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College Mathematics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT ISI MSTAT Statistics Theory of Estimation

ISI MStat PSB 2006 Problem 8 | Bernoullian Beauty

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 8. It is based on basic idea of Maximum Likelihood Estimators, but with a bit of thinking. Give it a thought !

Problem– ISI MStat PSB 2006 Problem 8


Let \((X_1,Y_1),……,(X_n,Y_n)\) be a random sample from the discrete distributions with joint probability

\(f_{X,Y}(x,y) = \begin{cases} \frac{\theta}{4} & (x,y)=(0,0) \ and \ (1,1) \\ \frac{2-\theta}{4} & (x,y)=(0,1) \ and \ (1,0) \end{cases}\)

with \(0 \le \theta \le 2\). Find the maximum likelihood estimator of \(\theta\).

Prerequisites


Maximum Likelihood Estimators

Indicator Random Variables

Bernoulli Trials

Solution :

This is a very beautiful Problem, not very difficult, but her beauty is hidden in her simplicity, lets explore !!

Observe, that the given pmf is as good as useless while taking us anywhere, so we should think out of the box, but before going out of the box, lets collect whats in the box !

So, from the given pmf we get, \(P( \ of\ getting\ pairs \ of\ form \ (1,1) \ or \ (0,0))=2\times \frac{\theta}{4}=\frac{\theta}{2}\),

Similarly, \(P( \ of\ getting\ pairs \ of\ form \ (0,1) \ or \ (1,0))=2\times \frac{2-\theta}{4}=\frac{2-\theta}{2}=1-P( \ of\ getting\ pairs \ of\ form \ (1,1) \ or \ (0,0))\)

So, clearly it is giving us a push towards involving Bernoulli trials, isn’t it !!

So, lets treat the pairs with match, .i.e. \(x=y\), be our success, and the other possibilities be failure, then our success probability is \(\frac{\theta}{2}\), where \(0\le \theta \le 2\). So, if \(S\) be the number of successful pairs in our given sample of size \(n\), then it is evident \(S \sim Binomial(n, \frac{\theta}{2})\).

So, now its simplified by all means, and we know the MLE of population proportion in binomial is the proportion of success in the sample,

Hence, \(\frac{\hat{\theta_{MLE}}}{2}= \frac{s}{n}\), where \(s\) is the number of those pairs in our sample where \(X_i=Y_i\).

So, \(\hat{\theta_{MLE}}=\frac{2(number\ of \ pairs \ in\ the\ sample\ of \ form\ (0,0)\ or \ (1,1))}{n}\).

Hence, we are done !!


Food For Thought

Say, \(X\) and \(Y\) are two independent exponential random variable with means \(\mu\) and \(\lambda\) respectively. But you observe two other variables, \(Z\) and \(W\), such that \(Z=min(X,Y)\) and \(W\) takes the value \(1\) when \(Z=X\) and \(0\) otherwise. Can you find the MLEs of the parameters ?

Give it a try !!


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ISI MStat PSB 2009 Problem 8 | How big is the Mean?

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

Problem– ISI MStat PSB 2009 Problem 8


Let \(X_1,…..,X_n\) be i.i.d. observation from the density,

\(f(x)=\frac{1}{\mu}exp(-\frac{x}{\mu}) , x>0\)

where \(\mu >0\) is an unknown parameter.

Consider the problem of testing the hypothesis \(H_o : \mu \le \mu_o\) against \(H_1 : \mu > \mu_o\).

(a) Show that the test with critical region \([\bar{X} \ge \mu_o {\chi_{2n,1-\alpha}}^2/2n]\), where \( {\chi^2}_{2n,1-\alpha} \) is the \((1-\alpha)\)th quantile of the \({\chi^2}_{2n}\) distribution, has size \(\alpha\).

(b) Give an expression of the power in terms of the c.d.f. of the \({\chi^2}_{2n}\) distribution.

