probability

[Devika]

Let {Xi} be a sequence of i.i.d random variables such that
 Xi = 1with probability p
     = 0 with probability 1 – p

Define y = 1 if i=1to 100 ∑Xi= 100
              =0 otherwise

  then E(y^2)=
 
a) infinity
b) n!/100! * (n-100)!  p^100 (1-p)^n-100
c) np
d) (np)^2

[Mauli]

is this ques from a past year paper,?
if yes please tell which year is it from?

[Devika]

No Mauli, its not.

[Sinistral]

If you could please check if the limit of ∑Xi is from (1 to 100) OR from (1 to n)??
If its the former I am getting my ans as p^100. For the latter case my answer is coming out to be option b.

[Devika]

Heyy Sinistral.
Yes, its summation 1 to n. Sorry for the mistake.

Your answer is right. Option b. Explain please?

[Sinistral]

Xi takes either 1(with prob p) or 0(with prob 1-p ).
Means Xi is nothing but Bernoulli Distribution. so the pdf of an Xi can be written as
f(xi)=p   , xi=1
     =1-p , xi=0

∑Xi= 100 implies any 100 Xi's out of total n Xi's should be 1.

f(y) = probability of any (exactly) 100 Xi's out of total n Xi's to be 1     ; y=1 (in such a case)
      = 1- probability of any(exactly) 100 Xi's out of total n Xi's to be 1  ; y=0 (in such a case)
 ie Y~bin(n,p)

So,
f(y)= nC100* p ^100 * (1-p)^(n-100)    ; y=1
     =1- nC100* p ^100 * (1-p)^(n-100) ; y=0

E(Y^2)= 1^2* nC100* p ^100 * (1-p)^(n-100)  + 0^2 * [1- nC100* p ^100 * (1-p)^(n-100)]
          = nC100* p ^100 * (1-p)^(n-100)
          option b

[Devika]

Thank youuuuu Sinistral :)