Consider the number of children in a family (poisson-distributed with parameter $\lambda$).
Suppose each child is equally likely to be male or female.
Let $X$ be the number of male children and let $Y$ be the number of female children.
$a)$ Find the joint distribution of $X$ and $Y$.
$b)$Find two different ways to calculate $\mathbb{E}$$(X)$.
Idea:
$a)$ So I want to have $P(X=x,Y=y)$, right? I know, that the probability for $x+y$ children is $e^{-\lambda} \cdot \frac{\lambda^{x+y}}{(x+y)!}$, because the number of children in a family is poisson distributed, right? Hm, if this right it is intuitively clear that the probability for $x$ male children and $y$ female children is: $\binom{x+y}{x} \cdot (\frac{1}{2})^{x+y}$. If this is correct I have to find a way to combine this. If this isn't correct, please tell me how I can calculate the joint distribution. Second question: $ X $ and $Y$ are independent if $P$($X \cap Y$)=$P$($X$)$P$($Y$), but how I can find these terms?
b) Obviously one way has to be $a)$. So we have to use $a)$ somehow. My idea for the second way is to Wald's equation. I hope this is right and I hope that you can help me to calculate $\mathbb{E}(X)$.
You are on the right track for the first one. The correct way to combine them is multiplication. We have $$ P(X=n,Y=m) = e^{-\lambda}\frac{\lambda^{n+m}}{(n+m)!}\frac{1}{2^{n+m}}{n+m\choose m}.$$
As an alternative way to compute $E(X)$ is to note that by symmetry $E(X) = E(Y)$ so we have $$2E(X) = E(X+Y)=E(N) =\lambda.$$