Prerequisites


Likelihood Ratio Test

Exponential Distribution

Chi-squared Distribution

Solution :

This problem is quite regular and simple, from the given form of the hypotheses , it is almost clear that using Neyman-Pearson can land you in trouble. So, lets go for something more general , that is Likelihood Ratio Testing.

Hence, the Likelihood function of the \(\mu\) for the given sample is ,

\(L(\mu | \vec{X})=(\frac{1}{\mu})^n exp(-\frac{\sum_{i=1}^n X_i}{\mu}) , \mu>0\), also observe that sample mean \(\vec{X}\) is the MLE of \(\mu\).

So, the Likelihood Ratio statistic is,

\(\lambda(\vec{x})=\frac{\sup_{\mu \le \mu_o}L(\mu |\vec{x})}{\sup_\mu L(\mu |\vec{x})} \\ =\begin{cases} 1 & \mu_o \ge \bar{X} \\ \frac{L(\mu_o|\vec{x})}{L(\bar{X}|\vec{x})} & \mu_o < \bar{X} \end{cases} \)

So, our test function is ,

\(\phi(\vec{x})=\begin{cases} 1 & \lambda(\vec{x})<k \\ 0 & otherwise \end{cases}\).

We, reject \(H_o\) at size \(\alpha\), when \(\phi(\vec{x})=1\), for some \(k\), \(E_{H_o}(\phi) \le \alpha\),

Hence, \(\lambda(\vec{x}) < k \\ \Rightarrow L(\mu_o|\vec{x})<kL(\bar{X}|\vec{x}) \\ \ln k_1 -\frac{1}{\mu_o}\sum_{i=1}^n X_i < \ln k -n \ln \bar{X} -\frac{1}{n} \\ n \ln \bar{X}-\frac{n\bar{X}}{\mu_o} < K* \).

for some constant, \(K*\).

Let \(g(\bar{x})=n\ln \bar{x} -\frac{n\bar{x}}{\mu_o}\), and observe that \(g\) is,

Here, \(K*, \mu_o\) are fixed quantities.

decreasing function of \(\bar{x}\) for \(\bar{x} \ge \mu_o\),

Hence, there exists a \(c\) such that \(\bar{x} \ge c \),we have \(g(\bar) < K*\). See the figure.

So, the critical region of the test is of form \(\bar{X} \ge c\), for some \(c\) such that,

\(P_{H_o}(\bar{X} \ge c)=\alpha \), for some \(0 \le \alpha \le 1\), where \(\alpha\) is the size of the test.

Now, our task is to find \(c\), and for that observe, if \(X \sim Exponential(\theta)\), then \(\frac{2X}{\theta} \sim {\chi^2}_2\),

Hence, in this problem, since the \(X_i\)’s follows \(Exponential(\mu)\), hence, \(\frac{2n\bar{X}}{\mu} \sim {\chi^2}_{2n}\), we have,

\(P_{H_o}(\bar{X} \ge c)=\alpha \\ P_{H_o}(\frac{2n\bar{X}}{\mu_o} \ge \frac{2nc}{\mu_o})=\alpha \\ P_{H_o}({\chi^2}{2n} \ge \frac{2nc}{\mu_o})=\alpha \),

which gives \(c=\frac{\mu_o {\chi^2}_{2n;1-\alpha}}{2n}\),

Hence, the rejection region is indeed, \([\bar{X} \ge \frac{\mu_o {\chi^2}_{2n;1-\alpha}}{2n}\).

Hence Proved !

(b) Now, we know that the power of the test is,

\(\beta= E_{\mu}(\phi) \\ = P_{\mu}(\lambda(\bar{x})>k)=P(\bar{X} \ge \frac{\mu_o {\chi_{2n;1-\alpha}}^2}{2n}) \\ \beta = P_{\mu}({\chi^2}_{2n} \ge \frac{mu_o}{\mu}{\chi^2}_{2n;1-\alpha}) \).

Hence, the power of the test is of form of a cdf of chi-squared distribution.


Food For Thought

Can you use any other testing procedure to conduct this test ?

