Let $N_t$ be a Poisson process with parameter $\lambda$ and let $T_i$ be the arrival of the $i^{th}$ event. Let $Y_i$ be i.i.d. with distribution $P(Y_i=1) = p$, $P(Y_i = 0) = 1-p$. Define the processes: $$Z_t := \sum_{i=1}^{\infty} Y_i 1_{T_i \leq t}$$ $$V_t := \sum_{i=1}^{\infty} (1-Y_i) 1_{T_i \leq t}$$
Compute $E[Z_tV_t]$.
My Attempt:
We could write $$Z_t = \sum_{i=1}^{X_t} Y_i $$ $$V_t = \sum_{i=1}^{W_t} (1-Y_i) $$ where $X_t = \sum_{i=1}^{\infty} 1_{\tau_i \leq t}$ and $W_t = \sum_{i=1}^{\infty} 1_{\tilde{\tau_i} \leq t}$.
I showed that $Z_t$ and $V_t$ are both Poisson process with parameters $\lambda p$, and $\lambda (1-p)$. This was determined simply by observing the MGF of these these processes.
Then, $E[X_tZ_t]= E[\sum_{i=1}^{X_t} Y_i \sum_{j=1}^{W_t} (1-Y_j)] = E[E[\sum_{i=1}^{X_t} Y_i \sum_{j=1}^{W_t} (1-Y_j)|X_t, W_t]]$.
Now the inner expectation is a function of $W_t$ and $X_t$.
$E[\sum_{i=1}^{X_t} Y_i \sum_{j=1}^{W_t} (1-Y_j)|X_t, W_t] = E[\sum_{i=1}^{X_t}\sum_{j=1}^{W_t}Y_i (1-Y_j)|X_t, W_t]$
but I am stuck evaluating it.