Maximum Likelihood Estimation for θ, where f(x)=1 and θ<x<θ+1

Question

Q) Suppose that X1, · · · , Xn form a random sample from a uniform distribution on the interval (θ, θ + 1).where the value of the parameter θ is unknown (−∞ < θ < ∞). Clearly, the density function is f(x|θ) = (1) , for θ ≤ x ≤ θ + 1

My attempt: I never faced a similar question before in all the previous examples that I did I took the product of the pdf/pmf for all Xs then differentiated and equated to 0 but in this question such approach doesn't work.I am thinking maybe I have to manipulate the domain for x.

Thanks in advance for any help

Answer 1

Let's look at an example. I have chosen a value of $\theta$, which I will not reveal to you. Then I have generated $n = 6$ IID observations $\boldsymbol x = (x_1, \ldots, x_6)$ from a $\operatorname{Uniform}(\theta,\theta+1)$ distribution. The values I have observed are: $$\boldsymbol x = \{3.57347, 4.06865, 4.22543, 3.68704, 3.73285, 3.85223 \}.$$ Now, given that data, what can you say about $\theta$? You know that it cannot possibly be larger than the smallest observation in the sample, nor can it be smaller than the largest observation in the sample minus one: i.e., you are constrained to choose $$3.22543 = x_{(n)} - 1 \le \theta \le x_{(1)} = 3.57347.$$ Conversely, any value of $\theta$ in this range is admissible, and moreover, because the likelihood is constant for any $\theta$ in this interval, it follows that any choice of $\hat \theta$ in this interval is a MLE for $\theta$.

Specifically, the likelihood should be properly written as $$\mathcal L(\theta \mid \boldsymbol x) = \mathbb 1(\theta \le x_{(1)} \le x_{(n)} \le \theta+1).$$ Note this likelihood is a function of the sufficient statistic $$\boldsymbol (\boldsymbol x) = (x_{(1)}, x_{(n)}),$$ which comprises the minimum and maximum order statistics of the sample. Then the likelihood is maximized for any $\hat \theta$ for which the indicator function equals $1$, namely whenever $x_{(n)} - 1 \le \hat \theta \le x_{(1)}.$

Answer 2

Letting $\mathcal{L}(x;\theta)$ be the likelihood function and $x=(x_i)_{1\leq i \leq n}$ your sample, note that:

$$ \mathcal{L}(x;\theta) = \prod_{i=1}^n{1_{\{\theta \leq x_i \leq \theta +1\}}} $$

This function take only 2 values, $0$ or $1$. For it to be $1$, and hence maximized, one possibility is that all $(x_i)_i$ must be at most within $1$ of their minimal value. Hence, letting $\text{ML}(x) = \{\theta^{\text{ML}}: \theta^{\text{ML}} = \text{argmax}_{\theta} \, \mathcal{L}(x;\theta)\}$ be the set of all maximum likelihood solutions to the problem $-$ see @heropup 's answer:

$$ \hat{\theta}^{\text{ML}} = \min_ix_i \, \in \, \text{ML}(x)$$