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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