Get inspired by the success stories of our students in IIT JAM 2021. Learn More

Content

[hide]

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!

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 \(0.05\).

Normal Distribution

Simple Hypothesis Testing

Bernoulli Trials

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 !!

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 !!

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google