Question

Let X1,X2,............Xn be a random sample from a distribution with density function
f(x, alpha)={ 1/ alpha ; 0 smaller than equal to x smaller than equal to alpha
and 0 ; otherwise
Obtain the maximum likelihood estimator of θ

Answer 1

Answer on Question #65385 - Math - Statistics and Probability

Question: Let $x_{1}, x_{2}, \ldots, x_{n}$ be a random sample from a distribution with density function

$$f(x|\alpha) = \begin{cases} \frac{1}{\alpha}, & 0 \leq x \leq \alpha; \\ 0, & \text{otherwise} \end{cases}$$

Obtain the maximum likelihood estimator of $\alpha$.

Solution: Let $x_{(1)} \leq x_{(2)} \leq \cdots \leq x_{(n)}$ be the order statistics. The likelihood function is given by

$$L(\alpha; x_1, x_2, \ldots, x_n) = f(x_1, x_2, \ldots, x_n | \alpha) = \prod_{k=1}^{n} f(x_k | \alpha) = \prod_{k=1}^{n} \frac{1}{\alpha} = \alpha^{-n}$$

for $0 \leq x_{(1)}$ and $\alpha \geq x_{(n)}$, and 0 otherwise.

Then, the log-likelihood function is equal to

$\ln L(\alpha; x_1, x_2, \ldots, x_n) = -n \ln \alpha$ for $0 \leq x_{(1)}$ and $\alpha \geq x_{(n)}$, and 0 otherwise.

Now taking the derivative of the log-likelihood wrt $\alpha$ gives:

$$\frac{\partial \ln L(\alpha; x_1, x_2, \ldots, x_n)}{\partial \alpha} = -\frac{n}{\alpha} < 0.$$

So, we can say that $L(\alpha; x_1, x_2, \ldots, x_n)$ is a decreasing function for $\alpha \geq x_{(n)}$. Therefore,

$L(\alpha; x_1, x_2, \ldots, x_n)$ is maximized at $\alpha = x_{(n)}$. Hence, the maximum likelihood estimator for $\alpha$ is given by

$$\hat{\alpha} = x_{(n)} = \max\{x_1, x_2, \ldots, x_n\}.$$

Answer: $\hat{\alpha} = x_{(n)} = \max\{x_1, x_2, \ldots, x_n\}$.

For maximum likelihood estimation (MLE) see

https://en.wikipedia.org/wiki/Maximum_likelihood_estimation or

https://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-fall-2006/lecture-notes/lecture2.pdf or

https://www.projectrhea.org/rhea/index.php/Maximum_Likelihood_Estimation_Analysis_for_various_Probability_Distributions

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