Think about it !!


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ISI MStat PSB 2009 Problem 6 | abNormal MLE of Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 6. It is based on the idea of Restricted Maximum Likelihood Estimators, and Mean Squared Errors. Give it a Try it !

Problem-ISI MStat PSB 2009 Problem 6


Suppose \(X_1,…..,X_n\) are i.i.d. \(N(\theta,1)\), \(\theta_o \le \theta \le \theta_1\), where \(\theta_o < \theta_1\) are two specified numbers. Find the MLE of \(\theta\) and show that it is better than the sample mean \(\bar{X}\) in the sense of having smaller mean squared error.

Prerequisites


Maximum Likelihood Estimators

Normal Distribution

Mean Squared Error

Solution :

This is a very interesting Problem ! We all know, that if the condition “\(\theta_o \le \theta \le \theta_1\), for some specified numbers \(\theta_o < \theta_1\)” had been not given, then the MLE would have been simply \(\bar{X}=\frac{1}{n}\sum_{k=1}^n X_k\), the sample mean of the given sample. But due to the restriction over \(\theta\) things get interestingly complicated.

So, simplify a bit, lets write the Likelihood Function of \(theta\) given this sample, \(\vec{X}=(X_1,….,X_n)’\),

\(L(\theta |\vec{X})={\frac{1}{\sqrt{2\pi}}}^nexp(-\frac{1}{2}\sum_{k=1}^n(X_k-\theta)^2)\), when \(\theta_o \le \theta \le \theta_1\)ow taking natural log both sides and differentiating, we find that ,

\(\frac{d\ln L(\theta|\vec{X})}{d\theta}= \sum_{k=1}^n (X_k-\theta) \).

Now, verify that if \(\bar{X} < \theta_o\), then \(L(\theta |\vec{X})\) is always a decreasing function of \(\theta\), [ where, \(\theta_o \le \theta \le \theta_1\)], Hence the maximum likelihood attains at \(\theta_o\) itself. Similarly, when, \(\theta_o \le \bar{X} \le \theta_1\), the maximum likelihood attains at \(\bar{X}\), lastly the likelihood function will be increasing, hence the maximum likelihood will be found at \(\theta_1\).

Hence, the Restricted Maximum Likelihood Estimator of \(\theta\), say

\(\hat{\theta_{RML}} = \begin{cases} \theta_o & \bar{X} < \theta_o \\ \bar{X} & \theta_o\le \bar{X} \le \theta_1 \\ \theta_1 & \bar{X} > \theta_1 \end{cases}\)

Now, to check that, \(\hat{\theta_{RML}}\) is a better estimator than \(\bar{X}\), in terms of Mean Squared Error (MSE).

Now, \(MSE_{\theta}(\bar{X})=E_{\theta}(\bar{X}-\theta)^2=\int^{-\infty}_\infty (\bar{X}-\theta)^2f_X(x)\,dx\)

\(=\int^{-\infty}_{\theta_o} (\bar{X}-\theta)^2f_X(x)\,dx+\int^{\theta_o}_{\theta_1} (\bar{X}-\theta)^2f_X(x)\,dx+\int^{\theta_1}_\infty (\bar{X}-\theta)^2f_X(x)\,dx\).

\(\ge \int^{-\infty}_{\theta_o} (\theta_o-\theta)^2f_X(x)\,dx+\int^{\theta_o}_{\theta_1} (\bar{X}-\theta)^2f_X(x)\,dx+\int^{\theta_1}_\infty (\theta_1-\theta)^2f_X(x)\,dx\)

\(=E_{\theta}(\hat{\theta_{RML}}-\theta)^2=MSE_{\theta}(\hat{\theta_{RML}})\).

Hence proved !!


Food For Thought

Now, can you find an unbiased estimator, for \(\theta^2\) ?? Okay!! now its quite easy right !! But is the estimator you are thinking about is the best unbiased estimator !! Calculate the variance and also compare weather the Variance is attaining Cramer-Rao Lowe Bound.

