Branching Process Poisson (process) variance

Question

I got the following question:

Let $\{X_n, n \in \mathbb{N} \}$ be a branching process with offspring distribution $Z$. Now let $\{N(t); t \geq 0 \}$ be a Poisson process with intensity $\lambda$.

For fixt $s,t$ such that $0 < s < t$ assume $Z \sim N(s) +N(t)$.

Give an expression for $Var(X_n)$.

There is a solution given

\begin{align*} Var(Z) = Var(N(t)+N(s)) &= Var(N(t) +N(s) +N(s) -N(s)) \\ &= Var(N(t) -N(s)) + 4Var(N(s)) \\ &= \lambda(t-s)+4\lambda s = \lambda(t+3s) \end{align*}

Maybe I'm just tired but I cant wrap my head around this. I understand every stop above and why we can do it but not why it is done.

Why is $Var(N(t)+N(s)) = Var(N(t)) +Var(N(s)) = \lambda(t+s)$ wrong?

Answer 1

This is because the increments of $(N(t))_{t\geqslant 0}$ are independent, but $N(t)$ is not independent of $N(s)$. We need to make appear $N(t)-N(s)$ in order to use the independence of $N(t)-N(s)$ with $N(s)$.

Answer 2

Because $N(t)$ and $N(s)$ are correlated. If they were independent, or even if they were not independent but uncorrelated, then that equality would be right.