Hi all,
I am preparing for an upcoming exam and I just don't seem to get this specific equation where I try to derive the variance of an estimator analytically.
I know the answer is var = sigma^2/(N*4) , however, I just cant see how that happens.
I may have made en error or missed something when taking the fractions out of the Variance.
In the image there are a few steps happening all at once and I suspect that there are two mistakes rolled into one step.
First, for any random variable $X$ it holds that $$ \text{Var}(X) = \text{Var}(X + c) $$ for any fixed $c \in \mathbb{R}$. This is shown by the following calculation \begin{align} \text{Var}(X+c) &= \text{Var}(Y) \\ &= \mathbb{E} \left [ ((X+c) - \mathbb{E}[X+c])^2 \right ] \\ &= \mathbb{E} \left [ ( X + c- c - \mathbb{E}[X])^2 \right ] \\ &= \mathbb{E} \left [ ( X - \mathbb{E}[X])^2 \right ] \\ &= \text{Var}(X) \end{align} Now letting $X = \frac{1}{2N} \sum_{i=1}^N x_i$ and $c = \frac{50}{N}$ this shows $$ \text{Var} \left (\frac{50}{N} + \frac{1}{2N} \sum_{i=1}^N x_i \right ) = \text{Var}(X + c) = \text{Var}(X) = \text{Var}\left ( \frac{1}{2N} \sum_{i=1}^N x_i \right ). $$ Second, For any $a \in \mathbb{R}$ and any random variable $X$ it holds $$\text{Var}(aX) = a^2\text{Var}(X)$$ and this can also be shown using the formula for variance, similarly to what was done above. Now let $a = \frac{1}{2N}$ and $X = \sum_{i=1}^N x_i$, then $$ \text{Var}\left ( \frac{1}{2N} \sum_{i=1}^N x_i \right ) = \text{Var}(aX) = a^2\text{Var}(X) = \frac{1}{4N^2}\text{Var}\left( \sum_{i=1}^N x_i \right). $$ I think that this may have been a small algebra mistake (happens to everyone!) since you squared the $N$ but not the 2.
After this use that the Variance of independent random variables is additive to see $$\frac{1}{4N^2}\text{Var}\left( \sum_{i=1}^N x_i \right) = \frac{1}{4N^2} \sum_{i=1}^N \text{Var}(x_i).$$
Putting all the pieces together and simplifying we arrive at $$ \text{Var} \left (\frac{50}{N} + \frac{1}{2N} \sum_{i=1}^N x_i \right ) = \frac{1}{4N^2} \sum_{i=1}^N \text{Var}(x_i) = \frac{1}{4N^2}(N \sigma^2) = \frac{\sigma^2}{4N} $$ where I'm using the fact that $\text{Var}(x_i) = \sigma^2$ and there are $N$ of these terms in the last summation. After that it's just more algebra.
To test your knowledge, I recommend proving for yourself that $\text{Var}(aX) = a^2\text{Var}(X)$ using only the definition of variance and simple properties of expectation.