I've already worked on a couple of random variables with their value in $\mathbb{N}$ $(X,Y)$

$p_{kn}=P(X=k,Y=n)= \left( \frac{λ^ke^{-y} α^n(1-α)^{k-n}}{n!(k-n)!} \right)\mathbb{1}_{\{0\le n\le k}\} (k,n)$

I had to find the law of X, of Y, trying to show that X and Y are independent...

Here is the point: the assumption is that the number of children for a random family follows a Poisson law with $E(Y)=2$. Assuming that for every birth, the probability to observe a boy's birth is equal to $1/2$. Show that, conditionally to the fact that the family has n children, the boys number follows a a Binomial law $B(n,1/2)$

I answered that conditionally to the fact that a family has $Y=n$ children, Y being the random variable representing the possible outcomes of the number of children inside a family, the boys number ,represented by X is a repetition of independent events identically distributed following a Bernouilli law with the probability $p=1/2$.

Thus $P(X=k|Y=n)~B(n,1/2)$ is the number of boys in a family conditonaly on the fact that it has n children.

But is that answer sufficient to prove P(X=k|Y=n)~B(n,1/2)?


Yes, that would be sufficient to argue that $X\mid Y=n \sim \mathcal {Bin}(n, \alpha)$.

However, the expression you provided is not quite right.   Where did the (lowercase) $y$ term come from?   Why are you raising $\lambda$ to the power of $k$?   You should have obtained:

$$\begin{align} p_{k,n;\alpha,\lambda} & = \mathsf P(X=k, Y=n; \alpha, \lambda) \\[1ex] & = \mathsf P(X=k \mid Y=n; \alpha)\cdot\mathsf P(Y=n; \lambda) \\[1ex] & = \binom{n}{k}\alpha^k(1-\alpha)^{n-k}\cdot\frac{\lambda^n \mathsf e^{-\lambda}}{n!}\;\mathbf 1_{0\leq k\leq n} \\[1ex] & = \frac{\alpha^k(1-\alpha)^{n-k}\lambda^n\mathsf e^{-\lambda}}{k!\,(n-k)!}\;\mathbf 1_{0\leq k\leq n} \end{align}$$

Where $\alpha = \tfrac 1 2$ is the success rate (of obtaining a boy each birth), and $\lambda=2$ is the average rate of births.

PS: You can now show that $X$ and $Y$ are not independent.


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