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The normal derivation of the MLE for the uniform is as follows:

Suppose $X_{1},X_{2},\ldots, X_{n}$ are iid and $X\sim U(a,b)$ where $x\in[a,b]$.

$f_{X}(x)=\begin{cases} 1, \,\,\,a\leq x\leq b \\ 0, \,\text{ otherwise} \end{cases}$

The likelihood function is:

$L(a,b; x)=\Big(\frac{1}{b-a}\Big)^{n}\prod_{i=1}^{n}1_{\{a\leq x_{i}\leq b\}}$

Now we can't use calculus to solve for a maximum. However, we know that:

$\prod_{i=1}^{n}1_{\{a\leq x_{i}\leq b\}}=\begin{cases} 1, \,\,\,a\leq x_{i}\leq b, \forall i \\ 0, \,\text{ otherwise} \end{cases}$

Therefore, to maximize, we need $b\geq x_{(n)}$ and $a\leq x_{(1)}$ (otherwise the product goes to zero).

Similarly, we know $\Big(\frac{1}{b-a}\Big)^{n}$ is a decreasing function of $b$ and increasing function of $a$. Thus, we need the smallest value of $b$ that is at least as large as $x_{(n)}$ and the largest value of $a$ that is smaller than or equal to $x_{(1)}$.

Hence we arrive at:

$\hat{b}^{\text{MLE}}=x_{(n)}$

$\hat{a}^{\text{MLE}}=x_{(1)}$.

Now, let $X\sim U(c\theta,d\theta)$ where $c,d$ are constants with $c<d$. So the upper and lower bounds of the uniform are now dependent or related.

Considering the above derivation, this would imply our MLE for $\theta$ is:

Lower bound: $\hat{\theta}^{\text{MLE}}=x_{(1)}/c$

Upper bound: $\hat{\theta}^{\text{MLE}}=x_{(n)}/d$

So, what is $\hat{\theta}^{\text{MLE}}$? Is it an interval? Does the derivation have to jointly consider both bounds?

statsplease
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2 Answers2

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I think the problem assumes $c$ and $d$ are known constants, and $\theta$ is the only unknown.

The likelihood is $(d-c)^{-n} \theta^{-n} \prod_{i=1}^n 1_{\{\theta c \le x_i \le \theta d\}}$.

Note that each indicator can be rewritten as $1_{\{x_i/d\le \theta \le x_i/c\}}$, so we need to maximize the likelihood over $\theta \in [x_{(n)}/d, x_{(1)}/c]$. Since $\theta^{-n}$ is decreasing in $\theta$, the MLE is $x_{(n)}/d$. The result is intuitive: the smaller the $\theta$, the smaller the interval $[\theta c, \theta d]$, so the density is larger on this interval. The MLE chooses the smallest $\theta$ such that all the points $x_i$ lie in the interval.

angryavian
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  • But what if $x_{(1)}/c<x_{(n)}/d$? This is feasible is it not? Then the interval would swap and the answer change? So would the true MLE not be $\text{max}(x_{(n)}/d,x_{(1)}/c)$? – statsplease Nov 04 '16 at 01:49
  • Apologies I meant $\text{min}(x_{(n)}/d,x_{(1)}/c)$? – statsplease Nov 04 '16 at 02:07
  • I'm still a bit confused, if you were trying to select the smallest $\theta$, wouldn't the interval be $\theta \in [x_{(1)}/d,x_{(n)}/c]$? – statsplease Nov 04 '16 at 02:11
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    I mean your solution makes sense given the general case of a $U(0,d\theta)$ which we know has $\hat{\theta}^{\text{MLE}}=x_{(n)}/d$. In your solution if we let $c=0$, then this also has $\hat{\theta}^{\text{MLE}}=x_{(n)}/d$. – statsplease Nov 04 '16 at 02:28
  • @StatsPlease It is impossible for $x_{(1)}/c < x_{(n)}/d$ to happen because $\theta c \le x_{(1)}$ and $x_{(n)} \le \theta d$ for some $\theta>0$. – angryavian Nov 04 '16 at 05:33
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The MLE is not well defined in your parametrization of the distribution, because if $c$, $d$, and $\theta$ are unknown such that $c < d$ and $\theta > 0$, then the system $$c\theta = x_{(1)}, \quad d\theta = x_{(n)}$$ is underdetermined: you can choose any $\hat \theta > 0$, and there is a corresponding choice of $\hat c, \hat d$ that results in $$(\hat d - \hat c) \hat \theta = x_{(n)} - x_{(1)}.$$ Put another way, your parametrization does not permit you to distinguish whether your sample was generated, for example, from a $\operatorname{Uniform}(c = 1, d = 3, \theta = 2)$ distribution, or a $\operatorname{Uniform}(c = 2, d = 6, \theta = 1)$ distribution, because both are uniform on $[2,6]$.

heropup
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