# Variance of an estimator?

I've estimated a parameter $\theta$ of a linear model as

$$\hat\theta = \frac{2 \sum x_i^2 Y_i}{\sum x_i^4}$$

Where $Y_i$ is the response variable.

I was wondering how does one find the variance of this estimator given normally distributed error terms with mean 0?

I've gotten to$$V(\hat\theta) = V \left( \frac{2 \sum x_i^2 Y_i}{\sum x_i^4} \right) = 4 V \left( \frac{\sum x_i^2 Y_i}{\sum x_i^4} \right)$$ but im not sure how to proceed. The book gives the result as

$$\frac{4 \sigma^2}{\sum x_i^4}$$

Since $V(aX)=a^2V(X)$ and persumably $Y_i$ are iid, we can simplify $4 V \left( \frac{\sum x_i^2 Y_i}{\sum x_i^4} \right)$ as follows:

$$4 V \left( \frac{\sum x_i^2 Y_i}{\sum x_i^4} \right)=4 \left( \frac{\sum x_i^4 V(Y_i)}{(\sum x_i^4)^2} \right)=4 \left( \frac{\sum x_i^4 \sigma^2}{(\sum x_i^4)^2} \right)$$

Factor out $\sigma^2$ and simplify to get to the final result.

• for the second inequality, why are you allowed to go from $\sum x_i^2$ to $\sum x_i^4$? – ketchup Oct 11 '17 at 10:27
• Use the fact that $V(aX+bY)=a^2V(X)+b^2V(Y)$. The above equality holds if $X$ and $Y$ are independent as otherwise, we need to account for covariances as well. See: en.wikipedia.org/wiki/Variance#Weighted_sum_of_variables – Anon Oct 11 '17 at 13:11
• I see how thats the case for 4, $(\sum x_i^4)^2$, but for the term in the numerator, doesn't $(\sum x_i^2)^2 \neq \sum x_i^4$? – ketchup Oct 11 '17 at 13:29
• The numerator is just a sum of individual terms. To expand that a bit, the numerator is: $x_1^2 Y_1 + x_2^2 Y_2 +...$. Thus, $V(x_1^2 Y_1 + x_2^2 Y_2 +...)=V(x_1^2 Y_1) + V(x_2^2 Y_2) +...$. – Anon Oct 11 '17 at 13:30