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 !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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 !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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