What is the Variance of Tossed Coins?

[anirudh dutta]

Homework Statement

Let Y denote the number of heads obtained when three fair coins are tossed.The
variance of Y² is

Homework Equations

The Attempt at a Solution

MY problem is understanding what Y² is. i have tried to calculate VAR(Y*Y) but my answer is wrong

 

[Ray Vickson]

What is the probability distribution of Y itself? You can use this distribution to get the distribution of Z = Y^2. Then you can compute the variance of Z. Alternatively: can you see how to determine Var (Y^2) from the moments of Y?

RGV

 

[anirudh dutta]

i tried that...but
var(x*y)= [E(x)]²var(Y)+[E(y)]²var(x)+var(x)var(y)...
putting x=y...E[y]=3*.5...var[y]=3*.5*.5 ( i have taken y be be a binomial variate ).
thr answer i get is wrong

 

[HallsofIvy]

Let Y denote the number of heads obtained when three fair coins are tossed.

The probabiity that there are 0 heads is 1/8 so the probability Y^2= 0 is 1/8.

The probability that there is 1 head is 3/8 so the probability Y^2= 1 is 3/8.

The probability that there are 2 heads is 3/8 so the probabilty Y^2= 4 is 3/8.

The probability that there are 3 heads is 1/8 so the probability Y^2= 9 is 1/8.

Now can you find the mean value, \mu, of Y^2?

Can you find the expected value of \sum (Y^2- \mu)^2, the variance?

 

[Ray Vickson]

i tried that...but
var(x*y)= [E(x)]²var(Y)+[E(y)]²var(x)+var(x)var(y)...
putting x=y...E[y]=3*.5...var[y]=3*.5*.5 ( i have taken y be be a binomial variate ).
thr answer i get is wrong

You say "I tried that". Which suggestion does "that" refer to? (I gave you two suggestions.) HallsohIvy has expanded on my first suggestion, so let me expand on my second one. We know that for any random variable Z we have Var(Z) = E(Z^2) - (EZ)^2. Apply this to Z = Y^2.

The equation Var(X*Y) = Var(X)*(EY)^2 + Var(Y)*(EX)^2 + Var(X)*Var(Y) is true if X and Y are independent (or, at least, uncorrelated); it does not apply to the case X = Y with your Y.

RGV

 

[HallsofIvy]

Personally, I think using the basic definition of variance, \sum P(Y^2)(Y^2- \mu)^2, is simpler than using that formula.