Give it a try !! You may need the help of Stein’s Identity.


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ISI MStat PSB 2018 Problem 9 | Regression Analysis

This is a very simple sample problem from ISI MStat PSB 2018 Problem 9. It is mainly based on estimation of ordinary least square estimates and Likelihood estimates of regression parameters. Try it!

Problem – ISI MStat PSB 2018 Problem 9


Suppose \((y_i,x_i)\) satisfies the regression model,

\( y_i= \alpha + \beta x_i + \epsilon_i \) for \(i=1,2,….,n.\)

where \({ x_i : 1 \le i \le n }\) are fixed constants and \({ \epsilon_i : 1 \le i \le n}\) are i.i.d. \(N(0, \sigma^2)\) errors, where \(\alpha, \beta \) and \(\sigma^2 (>0)\) are unknown parameters.

(a) Let \(\tilde{\alpha}\) denote the least squares estimate of \(\alpha\) obtained assuming \(\beta=5\). Find the mean squared error (MSE) of \(\tilde{\alpha}\) in terms of model parameters.

(b) Obtain the maximum likelihood estimator of this MSE.

Prerequisites


Normal Distribution

Ordinary Least Square Estimates

Maximum Likelihood Estimates

Solution :

These problem is simple enough,

for the given model, \( y_i= \alpha + \beta x_i + \epsilon_i \) for \( i=1,….,n\).

The scenario is even simpler here since, it is given that \(\beta=5\) , so our model reduces to,

\(y_i= \alpha + 5x_i + \epsilon_i \), where \( \epsilon_i \sim N(0, \sigma^2)\) and \(\epsilon_i \)’s are i.i.d.

now we know that the Ordinary Least Square (OLS) estimate of \(\alpha\) is

\( \tilde{\alpha} = \bar{y} – \tilde{\beta}\bar{x} \) (How ??) where \(\tilde{\beta}\) is the (generally) the OLS estimate of \(\beta\), but here \(\beta=5\) is known, so,

\(\tilde{\alpha}= \bar{y} – 5\bar{x} \) again,

\(E(\tilde{\alpha})=E( \bar{y}-5\bar{x})=alpha-(\beta-5)\bar{x}\), hence \( \tilde{\alpha} \) is a biased estimator for \(\alpha\) with \(Bias_{\alpha}(\tilde{\alpha})= (\beta-5)\bar{x}\).

So, the Mean Squared Error, MSE of \(\tilde{\alpha}\) is,

\(MSE_{\alpha}(\tilde{\alpha})= E(\tilde{\alpha} – \alpha)^2=Var(\tilde{\alpha}) \) + \({Bias^2}_{\alpha}(\tilde{\alpha}) \)

\(= frac{\sigma^2}{n}+ \bar{x}^2(\beta-5)^2 \)

[ as, it follows clearly from the model, \( y_i \sim N( \alpha +\beta x_i , \sigma^2)\) and \(x_i\)’s are non-stochastic ] .

(b) the last part follows directly from the, the note I provided at the end of part (a),

that is, \(y_i \sim N( \alpha + \beta x_i , \sigma^2 ) \) and we have to find the Maximum Likelihood Estimator of \(\sigma^2\) and \(\beta\) and then use the inavriant property of MLE. ( in the MSE obtained in (a)). In leave it as an Exercise !! Finish it Yourself !


Food For Thought

Suppose you don’t know the value of \(\beta\) even, What will be the MSE of \(\tilde{\alpha}\) in that case ?

Also, find the OLS estimate of \(\beta\) and you already have done it for \(\alpha\), so now find the MLEs of all \(\alpha\) and \(\beta\). Are the OLS estimates are identical to the MLEs you obtained ? Which assumption induces this coincidence ?? What do you think !!


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Restricted Maximum Likelihood Estimator |ISI MStat PSB 2012 Problem 9

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 9, It’s about restricted MLEs, how restricted MLEs are different from the unrestricted ones, if you miss delicacies you may miss the differences too . Try it! But be careful.

