MLE of continuous uniform distribution

Question

A series of $n$ geomagnetic readings are taken from a meter, but the readings are judged to be approximate and unreliable. The chief scientist does know however that the true values are all positive and she suggests that an approximate model for the readings is that they are independent observations of a random variable which is uniformly distributed on $(0,\theta)$, where $\theta >1$.

Suppose that the chief scientist knows only that the number, $M$, of the readings which are less than $1$ is $m$, with the remaining $n-m$ being greater than $1$ and that she adopts the model as suggested above.

Demonstrate that the maximum likelihood estimate of $\theta$ is $n/m$.

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I don't understand why the binomial distribution is used to find the MLE. That is what the method is on the answers. It even says "she adopts the model as suggested above" so the uniform should be used shouldn't it?

Answer 1

My guess is that this is being done along the following lines:

Mainly, we are using the invariance property of maximum likelihood estimators.

Let $X_i$ be i.i.d $U(0,\theta)$ variables for $i=1,2,\cdots,n$, i.e. our original sample of size $n$.

Define a $Y_i=\begin{cases}1&,\text{ if }X_i<1\\0&,\text{ otherwise }\end{cases}$ for each $i=1,2,\cdots,n$.

Then the $Y_i$'s are i.i.d $\text{Ber}(p)$ for each $i$ with $p=P(X_1<1)=1/\theta\,,\quad\theta>1$

If $\bar y$ is the sample mean of the $Y_i$'s, then we can show that

\begin{align}\hat p_{MLE}&=\overline{y}\\\implies \frac{1}{\hat\theta_{MLE}}&=\overline y\\\implies\hat\theta_{MLE}&=\frac{1}{\overline y} \end{align}

As the observed value of $\overline y$ is $m/n$, we get $$\hat\theta_{MLE}=\frac{n}{m}$$

Answer 2

Given $\theta\gt1$, the probability for a measurement to be less than $1$ is $\frac1\theta$. Thus the likelihood for $m$ out of $n$ measurements to be less than $1$ is

$$ \mathsf P(M=m\mid\theta)=\binom nm\left(\frac1\theta\right)^m\left(1-\frac1\theta\right)^{n-m}\;. $$

Setting the derivative with respect to $p=\frac1\theta$ to zero yields

$$ \frac mp-\frac{n-m}{1-p}=0\;, $$

with solution $p=\frac mn$ and thus $\theta=\frac nm$.