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Im having a bit of trouble finding the solution to this problem:

"Independent random variables X1 and X2 both come from a population with mean $µ$ and variance $σ^2$. Three estimators for the parameter µ are as follows: $$\hatµ_1 = \frac{2}{3}X_1+\frac{1}{3}X_2$$ $$\hatµ_2 = \frac{1}{3}X_1+\frac{3}{4}X_2$$ $$\hatµ_3 = \frac{1}{2}X_1+\frac{1}{2}X_2$$

Calculate the expectation and the variance of each estimator."

I do believe that I have to use point estimation to work out the expectation and variance of each estimator but I have idea how to work them out.

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2 Answers 2

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Just use the linearity of the expectation operator, for example

$$E[\hat \mu_1] = \frac{2}{3} E[X_1] + \frac{1}{3} E[X_2] = \mu$$

For the variance, use the fact that $X_1$ and $X_2$ are independent, for example:

$$\text{Var}[\hat \mu_1] = \frac{4}{9} \text{Var}[X_1] + \frac{1}{9} \text{Var}[X_2] = \frac{5}{9}\sigma^2$$

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  • $\begingroup$ But doesn't $E(X_1)=X_1$ and similarly for $X_2$ since they just a constants, meaning I would just be back where I started again? $\endgroup$
    – Sean
    Mar 6, 2016 at 11:56
  • $\begingroup$ @Sean $X_1$ is a random variable, not a constant. $E[X_1] = E[X_2] = E[X_3]= \mu$ $\endgroup$ Mar 6, 2016 at 12:16
  • $\begingroup$ OH, of course, thanks. I was getting confused there. $\endgroup$
    – Sean
    Mar 6, 2016 at 12:24
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The expectation of first and third is $\mu $.for the second it is $7\mu/12$ And the variances are $5\sigma^2/9 \;,$ $25\sigma^2/144\;,$ and $\sigma^2/4\;$ for the first second and third respectively. Just use the linearity of Expectation and the independence of variables because of which $Cov(X_i,X_j)=0\;$ for $i \ne j $

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