Applied Probability Question on Expectation and Variance .. Please Help?

Question

The discrete random variable N has distribution P(N=n) = 1/2^n , n = 1,2,...
Let X be a poisson variable with mean 3 and let X1,X2, ... be a sequence of independent observations of X.
Let S = Summation of Xj from 1 to N

Assuming that X and N are independent, find E(S) and var(S)

PLEASE EXPLAIN.

THANK U BUNCHES

Answer 1

Since Xj is Poisson with mean 3, E(Xj) = Var(Xj) = 3, for 1 <= j <= n.



Since P(N=n) = 1/2^n = (1/2)(1 - 1/2)^(n-1), n = 1,2,..., we recognize N as a geometric random variable with success probability p = 1/2.

So E(N) = 1/p = 1/(1/2) = 2; Var(N) = (1 - p)/p^2 = (1 - 1/2)/(1/2)^2 = 2.



E(S) = E(E(S|N))

= E(E(Summation of Xj from 1 to N|N))

= E(Summation from 1 to N of E(Xj)) since Xj's are independent of N

= E(3N)

= 3E(N)

= 3(2)

= 6.



Var(S) = Var(E(S|N)) + E(Var(S|N))

= Var(E(Summation of Xj from 1 to N|N)) + E(Var(Summation of Xj from 1 to N|N))



= Var(E(Summation of Xj from 1 to N)) + E(Var(Summation of Xj from 1 to N)), since the Xj's are independent of N



= Var(Summation from 1 to N of E(Xj)) + E(Summation from 1 to N of Var(Xj)), since the assumption that Xj's are independent justifies replacing the variance of the sum of Xj's by the sum of the variances of the Xj's



= Var(3N) + E(3N)

= 3^2 Var(N) + 3E(N)

= 9(2) + 3(2)

= 24.



Lord bless you today!

Answer 2

call p = 0.40 8 and q = a million-p = 0.fifty two provided that one in each and every of those walk has no memory and is translation invariant your 2 q's study for a walker commencing on the commencing place a) what's the possibility that the element ever reaches -5? b) same with 5 the 1st ( possibility that the element ever reaches -a million)^5 and in addition for the 2nd. call P(ok). the possibility of ever attaining ok. Then P(a million) = p + q*P(2). So P(a million) = 0.40 8 + 0.fifty two P(a million)^2. 0.fifty two x^2 - x + 0.40 8 = 0 has 2 roots, a million and 0.40 8 / 0.fifty two. in addition P(-a million) is the two a million or 0.fifty two / 0.40 8; consequently little question P( -a million ) = a million. If P(a million) have been additionally equivalent to a million, that would mean that with possibility a million the walk visits 0 infinitely in many cases. however the expectancy of the style of visits is sum 0 to inf (2n C n) (p(a million-p))^n which converges on the grounds that p(a million-p) < a million/4. subsequently the walk does visit 0 infinitely many cases with possibility 0. consequently P(a million) = 12/13; Your solutions are a) a million, b) (12/13)^5

Answer 3

Optimality and Clustering

Vol I: Single-Parameter

by Frank K Hwang (National Chiao-Tung University, Taiwan) & Uriel G Rothblum (Technion, Israel)



The need of optimal partition arises from many real-world problems involving the distribution of limited resources to many users. The "clustering" problem, which has recently received a lot of attention, is a special case of optimal partitioning. This book is the first attempt to collect all theoretical developments of optimal partitions, many of them derived by the authors, in an accessible place for easy reference. Much more than simply collecting the results, the book provides a general framework to unify these results and present them in an organized fashion.