7. (MoM, MLE, FI, DM) Let X1, X2, …, Xn be a random sample...

7. (MoM, MLE, FI, DM) Let $X_1, X_2, \dots, X_n$ be a random sample of $Beta(1, \theta)$ random variables. The density function $f_X(x|\theta) = c(\theta)(1 - x)^{\theta - 1}$, $0 \le x \le 1$, $\theta > 0$. (a) Find the value of $c(\theta)$ so that $f_X(x|\theta)$ is a density function. (b) Find the maximum likelihood estimator for $\theta$. (c) Find the method of moments estimator of $\theta$. (d) Use the delta method to estimate the standard deviations of the two estimators. (e) Find the Fisher information. How do the standard deviation estimates in the previous part relate to the Fisher information?

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00:01 Now here we haven't given questions.

00:05 Now the given equation is f of y by theta is equal to theta y theta minus one for zero less than y less than one and theta greater than zero right so in the a part we have l of theta by y that is equal to pi for i is equal to 1 to n theta y i theta minus one right that is given in the question only that is equal to theta 30 power n pi for i is equal to 1 to n y i theta minus 1 right this is the equation in the second part for the method of movements we have we have e of y is equal to 0 to 1 theta y divide that is equal to theta y theta plus 1 by theta plus 1 that is equal to this is for 0 to 1 right and this is equal to theta by theta plus 1 right so here theta pi theta plus 1 is equal to y bar that implies that theta is equal to theta y bar plus y bar that is equal to theta minus theta y bar that is equal to y bar that is equal to y bar this implies that theta m om is equal to y bar divide by one minus y bar right now now the third part we have l is equal to log l theta by y that is equal to n log theta plus theta minus one summation of i is equal to one to n log y i right so 2 l by two l 3 is equal to 0 that implies that n by theta plus sigma is equal 1 by 1 to n log y i is equal to 0 right that implies that theta hat m i is equal to n by summation of i is equal to 1 to n long yi right now in that equation we have so now method of moment's estimator depends on maximum likelihood on maximum likelihood depends on this momm estimator depends on sigma is equal to 1 2n yi maximum likelihood estimator depends on sigma i is equal to 1 2 n log yi right now now, this will give a different estimator of theta, right? since this depends on different functions of yi, right? now, we have to plot the graph, right? i have plot the graph in my excel sheet.

04:20 I'll just upload here.

04:35 So for plotting childhood as a function of theta, we need to have a sample from this distribution...

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