Problem 8 Consider a random sample X1, X2, …, Xn from a...

Problem 8 Consider a random sample $X_1, X_2, \dots, X_n$ from a distribution with pdf $f_x(x; \theta) = f_{x|\theta}(x|\theta) = 3\theta x^2 e^{-\theta x^3}$, $0 < x < \infty$. Let $\theta$ have a prior pdf that is gamma with $\alpha = 4$ and $\theta = 1/4$. Find the conditional mean of $\theta$, given that $X_1 = x_1, X_2 = x_2, \dots, X_n = x_n$. Note that the conditional mean that you find gives a Bayesian estimate of $\theta$ that minimizes the MSE risk.

Transcript

00:02 Here the random variables x1, x2, etc., xn represents independent identically distributed random variables with probability mass function p of x, theta equal to theta plus 1 x power theta where 0 less than x less than 1 and theta greater than minus 1.

00:23 So, now we have to find e of x1 power r which is integral 0 to 1 x power r theta plus 1 x power theta dx which is equal to theta plus 1 integral 0 to 1 x power r plus theta dx which is equal to theta plus 1 1 by r plus theta plus 1 upon solving this integration.

00:47 So, now substitute r equal to minus 1 so we can have e of 1 by x1 equal to theta plus 1 divided by theta which is equal to 1 plus 1 by theta and e of 1 by x1 minus 1 equal to 1 by theta.

01:13 So, from this we can have the method of moment estimation equal to 1 by summation i is equal to 1 to n 1 by xi minus 1.

01:32 Then the answer for the second question is we have to find the likelihood of theta which is log of product i is equal to 1 to n f of xi comma theta which is equal to log of theta plus 1 the whole power n product i is equal to 1 to n xi power theta which is equal to n times log of theta plus 1 plus theta times summation i is equal to 1 to n log xi.

02:02 Therefore, by differentiating this that is d by d theta of n of theta equal to 0 which is equal to n by theta plus 1 plus summation i is equal to 1 to n log xi.

02:16 So, now we can have theta hat that is maximum likelihood estimator equal to minus n by summation i is equal to 1 to n log xi is the maximum likelihood estimator of theta.

02:29 So, now consider y1 equal to minus log of x1 as 0 less than x1 less than 1 and 0 less than y1 less than infinity.

02:42 So, now we can have the jacobian for this transformation is dx by dx1 power minus 1 equal to minus e power minus x1 and the probability density function of y1 is equal to minus log of x1.

02:59 Hence, we can have f y1 of y1 equal to f x1 of e power minus y1 multiplied by modulus j which is equal to theta plus 1 e power minus theta y1 multiplied by e power minus y1 where 0 less than y less than infinity.

03:22 Hence, from this we can have f y1 of y1 is equal to theta plus 1 e power minus of theta plus 1 y1 where 0 less than y less than infinity and 0 otherwise...

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