Discussion Problem (Another one on finding expectation)

[Amit Goyal]

Solve the attached problem:
expectation.pdf

[anon_econ]

E(Xk) would be the same for all k because of symmetry. Let Tki=1 if a ball is placed in the kth box in the ith trial, and 0 if a ball isn't placed in the ith trial. Then Xk=Tk1+Tk2+..Tk100. E(Xk)=E(Tk1)+E(Tk2)+...E(Tk100). And E(Tki)=1/5 for all i. So E(Xk)=20.

[anon_econ]

And for part 2,
Cov(Xk,X5)=E[Xk-20][X5-20]=E[XkX5]-400
XkX5=(Tk1+Tk2+..+Tk100)(T51+T52+..+T5100)
When we open this up we get 100 product terms with the same i and 9900 product terms with different i.
For k=3 or 7, expectation of product terms with same i is 1/10 and for terms with different i is 1/25. So Covariance comes out be 6.
For k=5, expectation of product terms with the same i is 1/5 and for the other terms is 1/25. So covariance comes out to be 16.
And for other values of k, expectation of product terms with same i is 0, and for other terms is 1/25. So covariance comes out to be -4.

[rohan]

@vasudha

E(Tki)=1/5 ???

[Sinistral]

Vasudha wrote

And for part 2,
Cov(Xk,X5)=E[Xk-20][X5-20]=E[XkX5]-400
XkX5=(Tk1+Tk2+..+Tk100)(T51+T52+..+T5100)
When we open this up we get 100 product terms with the same i and 9900 product terms with different i.
For k=3 or 7, expectation of product terms with same i is 1/10 and for terms with different i is 1/25. So Covariance comes out be 6.
For k=5, expectation of product terms with the same i is 1/5 and for the other terms is 1/25. So covariance comes out to be 16.
And for other values of k, expectation of product terms with same i is 0, and for other terms is 1/25. So covariance comes out to be -4.

please tell me what am I missing in the following argument:

expected value of say X1*X5:
=E[X1X5] = Pr(putting a ball simultaneously in both Box1 and Box5 in 1st trial) * product of the number of balls put in Box1 and Box5 in 1st trial + same thing for 2nd trial + ... + same thing for 100th trial.

but we can never put a ball simultaneously in Box1 and Box5 in any trial. so isnt  Pr(putting a ball simultaneously  in both Box1 and Box5 in 1st trial)=0. wont that make E[X1X5]=0. for that matter for any k not equal to 3,5,7 E[XkX5] shud be zero.???

[rohan]

@sinistral
hey did u solve the first part by symmetry ???if nt den how??

[Sinistral]

first part is obvious:
we can put a ball in kth box by choosing k-1 th box or k+1 th box (with wrap around)
so E[Xk]= [ (1/10)*1  + (1/10) *1 ] + same thing 100 times= (2/10)*100 =20
in the above equation the first 1/10 is for choosing (k-1) box and we multiply it by 1 because it results in putting 1 ball in kth box.
similarly in the above equation the 2nd 1/10 is for choosing (k+1)th box and we multiply it by 1 because it results in putting 1 ball in kth box.
and by symmetry E[Xk] is 20 only for all k's

in vasudha's explanation (which u asked above) Tki ~ bernoulli(1/5). So E[Tki] =1/5

[rohan]

thank u ..

[rohan]

E[Tki] =1/5

as in 1/10(E(Tk|K+1TH BOX WAS CHOSEN)+E(Tk|K-1TH BOX WAS CHOSEN))=2/10=1/5 FOR EACH TRIAL

[Sinistral]

What is Tk?
 E[Tki] is the mean of a bernoulli distribution with p=1/5
f(tki)= (1-p)^(1-tki)*p^tki ; tki =0 or 1.
calculate the expected value of above distribution. it will come out to be p. which is 1/5 in this case.

waise what am I doing? I am explaining the solution of someone who is not present. :P Let her see ur and mine ques. in the morning. I am pretty sure she will explain this in a much simple way.

[Amit Goyal]

Well done, Vasudha.

[anon_econ]

Thanks sir.
Rohan, Tki is a random variable that can take values 0 or 1. So its expectation is the probability that it takes the value 1. Now this probability is 2/10 because Tki would be 1 when box k-1 or k+1 is selected.
And Sinistral, how are you getting that expression? If we want to write it as a sum of terms in this manner then we'll have to consider the entire range of values that XkX5 can take and the corresponding probabilities..