A random variable $X$ takes the value $\lambda_A$ with probability $\frac{1}{2}$ and $\lambda_B$ with probability $\frac{1}{2}$. Another random variable $Y$ is ~Poisson($\lambda_A$) if X = $\lambda_A$ and ~Poisson($\lambda_B$) if X = $\lambda_B$.
I need help calculating $Var(Y)$ and $Cov(X, Y)$. I first found $E(Y) = E(Y | X = \lambda_A)P(X = \lambda_A) + E(Y | X = \lambda_B)P(X = \lambda_B)$ = $\frac{1}{2} \lambda_A + \frac{1}{2} \lambda_B$. $E(X) = \frac{1}{2} \lambda_A + \frac{1}{2} \lambda_B$
$E(Y^2) = \frac{1}{2} (\lambda_A^2 + \lambda_A) + \frac{1}{2} (\lambda_B^2 + \lambda_B)$
$E(XY) = E(XY | X = \lambda_A)P(X = \lambda_A) + E(XY | X - \lambda_B)P(X = \lambda_B) = \frac{1}{2} \lambda_A^2 + \frac{1}{2} \lambda_B^2$
So then I can calculate $Var(Y) = E(Y^2) - E(Y)^2$ and $Cov(X, Y) = E(XY) - E(X)E(Y)$?
Am I on the right track for this problem?
Generally you are on the right track. Alternatively, you can use Eve's law to compute the variance quickly. Eve's law (or the variance decomposition formula) states:
$$\text{Var}(Y) = E[\text{Var}(Y|X)] + \text{Var}(E[Y|X]).$$
Now, because $Y \sim \text{Poi(X)},$ we have that
$$E[Y|X] = \text{Var(Y|X)} = X.$$
Thus,
$$\text{Var}(Y) = E(X)+\text{Var}(X).$$
Now, there is a similar covariance decomposition formula, but it is just as easy to compute it as you've stated.