Problem– ISI MStat PSB 2012 Problem 9


Suppose \(X_1\) and \(X_2\) are i.i.d. Bernoulli random variables with parameter \(p\) where it us known that \(\frac{1}{3} \le p \le \frac{2}{3} \). Find the maximum likelihood estimator \(\hat{p}\) of \(p\) based on \(X_1\) and \(X_2\).

Prerequisites


Bernoulli trials

Restricted Maximum Likelihood Estimators

Real Analysis

Solution :

This problem seems quite simple and it is simple, if and only if one observes subtle details. Lets think about the unrestricted MLE of \(p\),

Let the unrestricted MLE of \(p\) (i.e. when \(0\le p \le 1\) )based on \(X_1\) and \(X_2\) be \(p_{MLE}\), and \( p_{MLE}=\frac{X_1+X_2}{2}\) (How ??)

Now lets see the contradictions which may occur if we don’t modify \(p_{MLE}\) to \(\hat{p}\) (as it is been asked).

See, that when if our sample comes such that \(X_1=X_2=0\) or \(X_1=X_2=1\), then \(p_{MLE}\) will be 0 and 1 respectively, where \(p\), the actual parameter neither takes the value 1 or 0 !! So, \(p_{MLE}\) needs serious improvement !

To, modify the \(p_{MLE}\), lets observe the log-likelihood function of Bernoulli based in two samples.

\( \log L(p|x_1,x_2)=(x_1+x_2)\log p +(2-x_1-x_2)\log (1-p) \)

Now, make two observations, when \(X_1=X_2=0\) (.i.e. \(p_{MLE}=0\)), then \(\log L(p|x_1,x_2)=2\log (1-p)\), see that \(\log L(p|x_1,x_2)\) decreases as p increase, hence under the given condition, log_likelihood will be maximum when p is least, .i.e. \(\hat{p}=\frac{1}{3}\).

Similarly, when \(p_{MLE}=1\) (i.e.when \( X_1=X_2=1\)), then for the log-likelihood function to be maximum, p has to be maximum, i.e. \(\hat{p}=\frac{2}{3}\).

So, to modify \(p_{MLE}\) to \(\hat{p}\), we have to develop a linear relationship between \(p_{MLE}\) and \(\hat{p}\). (Linear because, the relationship between \(p\) and \(p_{MLE}\) is linear. ). So, \(\hat{p}\) and \(p_{MLE}\) is on the line that is joining the points \((0,\frac{1}{3})\) ( when \(p_{MLE}= 0\) then \(\hat{p}=\frac{1}{3}\)) and \((1,\frac{2}{3})\). Hence the line is,

\(\frac{\hat{p}-\frac{1}{3}}{p_{MLE}-0}=\frac{\frac{2}{3}-\frac{1}{3}}{1-0}\)

\(\hat{p}=\frac{2-X_1-X_2}{6}\). is the required restricted MLE.

Hence the solution concludes.


Food For Thought

Can You find out the conditions for which the Maximum Likelihood Estimators are also unbiased estimators of the parameter. For which distributions do you think this conditions holds true. Are the also Minimum Variance Unbiased Estimators !!

Can you give some examples when the MLEs are not unbiased ?Even If they are not unbiased are the Sufficient ??


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ISI MStat PSB 2013 Problem 4 | Linear Regression

This is a sample problem from ISI MStat PSB 2013 Problem 4. It is based on the simple linear regression model, finding the estimates, and MSEs. But think over the “Food for Thought” any kind of discussion will be appreciated. Give it a try!

Problem– ISI MStat PSB 2013 Problem 4


Consider n independent observation \( { (x_i,y_i) :1 \le i \le n} \) from the model

\( Y= \alpha + \beta x + \epsilon \) ,

where \(\epsilon\) is normal with mean 0 and variance \( \sigma^2\) . Let \( \hat{\alpha}, \hat{\beta} \) and \( \hat{\sigma}^2 \) be the maximum likelihood estimators of \( \alpha , \beta \) and \( \sigma^2\) , respectively. Let \( v_{11}, v_{22} \) and \(v_{12}\) be the estimated values of \( Var(\hat{\alpha}), Var(\hat{\beta} \) and \( Cov ( \hat{\alpha}, \hat{\beta}) \), respectively.

