Let \(x_{1} = 2.2,\ x_{2} = 4.3,\ x_{3} = 3.1,\ x_{4} = 4.5,\ x_{5} = 1.1\) and \(x_{6} = 5.7\) be the observed values of a random sample of size 6 from a \(U(\theta - 1, \theta + 4)\) distribution, where \(\theta \in (0, \infty)\) is unknown. Then a maximum likelihood estimate of \(\theta\) is equal to
GATE Mathematics - 2015,
Answer: A
A maximum likelihood estimate is 1.8.
The density of \(U(\theta - 1, \theta + 4)\) is \(\frac{1}{5}\) on an interval of fixed length 5, so the likelihood is \((1/5)^{6}\) whenever all observations lie in the interval, and 0 otherwise. Any \(\theta\) that keeps the sample inside is an MLE.
\[ \begin{aligned} \theta - 1 &\le x_{(1)} = 1.1 \Rightarrow \theta \le 2.1 \\ \theta + 4 &\ge x_{(6)} = 5.7 \Rightarrow \theta \ge 1.7 \end{aligned} \]
Every \(\theta\) in \([1.7, 2.1]\) maximises the likelihood, and 1.8 is the only option in that range. 2.3, 3.1 and 3.6 make \(\theta - 1\) exceed the smallest observation 1.1, giving likelihood 0.
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