Question

Let X1, X2, · · · , Xn ∼ N(µ, σ2 ). Show that the MLE of µ is efficient.

Answer 1

MLE of
L = pi^n_i = 1 1/sigma squareroot 2 pi exp[-1/2(x_i- mu/sigma)^2] = 1/sigma^n exp[-1/2 sigma^2 sigma^n_i = 1 (x_i - mu)^2] times constant Taking logs: log L = -n log sigma -1/2 sigma62 sigma^n_i = 1 (x_i -mu)^2 + constant Differentiating with respect to mu and sigma gives: partial differential/ partial differential mu log L = 1/2 sigma ^2 sigma^n_i = 1 2(x_i- mu) = 1/sigma^2 (sigma^n_i = 1 x_i -n mu) partial differential/partial differential sigma log L = -n/ sigma-1/2 (-2/sigma^3) sigma^n_i = 1 (x_i- mu)62 = 1/sigma (1/sigma^2 sigma^n I = 1 (x_i- mu)^2-n) Setting these to zero gives: mu = 1/n sigma^n_i = 1 x_i = x
will be the sample mean = $\frac{\sum_{i=1}^{n}Xi}{n}$

You can find this by taking a partial dervative of the log- likelihood function with respect to
L = pi^n_i = 1 1/sigma squareroot 2 pi exp[-1/2(x_i- mu/sigma)^2] = 1/sigma^n exp[-1/2 sigma^2 sigma^n_i = 1 (x_i - mu)^2] times constant Taking logs: log L = -n log sigma -1/2 sigma62 sigma^n_i = 1 (x_i -mu)^2 + constant Differentiating with respect to mu and sigma gives: partial differential/ partial differential mu log L = 1/2 sigma ^2 sigma^n_i = 1 2(x_i- mu) = 1/sigma^2 (sigma^n_i = 1 x_i -n mu) partial differential/partial differential sigma log L = -n/ sigma-1/2 (-2/sigma^3) sigma^n_i = 1 (x_i- mu)62 = 1/sigma (1/sigma^2 sigma^n I = 1 (x_i- mu)^2-n) Setting these to zero gives: mu = 1/n sigma^n_i = 1 x_i = x
and the simply equating it to 0, as follows:



Now, an estimator is said to be efficient if the mean squared error becomes 0, as $n\rightarrow \infty$ , where n is the sample size:

Proof:

MSE(
L = pi^n_i = 1 1/sigma squareroot 2 pi exp[-1/2(x_i- mu/sigma)^2] = 1/sigma^n exp[-1/2 sigma^2 sigma^n_i = 1 (x_i - mu)^2] times constant Taking logs: log L = -n log sigma -1/2 sigma62 sigma^n_i = 1 (x_i -mu)^2 + constant Differentiating with respect to mu and sigma gives: partial differential/ partial differential mu log L = 1/2 sigma ^2 sigma^n_i = 1 2(x_i- mu) = 1/sigma^2 (sigma^n_i = 1 x_i -n mu) partial differential/partial differential sigma log L = -n/ sigma-1/2 (-2/sigma^3) sigma^n_i = 1 (x_i- mu)62 = 1/sigma (1/sigma^2 sigma^n I = 1 (x_i- mu)^2-n) Setting these to zero gives: mu = 1/n sigma^n_i = 1 x_i = x
) = + $(Bias(\sum Xi/n))^{2}$


Now, as this is an unbiased estimator, it means that Bias=0

MSE =




= ${\frac{1}{n^{2}}}{Var({X1+X2...Xn})}$

=   ${\frac{1}{n^{2}}}{(Var({X1)+Var(X2)...Var(Xn)})}$

= ${\frac{1}{n^{2}}}{n* Var(X)}$

= $\frac{1}{n}{ Var(X)}$

= $\frac{1}{n}{\sigma^{2}}$


Now as $n\rightarrow \infty$

$\frac{1}{n}{\sigma^{2}}\rightarrow 0$

Thus, MLE of $\Mu$ $\Mu$
L = pi^n_i = 1 1/sigma squareroot 2 pi exp[-1/2(x_i- mu/sigma)^2] = 1/sigma^n exp[-1/2 sigma^2 sigma^n_i = 1 (x_i - mu)^2] times constant Taking logs: log L = -n log sigma -1/2 sigma62 sigma^n_i = 1 (x_i -mu)^2 + constant Differentiating with respect to mu and sigma gives: partial differential/ partial differential mu log L = 1/2 sigma ^2 sigma^n_i = 1 2(x_i- mu) = 1/sigma^2 (sigma^n_i = 1 x_i -n mu) partial differential/partial differential sigma log L = -n/ sigma-1/2 (-2/sigma^3) sigma^n_i = 1 (x_i- mu)62 = 1/sigma (1/sigma^2 sigma^n I = 1 (x_i- mu)^2-n) Setting these to zero gives: mu = 1/n sigma^n_i = 1 x_i = x
is efficient

=