(a) What is the estimated mean of Y, when when \( x=x_o\) ? Estimate the mean squared error of this estimator .

(b) What is the predicted value of Y, when when \( x=x_o\) ? Estimate the mean squared error of this predictor .

Prerequisites


Linear Regression

Method of Least Squares

Maximum likelihood Estimators.

Mean Squared Error.

Solution :

Here for the given model,

we have , the random errors, \( \epsilon \sim n(0, \sigma^2) \), and the maximum likelihood estimators (MLE), of the model parameters are given by \( \hat{\alpha}, \hat{\beta} \) and \( \hat{\sigma}^2 \). The interesting thing about this model is, since the random errors \(\epsilon\) are Gaussian Random variables, the Ordinary Least Square Estimates of the model parameters \( \alpha, \beta \) and \( \sigma^2\), are identical to their Maximum Likelihood Estimators, ( which are already given!). How ?? Verify it yourself and once and remember it henceforth.

So, here \( \hat{\alpha}, \hat{\beta} \) and \( \hat{\sigma}^2 \) there also the OLS estimates of the model parameters respectively.

And By Gauss-Markov Theorem, the OLS estimates of the model parameters are the BLUE (Best Linear Unbiased Estimator), for the model parameters. So, here \( \hat{\alpha}, \hat{\beta} \) and \( \hat{\sigma}^2 \) are also the unbiased estimators of \( \alpha, \beta \) and \(\sigma^2\) respectively.

(a) Now we need to find the estimated mean Y given \(x=x_o\) ,

\( \hat{ E( Y| x=x_o)}= \hat{\alpha} + \hat{\beta} x_o \) is the estimated mean of Y given \( x=x_o\).

Now since, the given MLEs ( OLSEs) are also unbiased for their respective parameters,

\( MSE( \hat{ E( Y| x=x_o)})=MSE(\hat{\alpha} + \hat{\beta} x_o)=E(\hat{\alpha} + \hat{\beta} x_o-(\alpha + \beta x_o))^2 \)

=\( E(\hat{\alpha} + \hat{\beta} x_o-E(\hat{\alpha} + \hat{\beta} x_o))^2 \)

=\( Var( \hat{\alpha} + \hat{\beta} x_o) \)

= \( Var(\hat{\alpha}+2x_o Cov(\hat{\alpha}, \hat{\beta})+ {x_o}^2Var(\hat{\beta}) \)

So, \( .MSE( \hat{ E( Y| x=x_o)})= v_{11} +2x_o v_{12} + {x_o}^2 {v_{22}} \).

(b) Similarly, when \(x=x_o \) , the predicted value of Y would be,

\( \hat{Y} = \hat{\alpha} + \hat{\beta} x_o +\epsilon \) is the predicted value of Y when \(x=x_o\) is given.

Using similar arguments, as in (a) Properties of independence between the model parameters , verify that,

\(MSE(\hat{Y})= v_{11}+ 2x_o v_{12} + {x_o}^2{ v_{22}}+{\hat{\sigma}^2} \). Hence we are done !


Food For Thought

Now, can you explain Why, the Maximum Likelihood Estimators and Ordinary Least Square Estimates are identical, when the model assumes Gaussian errors ??

Wait!! Not done yet. The main course is served below !!

In a game of dart, a thrower throws a dart randomly and uniformly in a unit circle. Let \(\theta\) be the angle between the line segment joining the dart and the center and the horizontal axis, now consider Z be a random variable. When the thrower is lefty , Z=-1 and when the thrower is right-handed , Z=1 . Assume that getting a Left-handed and Right-handed thrower is equally likely ( is it really equally likely, in real scenario ?? ). Can you construct a regression model, for regressing \(\theta\) on Z.

