How to show unbiased estimator of combination of bernoulli and normal variables? $X_1, X_2, \ldots, X_n$ is a random sample from $\mathrm{Bernoulli}(\theta)$, $\epsilon_1, \epsilon_2, \ldots, \epsilon_n$ are independent $\mathcal N(0, \sigma^2)$, independent of $X_i$. 
Define $Y_i = \theta X_i + \epsilon_i$, for $i = 1, 2, \ldots,n$. Define estimating function 
$$
\psi[\theta;(X,Y)] = \sum_{i =1}^n(Y_i - \theta X_i) \>.
$$
a) Show that $\psi[\theta;(X,Y)]$ is unbiased estimator of $\theta$ if $\mathbb E(\psi[\theta;(X,Y)])=0$.
b) Find the estimator $\hat\theta$ such that $\psi[\hat\theta;(X,Y)] = 0$. Is $\hat\theta$ unbiased?
 A: I'm still not quite sure how part (a) is different from part (b) but, from your comment above it appears you are now only asking about part (b), so: 
If $\psi[\hat{\theta};(X,Y)] = 0$, then
$$
\sum_{i =1}^n(Y_i - \hat{\theta} X_i) \> = 0 
$$
So, 
$$ \sum_{i=1}^{n} Y_i = \hat{\theta} \cdot \sum_{i=1}^{n} X_i $$ 
Therefore $\hat{\theta} = \overline{Y}/\overline{X}$, the ratio of the sample means, satisfies  $\psi[\hat{\theta};(X,Y)] = 0$. Regarding unbiasedness, it is easy to see that
$$ E( \hat{\theta} ) = E\left( \frac{ \sum_{i=1}^{n} \theta X_i + \varepsilon_{i} }{ \sum_{i=1}^{n} X_i }\right) = \theta + E \left( \frac{ \sum_{i=1}^{n} \varepsilon_i }{
\sum_{i=1}^{n} X_i }\right), $$ 
Edit: Based on the discussion in the comments, I've edited my answer. Let $B=\sum_{i=1}^{n} X_{i}$ and $Z = \sum_{i=1}^{n} \varepsilon_i$. Then, $B \sim {\rm Binomial}(n,\theta)$ and $Z \sim N(0, n \sigma^{2})$. 
Since the errors are independent of the $X_{i}$, 
$$ 
E \left( \frac{ \sum_{i=1}^{n} \varepsilon_i }{
\sum_{i=1}^{n} X_i }\right)
= E(Z) \cdot E \left( \frac{1}{B} \right) $$ 
Clearly $E(Z) = 0$. Assuming $\theta < 1$, $P(B = 0) = (1-\theta)^{n} > 0$. Therefore $E \left( \frac{1}{B} \right) = \infty$, so 
$E \left( \frac{ \sum_{i=1}^{n} \varepsilon_i }{
\sum_{i=1}^{n} X_i }\right)$ doesn't exist. Therefore $E(\hat{\theta})$ doesn't exist whenever $\theta < 1$, so $\hat{\theta}$ can't be unbiased (although it is consistent as long as $\theta > 0$). 
