Return to Answer

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 $Z \sim N(0, n^{2}\sigma^{2})$. 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 $\theta > 0$).

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 $\theta > 0$).

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 $Z \sim N(0, n^{2}\sigma^{2})$. 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).

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 $Z \sim N(0, n^{2}\sigma^{2})$. 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 $\theta > 0$).

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

$$ E( \hat{\theta} ) = E\left( \frac{ \sum_{i=1}^{n} \theta X_i + \varepsilon_{i} }{ \sum_{i=1}^{n} X_i }\right) = \theta + E \left( \frac{ \sum_{i=1}^{n} \varepsilon_i }{ \sum_{i=1}^{n} X_i }\right), $$

therefore $ \hat{\theta} $ is unbiased. Since this is homework I will leave it to you to explain why that would make $\hat{\theta}$ unbiased and why the independence between the $\varepsilon$'s and $X$'s (and that fact that $E(\varepsilon_i)=0$) implies that $E \left( \frac{ \sum_{i=1}^{n} \varepsilon_i }{ \sum_{i=1}^{n} X_i }\right) = 0$

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 $Z \sim N(0, n^{2}\sigma^{2})$. 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).