Assume that the random vector $(u_t, v_t)$ is sampled iid over time and that $E[u_t v_t] = 0$. We also assume that $E[u_t] = E[v_t] = 0$ and that $E[u_t^2] = \sigma^2_u$, $E[v_t^2] = \sigma^2_v$.
Define the following two $MA(1)$ processes:
$$x_t = u_t + \theta_1u_{t-1}\\ \epsilon_t = v_t + \theta_2 v_{t-1} $$
Now, define a process $m_{t+1} \equiv x_t \cdot \epsilon_{t+1}$. I am required to find the autocovariance function of $m_{t}$.
Let $L$ denote the lag operator. My approach was the following:
$$m_{t+1} = (1+\theta_1L)(1+\theta_2L)\cdot u_{t}\cdot v_{t+1} $$
which implies that $m_{t+1} = (1 + (\theta_1 + \theta_2)L + (\theta_1\theta_2)L^2)\eta_{t+1}$, with $\eta_{t+1} = u_t v_{t+1}$.
Now, I proceed to show that $\eta_t$ as defined is a white noise process. In fact, independence and mean zero implies that $E[\eta_{t+1}] = E[u_t]E[v_{t+1}] = 0$ for all $t$. Moreover, independence implies that $E[\eta_t \eta_{t-k}] = 0$ for $k \geq 2$ because there would be no terms from the same date anymore.
Finally, $E[\eta_t \eta_{t-1}] = E[u_{t-1}v_tu_{t-2}v_{t-1}] = E[u_{t-2}v_t]E[u_{t-1}v_{t-1}]$ due to independence. However, since $u_t$ and $v_t$ are uncorrelated, we have $E[\eta_t \eta_{t-1}] = 0$.
This proves that $\eta_t$ is White Noise since it's a zero-mean process, uncorrelated over time, with finite second moment: $E[\eta_t^2] = E[u_{t-1}^2] \cdot E[v_t^2] = \sigma_u^2 \sigma_v^2$.
So we can recognize $m_{t}$ as an MA(2) process: $$m_{t} = \eta_t + (\theta_1 + \theta_2)\eta_{t-1} + (\theta_1 \theta_2)\eta_{t-2}$$
That would imply that the autocovariance function $\{\gamma_k\}_{k = 0}^{\infty}$ would be given by:
$$\gamma_0 = (1 + (\theta_1 + \theta_2)^2 + (\theta_1\theta_2)^2)\sigma_u^2\sigma_v^2 \\ \gamma_1 = ((\theta_1 + \theta_2) + \theta_1 \theta_2)\sigma_u^2\sigma_v^2\\ \gamma_2 = (\theta_1 \theta_2)\sigma_u^2\sigma_v^2 $$ and $\gamma_k = 0$ for $k \geq 3$
My question is whether the community agrees with my computations, mainly with the argument that $\eta_t$ is white noise. Thanks in advance.