Maximum-likelihood-estimator of $f_a: \mathbb{R} \to [0, \infty), f_a(x) = (a + 1) \cdot x^a \cdot 1_{[0, 1]}(x)$ with $a > 0$

Question

Given the continuous density function $f_a: \mathbb{R} \to [0, \infty), f_a(x) = (a + 1) \cdot x^a \cdot 1_{[0, 1]}(x)$ with $a > 0$, what is the maximum-likelihood-estimator for a?

I am not sure about my current solution and would be grateful if somebody could answer my questions. First of all, I approach the problem with a small $\epsilon$ such that the like-lihood function would look like following

$$L(a; x) = P_a([X_1 \in [x_1 - \epsilon, x_1 + \epsilon], ... , X_n \in [x_n - \epsilon, x_n + \epsilon])$$ $$\approx (2 \epsilon)^n f_a(x1) ... f_a(x_n)$$ $$=(2\epsilon)^n \cdot (a + 1) \cdot x_1^a \cdot 1_{[0, 1]}(x_1) ...(a+1) \cdot x_n^a \cdot 1_{[0, 1]}(x_n)$$

From there I obtain the log-likelihood-function if no $x_i = 0$ (I am unsure about this step)

$$l(a; x) = n \log(2\epsilon) + \log(a + 1) + \log(x_1^a) + ... + \log(a + 1) + \log(x_n^a)$$

If I set the derivative of the log-likelihood-function equal to zero, I get

$$\cfrac{d}{da} l(a; x) = \cfrac{1}{a + 1} + ... + \cfrac{1}{a + 1} = \cfrac{n}{a + 1} = 0$$

I have two questions regarding this function. Am I dealing with the indicator function correctly? And am I correct that there is no maximum for $a$?

Answer 1

There is no need to introduce $\epsilon$ because as you can see, it becomes a constant multiplicative factor of the likelihood function that is independent of the parameter $\alpha$ and the sample $(x_1, \ldots, x_n)$. Any extremum of $L$ will therefore not depend on the choice of $\epsilon$.

I would write $$\mathcal L(a;x) = (a+1)^n \left(\prod_{i=1}^n x_i\right)^{\!a} \mathbb 1_{[0,1]}(\min_i x_i)\mathbb 1_{[0,1]}(\max_i x_i),$$ thaat is to say, the minimum and maximum of the sample must both lie in the interval $[0,1]$.

Then the log-likelihood becomes $$\ell(a;x) = n \log (a+1) + a \log \prod_{i=1}^n x_i,$$ where I have dropped the log of the indicators because all that part tells us is that the sample must be valid (no observations outside the support).

Then the critical points of the log-likelihood satisfy $$0 = \frac{n}{a+1} + \sum_{i=1}^n \log x_i,$$ which shows that there is in fact a well-defined extremum.

Answer 2

This is not correct. For continuous RVs, the likelihood is the product of the densities (assuming your sample is iid)

$$ L=(a+1)^n \left(\prod x_i\right)^a $$ So the log likelihood is $$ log \, L= n\, log(a+1)+ a\sum log \, x_i $$ FOC $$ \frac{n}{a+1}=-\sum log\, x_i $$ Hence $$ \hat{a}=\frac{n}{\sum log \, x_i} -1 $$