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# ISI MStat PSB 2013 Problem 7 | Bernoulli interferes Normally

This is a very simple and beautiful sample problem from ISI MStat PSB 2013 Problem 7. It is mainly based on simple hypothesis testing of normal variables where it is just modified with a bernoulli random variable. Try it!

## Problem- ISI MStat PSB 2013 Problem 7

Suppose $X_1$ and $X_2$ are two independent and identically distributed random variables with $N(\theta, 1)$. Further consider a Bernoulli random variable $V$ with $P(V=1)=\frac{1}{4}$ which is independent of $X_1$ and $X_2$ . Define $X_3$ as,

$X_3 = \begin{cases} X_1 & when & V=0 \\ X_2 & when & V=1 \end{cases}$

For testing $H_o: \theta= 0$ against $H_1=\theta=1$ consider the test:

Rejects $H_o$ if $\frac{(X_1+X_2+X_3)}{3} >c$.

Find $c$ such that the test has size $0.05$.

### Prerequisites

Normal Distribution

Simple Hypothesis Testing

Bernoulli Trials

## Solution :

These problem is simple enough, the only trick is that to observe that the test rule is based on 3 random variables, $X_1,X_2$ and $X_3$ but $X_3$ on extension is dependent on the the other bernoulli variable $V$.

So, here it is given that we reject $H_o$ at size $0.05$ if $\frac{(X_1+X_2+X_3)}{3}> c$ such that,

$P_{H_o}(\frac{X_1+X_2+X_3}{3}>c)=0.05$

So, Using law of Total Probability as, $X_3$ is conditioned on $V$,

$P_{H_o}(X_1+X_2+X_3>3c|V=0)P(V=0)+P_{H_o}(X_1+X_2+X_3>3c|V=1)P(V=1)=0.05$

$\Rightarrow P_{H_o}(2X_1+X_2>3c)\frac{3}{4}+P_{H_o}(X_1+2X_2>3c)\frac{1}{4}=0.05$ [ remember, $X_1$, and $X_2$ are independent of $V$].

Now, under $H_o$ , $2X_1+X_2 \sim N(0,5)$and $X_1+2X_2 \sim N(0,5)$ ,

So, the rest part is quite obvious and easy to figure it out which I leave it is an exercise itself !!

## Food For Thought

Lets end this discussion with some exponential,

Suppose, $X_1,X_2,....,X_n$ are a random sample from $exponential(\theta)$ and $Y_1,Y_2,.....,Y_m$ is another random sample from the population of $exponential(\mu)$. Now you are to test $H_o: \theta=\mu$ against $H_1: \theta \neq \mu$ .

Can you show that the test can be based on a statistic $T$ such that, $T= \frac{\sum X_i}{\sum X_i +\sum Y_i}$.

What distribution you think, T should follow under null hypothesis ? Think it over !!

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This is a very simple and beautiful sample problem from ISI MStat PSB 2013 Problem 7. It is mainly based on simple hypothesis testing of normal variables where it is just modified with a bernoulli random variable. Try it!

## Problem- ISI MStat PSB 2013 Problem 7

Suppose $X_1$ and $X_2$ are two independent and identically distributed random variables with $N(\theta, 1)$. Further consider a Bernoulli random variable $V$ with $P(V=1)=\frac{1}{4}$ which is independent of $X_1$ and $X_2$ . Define $X_3$ as,

$X_3 = \begin{cases} X_1 & when & V=0 \\ X_2 & when & V=1 \end{cases}$

For testing $H_o: \theta= 0$ against $H_1=\theta=1$ consider the test:

Rejects $H_o$ if $\frac{(X_1+X_2+X_3)}{3} >c$.

Find $c$ such that the test has size $0.05$.

### Prerequisites

Normal Distribution

Simple Hypothesis Testing

Bernoulli Trials

## Solution :

These problem is simple enough, the only trick is that to observe that the test rule is based on 3 random variables, $X_1,X_2$ and $X_3$ but $X_3$ on extension is dependent on the the other bernoulli variable $V$.

So, here it is given that we reject $H_o$ at size $0.05$ if $\frac{(X_1+X_2+X_3)}{3}> c$ such that,

$P_{H_o}(\frac{X_1+X_2+X_3}{3}>c)=0.05$

So, Using law of Total Probability as, $X_3$ is conditioned on $V$,

$P_{H_o}(X_1+X_2+X_3>3c|V=0)P(V=0)+P_{H_o}(X_1+X_2+X_3>3c|V=1)P(V=1)=0.05$

$\Rightarrow P_{H_o}(2X_1+X_2>3c)\frac{3}{4}+P_{H_o}(X_1+2X_2>3c)\frac{1}{4}=0.05$ [ remember, $X_1$, and $X_2$ are independent of $V$].

Now, under $H_o$ , $2X_1+X_2 \sim N(0,5)$and $X_1+2X_2 \sim N(0,5)$ ,

So, the rest part is quite obvious and easy to figure it out which I leave it is an exercise itself !!

## Food For Thought

Lets end this discussion with some exponential,

Suppose, $X_1,X_2,....,X_n$ are a random sample from $exponential(\theta)$ and $Y_1,Y_2,.....,Y_m$ is another random sample from the population of $exponential(\mu)$. Now you are to test $H_o: \theta=\mu$ against $H_1: \theta \neq \mu$ .

Can you show that the test can be based on a statistic $T$ such that, $T= \frac{\sum X_i}{\sum X_i +\sum Y_i}$.

What distribution you think, T should follow under null hypothesis ? Think it over !!

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