Question

Problem 3. Let X1​,X2​,…,Xn​ be a random sample from distributions with the given probability density function. In each case, find the maximum likelihood estimator θ^. (1) f(x;θ)=(1/θ2)xe−x/θ,0

Answer 1

The maximum likelihood estimator (MLE) θ^ for the given probability density function f(x; θ) =
`(1/θ^2) * x * e^(-x/θ) is θ^` =
`(1/n) * ΣXi`, where Xi represents the observed values in the sample.

Explanation:

The maximum likelihood estimator (MLE) seeks the parameter value that maximizes the likelihood function, representing the probability of observing the given sample. For the probability density function
`\( f(x`;
`\theta)` =
`(1)/(\theta^2) \cdot` x
`\cdot e`^
`{-(x)/(\theta)} \)`, we aim to find the parameter θ^ that maximizes this function.

The likelihood function L(θ) is given by
`\( L(\theta)` =
`\prod_(i=1)^(n) f(x_i; \theta) \)`, where x1, x2, ..., xn represent the observed values in the sample. Taking the logarithm to simplify the calculations, we get
`\( \ln(L(\theta))` =
`-n \cdot \ln(\theta)` -
`(1)/(\theta) \cdot \sum_(i=1)^(n)`x
`_i \)`. Maximizing this with respect to θ, we differentiate
`\( \ln(L(\theta)) \)` with respect to θ and set it to zero to find the maximum.

`\[ (d)/(d\theta) \ln(L(\theta))` =
`-(n)/(\theta)` +
`(1)/(\theta^2) \cdot \sum`_
`{i=1}^(n)` x
`_i = 0 \]`

Solving for θ, we get
`\( \theta^2`=
`(1)/(n) \cdot \sum`_
`{i=1}^(n)` x
`_i \)`, leading to the MLE θ^ =
`\( \sqrt{(1)/(n) \cdot \sum_(i=1)^(n) x_i} \)`.

The MLE for θ is calculated as θ^ =
`\( \sqrt{(1)/(n) \cdot \sum_(i=1)^(n) x_i} \)`, which is derived from maximizing the logarithm of the likelihood function. This estimator represents the parameter value that maximizes the likelihood of observing the given sample, aligning with the principles of maximum likelihood estimation.