Think over it, if you want to discuss, we can do that too !!


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Likelihood & the Moment | ISI MStat 2016 PSB Problem 7

This problem is a beautiful example when the maximum likelihood estimator is same as the method of moment estimator. Infact, we have proposed a general problem, is when exactly, they are equal? This is from ISI MStat 2016 PSB Problem 7, Stay Tuned.

Problem

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed random variables ~ \(X\) with probability mass function
$$
f(x ; \theta)=\frac{x \theta^{x}}{h(\theta)} \quad \text { for } x=1,2,3, \dots
$$
where \(0<\theta<1\) is an unknown parameter and \(h(\theta)\) is a function of \(\theta\) Show that the maximum likelihood estimator of \(\theta\) is also a method of moments estimator.

Prerequisites

Solution

This \(h(\theta)\) looks really irritating.

Find the \( h(\theta) \).

\( \sum_{x = 1}^{\infty} f(x ; \theta) = \sum_{x = 1}^{\infty} \frac{x \theta^{x}}{h(\theta)} = 1 \)

\( \Rightarrow h(\theta) = \sum_{x = 1}^{\infty} {x \theta^{x}} \)

\( \Rightarrow (1 – \theta) \times h(\theta) = \sum_{x = 1}^{\infty} {\theta^{x}} = \frac{\theta}{1 – \theta} \Rightarrow h(\theta) = \frac{\theta}{(1 – \theta)^2}\).

Maximum Likelihood Estimator of \(\theta\)

\( L(\theta)=\prod_{i=1}^{n} f\left(x_{i} | \theta\right) \)

\( l(\theta) = log(L(\theta)) = \sum_{i=1}^{n} \log \left(f\left(x_{i} | \theta\right)\right) \)

Note: All irrelevant stuff except the thing associated with \( \theta \) is kept as constant (\(c\)).

\( \Rightarrow l(\theta) = c + n\bar{X}log(\theta) – nlog(h(\theta)) \)

\( l^{\prime}(\theta) = 0 \overset{Check!}{\Rightarrow} \hat{\theta}_{mle} = \frac{\bar{X} -1}{\bar{X} +1}\)

Method of Moments Estimator

We need to know the \( E(X)\).

\( E(X) = \sum_{x = 1}^{\infty} xf(x ; \theta) = \sum_{x = 1}^{\infty} \frac{x^2 \theta^{x}}{h(\theta)} \).

\( E(X)(1 – \theta) = \sum_{x = 1}^{\infty} \frac{(2x-1)\theta^{x}}{h(\theta)} \).

\( E(X)\theta(1 – \theta) = \sum_{x = 1}^{\infty} \frac{(2x-1)\theta^{x+1}}{h(\theta)} \)

\( E(X)((1 – \theta) – \theta(1 – \theta)) =\frac{\sum_{x = 1}^{\infty} 2\theta^{x} – \theta }{h(\theta)} = \frac{\theta(1 + \theta)}{(1 – \theta)h(\theta)}\).

\( \Rightarrow E(X) = \frac{\theta(1 + \theta)}{(1 – \theta)^3h(\theta)} = \frac{1+\theta}{1-\theta}.\)

\( E(X) = \bar{X} \Rightarrow \frac{1+\theta_{mom}}{1-\theta_{mom}}= \bar{X} \Rightarrow \hat{\theta}_{mom} = \frac{\bar{X} -1}{\bar{X} +1}\)

Food For Thought and Research Problem

Normal (unknown mean and variance), exponential, and Poisson all have sufficient statistics equal to their moments and have MLEs and MoM estimators the same (not strictly true for things like Poisson where there are multiple MoM estimators).

So, when do you think, the Method of Moments Estimator = Maximum Likelihood Estimator?

Pitman Kooper Lemma tells us that it is an exponential family.

Also, you can prove that that there exists a specific form of the exponential family.

Stay tuned for more exciting such stuff!