Reply to queries

[Aritra Mandal]

Hello learners,

I am Aritra Mandal, your TA. I have tried to address all the queries. 

Please go through the following. Feel free to reply if there is any doubt.

Question 1> I have a question about what is happening in this video at 00:14:00 -

http://www.youtube.com/watch?v=SRf5pxmq8Oo#t=840s.

I have a doubt in the proof of the reverse implication ( E[|X|] finite implies E[X] converges absolutely or finite). I think that it is true only when E[|X|] = 0. Because in real analysis, I have studied that if a sequence |xn| converges to a real number not equal to zero, then xn converges to a real number may not always be true. Ex - xn = (-1)^n. The sequence xn takes value -1 and 1 thus does not converge to any real number but sequence |xn|=1 which coverges to 1. I don't understand how to prove the reverse implication?.

Answer> Dear learner, in the proof "Convergence of series" is used. Whatever you said is correct. But those are related to "Convergence of sequences" not "Convergence of series".

So the proof is correct. I suggest you to go through "Convergence of series".

Question 2> I have a doubt in the interpretation of the result of Ex. 3.2.1 (c) of the book "Probability and Statistics with examples using R". I have attached the screenshort of the problem.     

As we are first choosing a ball and then put it aside and then choosing the ball from the remaining one second time. So the sample space S is {(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)} with the probability of each outcome = 1/12.            

P(Y=2) = P({(1,2),(3,2),(4,2)}) = 3/12 = 1/4       

We can also solve it as P(Y=2)=P(Y=2|X=1)P(X=1)+P(Y=2|X=3)P(X=3)+P(Y=2|X=4)P(X=4) = 3 X (1/3 * 1/4) = 1/4 

The result P(Y=2) = 1/4 is surprising. If we don't have the information of what we get while choosing the ball for first time, then choosing the ball for second time is like putting the ball chosen first back into the urn and choosing the ball during second time out of all the four balls. As choosing the all the balls are equally likely, So, P(Y=2) = 1/4.  

Computation-wise, the result which we arrived, I can understand. But I am not able to interpret the result physically, like how the problem reduced to choosing the ball out of 4. If we are doing the experiment and I don't know what ball I get for the first time, how likelihood of choosing the ball-2 during second time is 1/4. Intuitively, I don't think the probability to be 1/4.                  

Intuitively, I am not able to interpret the result. Please help me in clarifying the doubt.   

Answer>Outcomes of getting a number out of {1,2,3,4} in first draw are equally likely.

Similarly outcomes of getting a number out of {1,2,3,4} in second draw are equally likely. Think about it.  

We know P(Y=1)+P(Y=2)+P(Y=3)+P(Y=4) = 1. Now the outcomes of getting any number out of {1,2,3,4} in second draw are equally likely. 

So, P(Y=1)=P(Y=2)=P(Y=3)=P(Y=4)=1/4

I hope it clears your doubt.

Thanks and regards

[Himanshu]

@Aritra mam, thanks for the replies.   

I have a doubt in question-2. 

I am not able to understand the statement "Similarly outcomes of getting a number out of {1,2,3,4} in second draw are equally likely."

I am choosing a ball from an urn with my eyes being covered with a black cloth and then the chosen ball has been kept aside. So, I won't know which ball I have chosen for the first time. Then i am going to choose second ball with the my eyes being covered. In that case I know that the urn has 3 balls instead of 4 because I also know that I have chosen a ball for the first time which has been kept aside.    

So, if I think from this point of view - 

Event of choosing ball-2 (Y=2) = Choosing ball-2 if it has been chosen initially U choosing ball-2 if it has not been chosen initially     

Choosing ball-2 if it has been chosen initially = phi   

Choosing ball-2 if it has not been chosen initially is same as like 3 having balls in the urn with anyone of them being ball-2 --------(1)

P(Y=2) = P(phi) + P(choosing ball-2 out of 3 balls) = 0+1/3 = 1/3         

I don't understand, with that reasoning, why am I getting answer 1/3 instead of 1/4? I know that the whole problem lies in the argument (1). But one can naively make the argument (1). I am not able to understand how to prove it to be wrong in words not mathematically. Because mathematically, one can take sample space to be {(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)} and get 1/4 as answer. But still intuitively or say while doing experiment physically, I am not able to disprove the argument (1) and not get 1/4 as answer?      

The answer 1/4 seems to suggest that having the information of 3 balls in the urn while choosing them second time counterbalances not having the information of type of ball in the urn, leaving the sample space to remain same.       

That is my exact question. Initially I was not able to phrase it properly.      

That might be a silly question. But mam please help me in clarifying the doubt. I am very very confused.   

Thanks!  

[Vinay M]

Nice question, just for a moment forget about what we choose first, just completely shut your awareness on that X, now just focus on only Y, here what all numbers can come? It's definitely 1,2,3,4 and all of them have the same probability to come there to the second place, so therefore thinking in this way we say, hey all o these four balls can come equally like in the second place Y, so for Y=2 it should be 1/4. Thats what I think is meant by "Similarly outcomes of getting a number out of {1,2,3,4} in second draw are equally likely". Thank you. 

[Himanshu]

@Vinay, thanks for the reply.   

But how we forget about what comes first, we have the information that we have chosen the ball for the first time, so though we don't know about which ball we choose in the first trial, we know for sure that after choosing a ball now we have 3 balls in the urn while choosing second time.    

The problem is that mathematically we get 1/4 as an answer which is pretty much clear to me. But intuitively or physically, I get different answer. That's the major problem. I am not able to understand how to solve it at that level.

[Vinay M]

Please tell me that intuitive answer, then maybe I can say something. 

[Himanshu]

@Vinay I got 1/3. I have explained the reason in the first reply to Aritra mam in this conversation.    

But don't you think there is some sort of mess in this problem if we think intuitively? 

I think reply of TA will also help. 

By the way thanks for the discussion.

[Vinay M]

Ah that's a classic pitfall, I suggest you spend some more time thinking about it and try to tease and play with that intuitive idea and see if anything cracks. 

[Himanshu]

@Vinay, if you have any intuitive idea, can you please tell? I have spend a good time thinking intuitively over that problem, but can't find any way.