How did the author get that $\mathbb{E}(N) = 2$ and $\text{Var}(N) = 2$?

Question

I have the following problem:

In the country Maxvierk the number $N$ of children of a randomly selected family, and the number $G$ of girls among the children, are given by the following joint probability mass function:

Determine $\mathbb{E}[N|G = g]$ and $\text{Var}[N|G = g]$.

The solution begins as follows:

We know already that $\mathbb{E}(N) = 2$ and $\text{Var}(N) = 2$.

How did the author get that $\mathbb{E}(N) = 2$ and $\text{Var}(N) = 2$?

I would greatly appreciate it if someone could please take the time to clarify this.

Answer 1

For $\Bbb E(N)$, notice that the sums across each row are equal to $1/5$. Thus, $N=0,\cdots,4$ are all equally likely. Thus, the average value of $0,\cdots,4$ is the expectation, giving $2$.

More rigorously, consider the definition:

$$\Bbb E(N) = \sum_{N=0}^4 N \cdot \Bbb P(N)$$

Since $\Bbb P(N) = 1/5$ for all $N$ here, since the sums across the rows each are that, we simply end up with $1/5 \sum N$, which is again visibly $2$.

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For $\text{Var}(N)$, again, a definition:

$$\text{Var}(N) = \underbrace{\sum_{N=0}^4 N^2 \cdot \Bbb P(N)}_{\Bbb E(N^2)} - \underbrace{\left( \sum_{N=0}^4 N \cdot \Bbb P(N) \right)^2}_{\Bbb E(N)^2} = \frac 1 5 \sum_{N=0}^4 N^2 - 4$$

Taking the sum by your method of your choice gives $6$, so $\text{Var}(N) = 2$.

Answer 2

Hint: $$\mathbb{E}(N)=\mathbb{E}[\mathbb{E}(N\mid G=g)]=\sum_g \mathbb{E}(N\mid G=g)\mathbb{P}(G=g).$$

Also, $$Var(N)=Var(\mathbb{E}(N\mid G=g))+\mathbb{E}(Var(N\mid G=g)).$$