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ISI MStat PSB 2014 Problem 9 | Hypothesis Testing

This is a another beautiful sample problem from ISI MStat PSB 2014 Problem 9. It is based on testing simple hypothesis, but reveals and uses a very cute property of Geometric distribution, which I prefer calling sister to Loss of memory . Give it a try !

Problem- ISI MStat PSB 2014 Problem 9


Let X_i \sim Geo(p_1) and X_2 \sim Geo(p_2) be independent random variables, where Geo(p) refers to Geometric distribution whose p.m.f. f is given by,

f(k)=p(1-p)^k,   k=0,1,.....

We are interested in testing the null hypothesis H_o : p_1=p_2 against the alternative H_1: p_1<p_2. Intuitively it is clear that we should reject if X_1 is large, but unfortunately, we cannot compute the cut-off because the distribution of X_1 under H_o depends on the unknown (common) value p_1 and p_2.

(a) Let Y= X_1 +X_2. Find the conditional distribution of X_1|Y=y when p_1=p_2.

(b) Based on the result obtained in (a), derive a level 0.05 test for H_o against H_1 when X_1 is large.

Prerequisites


Geometric Distribution.

Negative binomial distribution.

Discrete Uniform distribution .

Conditional Distribution . .

Simple Hypothesis Testing.

Solution :

Well, Part (a), is quite easy, but interesting and elegant, so I'm leaving it as an exercise, for you to have the fun. Hint: verify whether the required distribution is Discrete uniform or not ! If you are done, proceed .

Now, part (b), is further interesting, because here we will not use the conventional way of analyzing the distribution of X_1 and X_2, whereas we will be concentrating ourselves on the conditional distribution of X_1 | Y=y ! But why ?

The reason behind this adaptation of strategy is required, one of the reason is already given in the question itself, but the other reason is more interesting to observe , i.e. if you are done with (a), then by now you found that , the conditional distribution of X_1|Y=y is independent of any parameter ( i.e. ithe distribution of X_1 looses all the information about the parameter p_1 , when conditioned by Y=y , p_1=p_2 is a necessary condition), and the parameter independent conditional distribution is nothing but a Discrete Uniform {0,1,....,y}, where y is the sum of X_1 and X_2 .

so, under H_o: p_1=p_2 , the distribution of X_1|Y=y is independent of the both common parameter p_1 and p_2 . And clearly as stated in the problem itself, its intuitively understandable , large value of X_1 exhibits evidences against H_o. Since large value of X_1 is realized, means the success doesn't come very often .i.e. p_1 is smaller.

So, there will be strong evidence against H_o if X_1 > c , where , for some constant c \ge y, where y is given the sum of X_1+X_2.

So, for a level 0.05 test , the test will reject H_o for large value of k , such that,

P_{H_o}( X_1 > c| Y=y)=0.05  \Rightarrow  \frac{y-c}{y+1} = 0.05  \Rightarrow c= 0.95 y - 0.05 .

So, we reject H_o at level 0.05, when we observe X_1 > 0.95y - 0.05 , where it is given that X_1+X_2 =y . That's it!


Food For Thought

Can you show that for this same X_1 and X_2 ,

P(X_1 \le n)- P( X_1+X_2 \le n)= \frac{1-p}{p}P(X_1+X_2= n)

considering p_1=p_2=p , where n=0,1,.... What about the converse? Does it hold? Find out!

But avoid loosing memory, it's beauty is exclusively for Geometric ( and exponential) !!


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


This is a another beautiful sample problem from ISI MStat PSB 2014 Problem 9. It is based on testing simple hypothesis, but reveals and uses a very cute property of Geometric distribution, which I prefer calling sister to Loss of memory . Give it a try !

Problem- ISI MStat PSB 2014 Problem 9


Let X_i \sim Geo(p_1) and X_2 \sim Geo(p_2) be independent random variables, where Geo(p) refers to Geometric distribution whose p.m.f. f is given by,

f(k)=p(1-p)^k,   k=0,1,.....

We are interested in testing the null hypothesis H_o : p_1=p_2 against the alternative H_1: p_1<p_2. Intuitively it is clear that we should reject if X_1 is large, but unfortunately, we cannot compute the cut-off because the distribution of X_1 under H_o depends on the unknown (common) value p_1 and p_2.

(a) Let Y= X_1 +X_2. Find the conditional distribution of X_1|Y=y when p_1=p_2.

(b) Based on the result obtained in (a), derive a level 0.05 test for H_o against H_1 when X_1 is large.

Prerequisites


Geometric Distribution.

Negative binomial distribution.

Discrete Uniform distribution .

Conditional Distribution . .

Simple Hypothesis Testing.

Solution :

Well, Part (a), is quite easy, but interesting and elegant, so I'm leaving it as an exercise, for you to have the fun. Hint: verify whether the required distribution is Discrete uniform or not ! If you are done, proceed .

Now, part (b), is further interesting, because here we will not use the conventional way of analyzing the distribution of X_1 and X_2, whereas we will be concentrating ourselves on the conditional distribution of X_1 | Y=y ! But why ?

The reason behind this adaptation of strategy is required, one of the reason is already given in the question itself, but the other reason is more interesting to observe , i.e. if you are done with (a), then by now you found that , the conditional distribution of X_1|Y=y is independent of any parameter ( i.e. ithe distribution of X_1 looses all the information about the parameter p_1 , when conditioned by Y=y , p_1=p_2 is a necessary condition), and the parameter independent conditional distribution is nothing but a Discrete Uniform {0,1,....,y}, where y is the sum of X_1 and X_2 .

so, under H_o: p_1=p_2 , the distribution of X_1|Y=y is independent of the both common parameter p_1 and p_2 . And clearly as stated in the problem itself, its intuitively understandable , large value of X_1 exhibits evidences against H_o. Since large value of X_1 is realized, means the success doesn't come very often .i.e. p_1 is smaller.

So, there will be strong evidence against H_o if X_1 > c , where , for some constant c \ge y, where y is given the sum of X_1+X_2.

So, for a level 0.05 test , the test will reject H_o for large value of k , such that,

P_{H_o}( X_1 > c| Y=y)=0.05  \Rightarrow  \frac{y-c}{y+1} = 0.05  \Rightarrow c= 0.95 y - 0.05 .

So, we reject H_o at level 0.05, when we observe X_1 > 0.95y - 0.05 , where it is given that X_1+X_2 =y . That's it!


Food For Thought

Can you show that for this same X_1 and X_2 ,

P(X_1 \le n)- P( X_1+X_2 \le n)= \frac{1-p}{p}P(X_1+X_2= n)

considering p_1=p_2=p , where n=0,1,.... What about the converse? Does it hold? Find out!

But avoid loosing memory, it's beauty is exclusively for Geometric ( and exponential) !!


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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