Asymptotic approach

As above the likelihood is

$$l(\theta \vert {\mbox{\boldmath$x$}}) \propto \exp \left( \frac{-n}{2\sigma^2} (\mu - \bar x)^2 \right)$$ (34)

so the log-likelihood is

$$L(\theta \vert {\mbox{\boldmath$x$}}) = C - \frac{n}{2\sigma^2} (\mu - \bar x)^2$$ (35)

where $C$ is a constant. Differentiating and equating to zero gives

$$\frac{n}{\sigma^2} (\theta - \bar x) = 0$$ (36)

so that

$$\hat \theta = \bar x$$ (37)

and a second differentiation gives

$$\sigma_n^{-2}=\frac{n}{\sigma^2}$$ (38)

giving, in this case, $\mu = 3.508$ and $\sigma_n = 0.00316$.

A histogram of 1000 samples from $N(3.508, 0.00316^2)$ is shown in figure 3. While not a good density estimator, a histogram is useful for the gross comparison of two samples needed here. Note that this histogram appears to come from a distribution with a smaller standard deviation than that shown in figure 1; the predictive sample is a safer estimate as it allows for the uncertainty in the point density estimation.

Figure:Histogram of estimative density (Maximum likelihood approximation).