I calculate covariance incorrectly and would like some help. Assume $\{N(t),\,t\ge 0\}$ is a Poisson process with parameter $\lambda$. $\{N_i(t),\,t\ge 0\}$ for $i=1,\,2$ are Type 1 and Type 2 events of $N(t)$ with probability $p$ for Type 1 and $1-p$ for Type 2. Event type classification is independent.
Now, with $s>t$, compute $Cov(N_1(t),\,N(s))$.
I argue $Cov(N_1(t),\,N(s)) = E[N_1(t)\cdot N(s)] - E[N_1(t)]\cdot E[N(s)]$ by definition. The second term is easy since both $N_1(t)$ and $N(s)$ are Poisson processes; they will have expected value $p \lambda t$ and $\lambda s$ respectively.
For the first term I use iterated conditional expectation and the fact that events are uniformly distributed on $[0,\,s]$ if $N(s)$ is considered to be known (*).
$E[N_1(t)\cdot N(s)] = E[E[N_1(t)\cdot N(s)\,|\,N(s)]] \stackrel{(*)}{=} E[\frac{t}{s} p N(s)]=\frac{t}{s}p \lambda s= p\lambda t.$
Taken together, we have $Cov(N_1(t),\,N(s)) = p \lambda^2 t s - \lambda t=p\lambda t(\lambda s-1)$.
This does not agree with the provided solution that results in $p \lambda t$, that exploits that a Poisson process has independent increments. What am I doing wrong?
First note $N(t)=N_1(t)+N_2(t)$, also the process $N_1$ is independent of $N_2$. Then
\begin{align} Cov(N_1(t),\,N(s)) = Cov(N_1(t),\,N_1(t)+N_2(t))= Cov(N_1(t),\,N_1(t))= Var(N_1(t)) = p\lambda t. \end{align}