Question

Let X1 and X2 be independent random variables with mean μand variance σ².

Suppose that we have 2 estimators of μ:

θ₁^ = (X1+X2)/2
θ₂^ = (X1+3X2)/4

a) Are both estimators unbiased estimators ofμ?
b) What is the variance of each estimator?

Answer 1

a)
`E(\hat \theta_1) =(1)/(2) [E(X_1) +E(X_2)]= (1)/(2) [\mu + \mu] = \mu`

So then we conclude that
`\hat \theta_1` is an unbiased estimator of
`\mu`

`E(\hat \theta_2) =(1)/(4) [E(X_1) +3E(X_2)]= (1)/(4) [\mu + 3\mu] = \mu`

So then we conclude that
`\hat \theta_2` is an unbiased estimator of
`\mu`

b)
`Var(\hat \theta_1) =(1)/(4) [\sigma^2 + \sigma^2 ] =(\sigma^2)/(2)`

`Var(\hat \theta_2) =(1)/(16) [\sigma^2 + 9\sigma^2 ] =(5\sigma^2)/(8)`

Explanation:

For this case we know that we have two random variables:

`X_1 , X_2` both with mean
`\mu = \mu` and variance
`\sigma^2`

And we define the following estimators:

`\hat \theta_1 = (X_1 + X_2)/(2)`

`\hat \theta_2 = (X_1 + 3X_2)/(4)`

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

`E(\hat \theta_i) = \mu , i = 1,2`

So let's find the expected values for each estimator:

`E(\hat \theta_1) = E((X_1 +X_2)/(2))`

Using properties of expected value we have this:

`E(\hat \theta_1) =(1)/(2) [E(X_1) +E(X_2)]= (1)/(2) [\mu + \mu] = \mu`

So then we conclude that
`\hat \theta_1` is an unbiased estimator of
`\mu`

For the second estimator we have:

`E(\hat \theta_2) = E((X_1 + 3X_2)/(4))`

Using properties of expected value we have this:

`E(\hat \theta_2) =(1)/(4) [E(X_1) +3E(X_2)]= (1)/(4) [\mu + 3\mu] = \mu`

So then we conclude that
`\hat \theta_2` is an unbiased estimator of
`\mu`

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

`Var(aX) = a^2 Var(X)`

For the first estimator we have:

`Var(\hat \theta_1) = Var((X_1 +X_2)/(2))`

`Var(\hat \theta_1) =(1)/(4) Var(X_1 +X_2)=(1)/(4) [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]`

Since both random variables are independent we know that
`Cov(X_1, X_2 ) = 0` so then we have:

`Var(\hat \theta_1) =(1)/(4) [\sigma^2 + \sigma^2 ] =(\sigma^2)/(2)`

For the second estimator we have:

`Var(\hat \theta_2) = Var((X_1 +3X_2)/(4))`

`Var(\hat \theta_2) =(1)/(16) Var(X_1 +3X_2)=(1)/(4) [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]`

Since both random variables are independent we know that
`Cov(X_1, X_2 ) = 0` so then we have:

`Var(\hat \theta_2) =(1)/(16) [\sigma^2 + 9\sigma^2 ] =(5\sigma^2)/(8)`