Problem 7 Let X1, X2, …, Xn be a random sample from the...

Problem 7 Let $X_1, X_2, \dots, X_n$ be a random sample from the distribution $B(1, p)$, that is, $n$ Bernoulli trials. Let $p$ have a prior pdf that is uniform, $U(0, 1)$. Find the Bayes estimator of $p$ that minimizes the MSE risk. Hint: You will need to use the following integral formula: $\int_0^1 \theta^m (1 - \theta)^r d\theta = \frac{m! r!}{(m + r + 1)!}$

Transcript

00:01 Hello students for the given problem first we have to calculate the maximum likelihood estimator that is mle of each.

00:07 For that we will first find out the likelihood function that is l of theta which is given by the product of the individual probability for the each xi which is given as product of theta to the power minus 1 exponential of minus of xi upon theta where i taking value from 1 to n.

00:28 Now we will take the natural logarithm of the given likelihood function for the further convenience and then we will get logarithm of l of theta is equal to the product turns into the summation over i logarithm of theta to the power minus 1 into exponential of minus of xi theta where i taking value from 1.

00:50 Now we will differentiate the logarithm function with respect to theta and after differentiating it and equaling it to the 0 we will get the value of theta to be summation of xi upon n.

01:10 Now we can say that the mle of theta that is t of theta.

01:16 So mle of t of theta is 1 upon theta is given to be t hat is equal to n upon summation xi.

01:28 Next we need to find the asymptotic distribution of square root of n t hat minus t theta as n tends to infinity.

01:41 Next to find the asymptotic distribution we can use the tlt.

01:45 Since t hat is the mle it is asymptotically normally distributed as n tends to infinity and the mean and variance of square root of t hat minus t hat can be found out as mean to be expected value of under root of t hat minus t theta which is equal to 0 and variance is equal to variance of this function that is under root of n t hat minus t of theta which is comes out to be as n tends to the infinity.

02:27 We can say that our variance tends to the any finite value and the distribution become approximately normal.

02:40 Next we need to show that the t hat obtained in the part a is the bias estimator for t theta.

02:49 Therefore here we will derive an unbiased estimator t tilde using t hat.

02:54 To show t hat is biased we need to calculate the expected value of t hat which is equal to expectation of n upon summation of xi which is equal to n upon expectation of summation of xi.

03:11 Now since xi follows exponential distribution with parameter theta then its expected value becomes theta and therefore expected value of t hat is equal to n upon n theta which comes out to be 1 by theta which is not equal to t of theta...

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