Question

True or false.
X1, X2, ... Xn is a random sample from a uniform distribution with probability density function
f(x) = 1 / b-a, a < x < b , 0,otherwise
Then the maximum likelihood estimator of a is min (X1, X2, ... Xn).

Answer 1

The answer is TRUE

Solution:

Case 1: When the value of a is positive

For Uniform (a, b) the likelihood function is $L(x_1, ..., x_n|a, b) = (\frac{1}{b−a})^n$ for any sample. To maximize this we must minimize the value of (b − a) (interval length), yet we must keep all samples with in the range, i.e. $∀x_i , x_i ∈ (a, b)$.

An MLE for a and b would then be $\hat a = min(X_i)$ and $\hat b = max(X_i)$.

These values yield the minimal length since it’s the smallest interval to include all sampled points.

Case 2: When the value of first parameter is negative (let's denote it as "-a")

For Uniform (-a, b) the likelihood function is $L(x_1, ..., x_n|a, b) = (\frac{1}{b+a})^n$ for any sample. To maximize this we must minimize the value of (b + a), yet we must keep all samples with in the range, i.e. $∀x_i , −a ≤ x_i ≤ b$ . An MLE for A would be $-\hat a = min(X_i)$. This is the smallest value that promises that all sampled points are in the required range. Note here that the negative sign before a is just to denote that it's a negative value. So again the MLE for the first parameter is min(X_i).

Thus, the statement provided in the question holds true for both cases. Hence it is a TRUE statement.