall the LoTP: Let $\{B_n\}_{n \in I}$ be disjoint events with union equal to the whole sample space (where $I$ is a finite or countably infinite set that contains all the $n$ subscripts). The law of total probability states that for any event $A$ we have $$ P[A] = \sum_{n\in I} P[A|B_n]P[B_n]$$ where we remove any terms on the right-hand-side with $P[B_n]=0$.

Application: In your problem you fix $k$ as a nonnegative integer and define events $A$ and $B_n$ by $$ A = \{X=k\}$$ $$B_n=\{N=n\} \quad \forall n \in \{0, 1, 2, ...\}$$ So the set of subscripts in this case is $I=\{0,1,2,...\}$.

Other problems: In other problems, say, when you partition on a dice roll, you might use $I=\{1,2,3,4,5,6\}$. Or if you partition on the result of two coin flips you might use $I=\{HH,HT,TH,TT\}$. Or if you partition on the value of a random variable $G$ that has a geometric distribution then $I=\{1,2,3,...\}$. \begin{align*} P[A]&=\sum_{i=1}^6P[A|\mbox{Roll is $i$}]P[\mbox{Roll is $i$}]\\ P[A]&= P[A|HH]P[HH]+P[A|HT]P[HT]\\ &\quad +P[A|TH]P[TH]+P[A|TT]P[TT]\\ P[A]&=\sum_{i=1}^{\infty}P[A|G=i]P[G=i] \end{align*}


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