Not understanding Maximum Likelihood last steps

Question

In finding the maximum likelihood estimate for the function $$f_X(x) = 2\beta x(1-x^2)^{\beta - 1}, x \in (0, 1)$$ where $\beta$ is unknown.

I'm asked to find the maximum likelihood estimation of $P(X > 0.5)$ based on the sample $(x_1, x_2, ..., x_6)$ from $X$ such that $\sum_{i=1}^6ln(1-x_i^2) = -3.698$

I correctly calculated from the product the conditions to find maximum likelihood, meaning $\frac{n}{\hat{\beta}} + \sum_{i=1}^nln(1-x_i^2) = 0$ and $-\frac{n}{\hat{\beta}} < 0$ which was expected.

Now I don't understand this part very well:

To find the estimate at $\beta$ I substituted $\hat{\beta} = -\frac{6}{-3.698} = 1.6225$. Why is this not the estimate at $\beta$? The solution from the textbook goes on without providing any indication on why is doing this (and I can't find info on this next step, so I don't really understand what is doing):

$$h(\beta) = P(X>0.5) = \int^{1}_{0.5} 2\beta x(1-x^2)^{\beta -1} = 0.75^\beta$$ $$\hat{h(\beta)} = h(\hat{\beta}) = 0.75^\hat{\beta} = 0.6270$$.

What do these last three steps mean? What exactly is the $h$ function that suddenly appears? What does $\beta = 1.6225$ mean in the context?

Answer 1

The key insight is that you are not interested in estimating the parameter $\beta$ itself, but rather, a function of that parameter, which the solution calls $h$. Getting an estimate of $\beta$ is merely a means to an end of estimating $h(\beta) = \Pr[X > 0.5]$.

For an analogy, suppose we have an IID sample from a normal distribution with unknown mean $\mu$ and variance $1$. The sample mean $\bar X$ is the MLE of $\mu$; i.e., $$\hat \mu = \bar X = \frac{1}{n} \sum_{i=1}^n X_i.$$ But what if we wanted the MLE of $\mu^2$, the square of the mean? By the invariance principle this would just be $\widehat{\mu^2} = (\bar X)^2$. In a similar fashion, we could also ask for the MLE of $\Pr[X > 1]$: this probability is also a function of the unknown parameter $\mu$, namely $$\Pr[X > 1] = \int_{x=1}^\infty \frac{1}{\sqrt{2\pi}} e^{-(x-\mu)^2/2} \, dx.$$ Again by the invariance principle, $$\widehat{\Pr}[X > 1] = \int_{x=1}^\infty \frac{1}{\sqrt{2\pi}} e^{-(x-\hat \mu)^2/2} \, dx.$$

So this gives us a convenient strategy to evaluate the desired estimator, by first calculating the MLE for $\beta$, then plugging it into the corresponding probability statement to get the desired MLE, which is what the last part of the solution is doing. In actuality, it first explicitly evaluates $h(\beta) = \Pr[X > 0.5] = (0.75)^\beta$ as a function of the parameter $\beta$, then the substitution takes place in the final line.

Answer 2

The result given in the book is based on a property that is called "functional equivariance". Or functional invariance.

See

So as you have correctly calculated $\hat{\beta}=1.6225$. Now consider the function $h(\beta)=P_{\beta}(X>0.5)$.

The invariance principle says that $\hat{h(\beta)}=h(\hat{\beta})$ and $P_{\hat{\beta}}(X>0.5)$.

But the last probability is

$\int_{0.5}^{1}2\hat{\beta}x(1-x^{2})^{\hat{\beta}-1}dx$=$0.75^{\hat{\beta}}$=$0.75^{1.6225}=0,6702795$