# Why is MLE reasonable?

I have always been convinced that maximum likelihood estimation is a very nice method to estimate the parameter of a probability distribution, since the first day I learn about it. However, when I was thinking about the intuition behind Fisher information, I ran into this What kind of information is Fisher information? and found myself a little bit confused. It says we can think Fisher information of a way to measure the curvature of the log likelihood, if we write it as \begin{aligned} I(\theta^{\star})&=-[\mathbb{E}_{X\sim\theta^{\star}}\frac{\partial^2}{\partial\theta^2}\log f(X;\theta)]\mid_{\theta=\theta^{\star}} \end{aligned} Adding some conditions, we may interchange integration and differentiation to get \begin{aligned} I(\theta^{\star})&=-[\frac{\partial^2}{\partial\theta^2}\mathbb{E}_{X\sim\theta^{\star}}\log f(X;\theta)]\mid_{\theta=\theta^{\star}} \end{aligned} Then we can interpret $$I(\theta^{\star})$$ as the curvature of the expected log likelihood around $$\theta^{\star}$$ when data actually follows $$\theta^{\star}$$.

So here comes my question: is the expected log likelihood $$\mathbb{E}_{X\sim\theta^{\star}}\log f(X;\theta)$$ (a function of $$\theta$$) maximized at $$\theta=\theta^{\star}$$? If not, I think it may weaken the "reasonability" of MLE. Is this just my needless worry? Or it's just that I get something wrong about Fisher information.

• – kjetil b halvorsen Sep 6 '20 at 13:18
• Yes, the expected log-likelihood is maximised at the true value $\theta^*$. This is why the Kullback-Leibler divergence is always non-negative. – Xi'an Sep 6 '20 at 15:05
• Furthermore, under regularity conditions, the expected score$$\mathbb{E}_{X\sim\theta^{\star}}\left[\frac{\partial}{\partial\theta}\log f(X;\theta)\right]$$is also equal to $0$ when $\theta=\theta^*$. – Xi'an Sep 6 '20 at 15:09
• @Xi'an, I suppose the answer I'm looking for is exactly what you say about Kullback-Leibler divergence! Besides, I appreciate all those insights provided by other comments and answers. – Chris Cloud Sep 7 '20 at 2:33

As an illustration, consider trying to estimate binomial $$p$$ with a known number $$n$$ of Bernoulli trials of which $$x$$ turn out to be successes. First, use small $$n = 10,$$ so that the MLE $$\hat p = x/n$$ may not be very accurate.

The likelihood function is the PDF considered as a function of $$p,$$ for observed data. Let's use R to plot the likelihood function for $$x =4.$$ likelihood function:

x = 4;  n = 10;  p=seq(.1, .9, by = 0.001)
like = dbinom(x, n, p)
plot(p, like, type="l", lwd=2)
abline(h=0, col="green2")

mle = mean(p[like == max(like)])  #'mean' in case of ties with discrete p
mle
 0.4
abline(v = mle, col="orange")


It seems clear that the likelihood function attains its maximum value at $$\hat p = 4/10 = 0.4.$$ Also, the curvature of the likelihood function at $$\hat p$$ is relatively gentle its maximum, so so we cannot expect the MLE to be extremely accurate. Of course, values other than $$x = 4$$ can occur and a Jeffries 95% interval estimate of $$p$$ is $$(0.153, 0.696).$$

qbeta(c(.025,.975), 4.5, 6.5)
 0.1530671 0.6963205


By contrast, if $$n = 100,$$ then the maximum if much more precisely determined.

x = 42;  n = 100;  p=seq(.1, .9, by = 0.001)
like = dbinom(x, n, p)
plot(p, like, type="l", lwd=2)
abline(h=0, col="green2")

mle = mean(p[like == max(like)])  #'mean' in case of ties with discrete p
mle
 0.42
abline(v = mle, col="orange") Here he Jeffries interval estimate is $$(0.317, 0.508).$$

qbeta(c(.025,.975), 41.5, 59.5)
 0.3172977 0.5078283


For the best estimation we need for the curvature of the likelihood curve to be as tight, on average, as possible at the value of the MLE.