Just a quick question:
I know a $U(0, A)$ with density of $1/A$ has as MLE of $X_{max}$, but would a $U(1,1+A)$ have the same MLE that of $X_{max}$?
I'm assuming so but just for clarity.
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2 Answers
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If $X \sim U(c, c + A)$. Then having observed $n$ independent observations, we can write the likelihood as,
$$L(A) = \dfrac{1}{A^n}\prod_{i=1}^{n}\mathbb{I}(c < X_i < c+A) = \dfrac{1}{A^n}\mathbb{I}(\min X_i \geq c) \mathbb{I}(\max X_i \leq c+A). $$
Since $1/A^n$ is a decreasing function of $A$, the MLE will be the smallest value possible such that $c+A \geq \max X_i$. Thus we want smallest value such that $A \geq (\max X_i - c)$.
Thus the MLE is $(\max X_i - c)$.
answered May 8, 2016 at 13:53
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It looks like you want to estimate $A$ which I'll assume is greater than zero. The estimate should be: $\hat{A} = X_{max} - 1$. For example suppose $A = 5$, so the upper bound on the uniform distribution is $6$. If your sample contained $X_{max}=5.97$, then your estimate would be: $\hat{A} = X_{max} - 1 = 4.97$.
