Finding variance?

Question

$\newcommand{\Var}{\mathrm{Var}}$ Consider $Z_i$ as a binary random variable with $\mathrm{Pr}[Z_i = 1] = \pi$. Also, consider $Y_i$ as:

$Y_i|Z_i = 0 \sim \mathrm{Poisson} (\lambda_0) $

$Y_i|Z_i = 1 \sim \mathrm{Poisson} (\lambda_1) $

My question is how we can find $\Var(Y_i)$.

Here is what I think I should do, but I've not had any success till now:

$\Var(Y_i) = E(\Var(Y_i|Z_i)) + \Var(E(Y_i|Z_i)) = E(\pi\cdot\lambda_1 + (1 - \pi)\cdot\lambda_0) + \Var(\pi\cdot\lambda_1 + (1 - \pi)\cdot\lambda_0) = \pi\cdot\lambda_1 + (1 - \pi)\cdot\lambda_0 + 0 $

What I've got above is different from what my instructor has in his lecture notes. He has: $\Var(Y_i) = \pi\cdot\lambda_1 + (1 - \pi)\cdot\lambda_0 + (\lambda_1 - \lambda_0)^2\cdot\pi\cdot(1 - \pi)$

I appreciate your help.

Answer 1

\begin{align} \textrm{Var}\left[Y\right] &= \textrm{E}\left[\textrm{Var}\left[Y\mid Z\right]\right]+\textrm{Var}\left[\textrm{E}\left[Y\mid Z\right]\right]\\ &= \textrm{E}\left[\left(1-Z\right)\lambda_0+Z\lambda_1\right]+\textrm{Var}\left[\left(1-Z\right)\lambda_0+Z\lambda_1\right]\\ &= \left(1-\pi\right)\lambda_0+\pi\lambda_1+\textrm{Var}\left[\lambda_0+Z\left(\lambda_1-\lambda_0\right)\right]\\ &= \left(1-\pi\right)\lambda_0+\pi\lambda_1+\left(\lambda_1-\lambda_0\right)^2\pi\left(1-\pi\right) \end{align}

Just a side note: If you ever use this law of total variance and find that your conditional expectation and variance do not depend on the random variable on which you've conditioned, something probably went wrong.