Given a normally distributed random variable X ∼ N(µ, 1) with an unknown mean, µ
but known variance = 1. Consider an IID random sample (n = 2) from this distribution, {X1, X2}. Now, consider the following two estimators of µ based on this sample:
µ1 = (X1 + X2)/2
µ2 =(X1 + 2X2)/3
1. Calculate each estimators bias
2. Calculate each estimator's variance
3. Which estimator should you select and why?
Hi everyone! Although I think I've managed to solve question 1 (I believe both estimators are unbias, but can't seem to figure out how I might solve the variance for these estimators, I don't believe I can integration is possible here. I also believe that 3 is reliant on MSE, but can't solve anything until I've figured out 2! Any help would be greatly appreciated =)
Best Answer
Use the fact that the $X_i$ are independent and identically distributed; therefore, the variance of a linear combination of them is equal to the linear combination of the individual variances; i.e., $$\operatorname{Var}[c_1 X_1 + c_2 X_2] \overset{\text{ind}}{=} \operatorname{Var}[c_1 X_1] + \operatorname{Var}[c_2 X_2] = c_1^2 \operatorname{Var}[X_1] + c_2^2 \operatorname{Var}[X_2],$$ where $c_1, c_2$ are scalar constants. The "ind" over the equals sign means the equality holds under the assumption that $X_1$ and $X_2$ are independent random variables. If they were not independent, then there would be an additional covariance term required.
In your case, for the first estimator of $\mu$, your choice of constants is $c_1 = c_2 = 1/2$, and for the second estimator, $c_1 = 1/3$, $c_2 = 2/3$.
Alternatively, you may compute $$\operatorname{E}[(c_1 X_1 + c_2 X_2)^2],$$ the second moment of the linear combination of $X_1$ and $X_2$, by expanding the binomial and using linearity of expectation to obtain $$c_1^2 \operatorname{E}[X_1^2] + 2c_1 c_2 \operatorname{E}[X_1 X_2] + c_2^2 \operatorname{E}[X_2^2].$$ Then use the fact that $X_1$ and $X_2$ are independent to write $\operatorname{E}[X_1 X_2] = \operatorname{E}[X_1]\operatorname{E}[X_2]$, and use $\operatorname{E}[X_1^2] = \operatorname{Var}[X_1] + \operatorname{E}[X_1]^2$. Then finally use $$\operatorname{Var}[c_1 X_1 + c_2 X_2] = \operatorname{E}[(c_1 X_1 + c_2 X_2)^2] - \operatorname{E}[c_1 X_1 + c_2 X_2]^2.$$ This of course is much more tedious than directly using the first equation, because in essence this second approach is a proof of the validity of the first equation.