Probability Quiz: Variables X1 to X46, Expectation E(Xj)=0

[ParisSpart]

the variables X1,X2,... are independents and taking values 1 and -1 and their expected value E(Xj)=0 and we have Y=X1+X2+X3+...+Xn AND Z=X1+X2+X3+...+Xn+1 find the ρ(Y,Z) for n=46

i know that ρ(Υ,Ζ)=COV(Y,Z)/(σΥ*σZ)

where σY = sqrt(varY) and σZ=sqrt(varZ) how i can find them because we don't have any sum or probability to estimate them, For the cov(Y,Z) i think tha is 0 because Xj are indepents and expected value still 0 but is says tha its not true what i am doing wrong?

 

[HallsofIvy]

the variables X1,X2,... are independents and taking values 1 and -1 and their expected value E(Xj)=0

In other words, for all i, P(Xi= -1)= 1/3, P(Xi= 0)= 1/3, and P(Xi= 0)= 1/3.
(Unless you are missing the word "is": "and their expected value is E(Xj)= 0". In that case, P(Xi= -1)= 1/2, P(Xi= 1)= 1/2.)

and we have Y=X1+X2+X3+...+Xn AND Z=X1+X2+X3+...+Xn+1 find the ρ(Y,Z) for n=46

i know that ρ(Υ,Ζ)=COV(Y,Z)/(σΥ*σZ)

where σY = sqrt(varY) and σZ=sqrt(varZ) how i can find them because we don't have any sum or probability to estimate them, For the cov(Y,Z) i think tha is 0 because Xj are indepents and expected value still 0 but is says tha its not true what i am doing wrong?

 

[haruspex]

In other words, for all i, P(Xi= -1)= 1/3, P(Xi= 0)= 1/3, and P(Xi= 0)= 1/3.
(Unless you are missing the word "is": "and their expected value is E(Xj)= 0". In that case, P(Xi= -1)= 1/2, P(Xi= 1)= 1/2.)

I read it as "their expected value, E(Xj), = 0". So P(Xi= -1) = P(Xi= 1)= 1/2.

For the cov(Y,Z) i think tha is 0 because Xj are indepents

But Y and Z depend on n of the same samples, so will not be independent. On an occasion when Y turns out to be higher than normal, Z likely will be too.

 

[ParisSpart]

how i can find E(YZ)=? i can't think how to find it

 

[haruspex]

how i can find E(YZ)=? i can't think how to find it

Write the expression for YZ in terms of the Xi (using Ʃ). The E() of a sum is the sum of the E()s.