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# ISI MStat PSB 2008 Problem 10 | Hypothesis Testing

This is a really beautiful sample problem from ISI MStat PSB 2008 Problem 10. It is based on testing simple hypothesis. This problem teaches me how observation, makes life simple. Go for it!

## Problem- ISI MStat PSB 2008 Problem 10

Consider a population with three kinds of individuals labelled 1,2 and 3. Suppose the proportion of individuals of the three types are given by $f(k, \theta)$, k=1,2,3 where 0< $\theta$<1.

$f(k, \theta) = \begin{cases} {\theta}^2 & k=1 \\ 2\theta(1-\theta) & k=2 \\ (1-\theta)^2 & k=3 \end{cases}$

Let $X_1,X_2,....,X_n$ be a random sample from this population. Find the most powerful test for testing $H_o : \theta =\theta_o$ versus $H_1: \theta = \theta_1$. ($\theta_o< \theta_1< 1$).

### Prerequisites

Binomial Distribution.

Neyman-Pearson Lemma.

Test function and power function.

Hypothesis Testing.

## Solution :

This is a quite beautiful problem, only when you observe it closely. Here the distribution of X may seem non-trivial ( non-theoretical), but if one observes the distribution of Y=X-1 (say), instead of X , one will find that $Y \sim binomial( 2, 1-\theta)$ .

so, now let, p= 1-$\theta$ , so, 0<p<1, and let, $p_o= 1-\theta_o$ and $p_1=1-\theta_1$.

and since , $\theta_o< \theta_1 so, p_0>p_1$, and our hypotheses, reduces to,

$H_o : p = p_o$ versus $H_1: p = p_1, where 1> p_o> p_1$.

so, under $H_o$ , our joint pmf ( of Y=X-1), is $f_o( \vec{y}) = \prod_{i=1}^n {2 \choose y_i} {(p_o)^{y_i}(1-p_0)^{2-y_i}}$ ; where $y_i=x_i-1 , i=1,...,n$

and under $H_1$, our joint pmf is, $f_o( \vec{y}) = \prod_{i=1}^n{2 \choose y_i}{(p_1)^{y_i}(1-p_1)^{2-y_i}}$ ; where $y_i=x_i-1, i=1,...,n$

So, now we can use, widely used Neyman-Pearson Lemma , and end up with,

$\lambda (\vec{y})$=$\frac{f_1(\vec{y})}{f_o(\vec{y})}$=$\frac{\prod_{i=1}^{n} {2 \choose y_i} {p_1}^{y_i} {(1-p_0)}^{2-y_i}}{\prod_{i=1}^n {2 \choose y_i}{p_1}^{y_i}{(1-p_1)}^{2-y_i}}$=${(\frac{p_1}{p_0})}^{\sum{y_i}} {(\frac{1-p_1}{1-p_o})}^{2-\sum{y_i}}$ .

now we define a test function, $\phi(\vec{x})= \begin{cases} 1& \lambda*(\vec{x})> k \\ 0 &\lambda*(\vec{x}) \le k \end{cases}$. for some positive constant k.

Where $\lambda(\vec{y})=\lambda*(\vec{x}), \vec{x}= ( X_1,....,X_n)$

so, our test rule is, we reject $H_o$ if $\phi(\vec{x})=1$, and we choose k such that the for a give level $\alpha$,

$E_{H_o}(\phi(\vec{x})) \le \alpha$, for a given $0<\alpha<1$,

with a power function , $\beta(\theta)= E(\phi(\vec{x}))$. Can you find the more subtle condition when,$\lambda^*(\vec{x}) \le k$ ? Try It!

## Food For Thought

Suppose, $\theta_o \le \theta_1$, can you verify, that there for any constant c, $P_{\theta_1}(X>c) \le P_{\theta_1}(X>c)$ . Can you generalize the situation, what kind distribution must X follow ?? Think over it, until we meet again !

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This is a really beautiful sample problem from ISI MStat PSB 2008 Problem 10. It is based on testing simple hypothesis. This problem teaches me how observation, makes life simple. Go for it!

## Problem- ISI MStat PSB 2008 Problem 10

Consider a population with three kinds of individuals labelled 1,2 and 3. Suppose the proportion of individuals of the three types are given by $f(k, \theta)$, k=1,2,3 where 0< $\theta$<1.

$f(k, \theta) = \begin{cases} {\theta}^2 & k=1 \\ 2\theta(1-\theta) & k=2 \\ (1-\theta)^2 & k=3 \end{cases}$

Let $X_1,X_2,....,X_n$ be a random sample from this population. Find the most powerful test for testing $H_o : \theta =\theta_o$ versus $H_1: \theta = \theta_1$. ($\theta_o< \theta_1< 1$).

### Prerequisites

Binomial Distribution.

Neyman-Pearson Lemma.

Test function and power function.

Hypothesis Testing.

## Solution :

This is a quite beautiful problem, only when you observe it closely. Here the distribution of X may seem non-trivial ( non-theoretical), but if one observes the distribution of Y=X-1 (say), instead of X , one will find that $Y \sim binomial( 2, 1-\theta)$ .

so, now let, p= 1-$\theta$ , so, 0<p<1, and let, $p_o= 1-\theta_o$ and $p_1=1-\theta_1$.

and since , $\theta_o< \theta_1 so, p_0>p_1$, and our hypotheses, reduces to,

$H_o : p = p_o$ versus $H_1: p = p_1, where 1> p_o> p_1$.

so, under $H_o$ , our joint pmf ( of Y=X-1), is $f_o( \vec{y}) = \prod_{i=1}^n {2 \choose y_i} {(p_o)^{y_i}(1-p_0)^{2-y_i}}$ ; where $y_i=x_i-1 , i=1,...,n$

and under $H_1$, our joint pmf is, $f_o( \vec{y}) = \prod_{i=1}^n{2 \choose y_i}{(p_1)^{y_i}(1-p_1)^{2-y_i}}$ ; where $y_i=x_i-1, i=1,...,n$

So, now we can use, widely used Neyman-Pearson Lemma , and end up with,

$\lambda (\vec{y})$=$\frac{f_1(\vec{y})}{f_o(\vec{y})}$=$\frac{\prod_{i=1}^{n} {2 \choose y_i} {p_1}^{y_i} {(1-p_0)}^{2-y_i}}{\prod_{i=1}^n {2 \choose y_i}{p_1}^{y_i}{(1-p_1)}^{2-y_i}}$=${(\frac{p_1}{p_0})}^{\sum{y_i}} {(\frac{1-p_1}{1-p_o})}^{2-\sum{y_i}}$ .

now we define a test function, $\phi(\vec{x})= \begin{cases} 1& \lambda*(\vec{x})> k \\ 0 &\lambda*(\vec{x}) \le k \end{cases}$. for some positive constant k.

Where $\lambda(\vec{y})=\lambda*(\vec{x}), \vec{x}= ( X_1,....,X_n)$

so, our test rule is, we reject $H_o$ if $\phi(\vec{x})=1$, and we choose k such that the for a give level $\alpha$,

$E_{H_o}(\phi(\vec{x})) \le \alpha$, for a given $0<\alpha<1$,

with a power function , $\beta(\theta)= E(\phi(\vec{x}))$. Can you find the more subtle condition when,$\lambda^*(\vec{x}) \le k$ ? Try It!

## Food For Thought

Suppose, $\theta_o \le \theta_1$, can you verify, that there for any constant c, $P_{\theta_1}(X>c) \le P_{\theta_1}(X>c)$ . Can you generalize the situation, what kind distribution must X follow ?? Think over it, until we meet again !

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