Let X1, ..., Xn be a random sample from a population with...

Let X_1, ..., X_n be a random sample from a population with probability density function : f(x) = 1/? e^(-x/?) for x > 0. The minimum variance unbiased estimator (MVUE) of ?^2 a. (n+1)/n max(X_1,...,X_n) b. n/(n+1) X?^2 c. (n+1)/n min(X_1,...,X_n) d. n/(n+1) X?

Transcript

00:01 Hello students, from the question here they given that the probability density function f of x which is 1 by theta into e power minus x by theta for x greater than 0.

00:17 Then now the likelihood function for a random sample of x1, x2 up to xn is the product of the individual probabilities.

00:30 So the likelihood function of theta of x1, x2 up to xn which is equal to f of x1, f of x2 into f of xn which is equal to 1 by theta to the power of n into e power minus sum of del divided by 0 theta.

01:02 Now here we need to find the log likelihood function taking logarithmic on both sides that is the log of l of theta of x1 to x2 up to xn which is equal to here the log of 1 by theta to the power of n into e to the power of minus sum of del divided by theta which is equal to here n into log of 1 by theta minus sum of del divided by theta.

01:46 We know that the log of a of b which is equal to b into log a.

01:56 In this case it can be taking log on outside.

02:01 Now here differentiate with respect to theta.

02:08 So dou d divided by d theta of log of l theta of x1 to x2 up to xn which is equal to here d divided by d theta n into log of 1 by theta minus sum of l log by theta which is equal to here 0 because to differentiate apply the chain rule that is n into d divided by d theta into log of 1 by theta minus d divided by d theta into sum of epsilon by theta which is equal to 0.

02:56 Now differentiate log of 1 by theta n into minus 1 by theta plus sum of epsilon by theta square which is equal to 0.

03:12 Therefore here n by theta minus sum of epsilon by theta square is equal to 0.

03:22 Now by solving this we get theta which is equal to sum of epsilon by n.

03:29 Therefore now here the mle of theta which is mle is equal to sum of epsilon divided by n.

03:43 We need to find an unbiased estimator for theta square...

Question

Let X_1, ..., X_n be a random sample from a population with probability density function : f(x) = 1/? e^(-x/?) for x > 0. The minimum variance unbiased estimator (MVUE) of ?^2 a. (n+1)/n max(X_1,...,X_n) b. n/(n+1) X?^2 c. (n+1)/n min(X_1,...,X_n) d. n/(n+1) X?

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