Mathematics 376 – Probability and Statistics II Problem Set 6 Due: In class March 26, 2004

A) In this problem you will complete the derivation of the maximum likelihood estimators for the parameters μ and σ^{2} of a normal distribution that we started in class on Wednesday, March 19. Recall that we saw the likelihood function is

L = L(y_{1}, …, y_{n} | μ, σ^{2}) = (\frac{1}{\sqrt{2πσ^{2}}} )^{n} exp (− ∑_{i=1}^{n}(y_{i} − μ)^{2}/(2σ^{2}))

so

ln(L) = \frac{−n}{2} ln(2πσ^{2}) − ∑_{i=1}^{n}(y_{i} − μ)^{2}/(2σ^{2})

We saw that

\frac{∂ ln(L)}{∂μ} = ∑_{i=1}^{n}(y_{i} − μ)/σ^{2}

and

\frac{∂ ln(L)}{∂σ^{2}} = \frac{−n}{2σ^{2}} + \frac{∑_{i=1}(y_{i} − μ)^{2}/2}{(σ^{2})^{2}}

Setting \frac{∂ ln(L)}{∂μ} = 0, we got

μ̂_{ML} = \overline{Y}

(the sample mean).

1) Substituting μ̂_{ML} = \overline{Y} for μ, solve the equation \frac{∂ ln(L)}{∂σ^{2}} = 0 for σ^{2} to derive σ̂^{2}_{ML}. Is this estimator biased or unbiased?

2) Check that the values of μ̂_{ML} and σ̂^{2}_{ML} are actually give a maximum (not just a critical point) of ln(L) using the second derivative test for functions of two variables.

From the text

9.78, 9.79, 9.92 (in part c, the efficiency is the ratio of the variances of the estimators – see Definition 9.1), 10.4, 10.7, 10.9, 10.23.