Find the mean and variance of Y^2

[BrownianMan]

Let Y by the number of heads obtained if a coin is tossed three times. Find the mean and variance of Y^2.

For the mean I get, (0+1+4+9)/4=7/2, and for variance I get (0+1+16+81)/4 - (7/2)^2 = 49/4. Is this correct?

For the following question, I'm not sure how to begin:

Show that if T has exponential distribution with rate lambda, then int(T), the greatest integer less than or equal to T, has geometric (p) distribution on {0, 1, 2,...}, and find p in terms of lambda.

 

[lanedance]

Let Y by the number of heads obtained if a coin is tossed three times. Find the mean and variance of Y^2.

For the mean I get, (0+1+4+9)/4=7/2

not quite, the definition of mean of a discrete RV is
E(Y) = \Sigma p_i(Y=y_i) y_i

so you need to include the probability of each outcome. The values you worte down would only be true if the probability of a number of heads is the same in each case, p(H=0) = p(H=1) = p(H=2) = p(H=3) which is clearly not true

 

[BrownianMan]

It's suppose to be the expectation of Y^2 though, not Y...

 

[BrownianMan]

Ok, so would E[Y^2] = 3, and Var[Y^2] = 7.5?

 

[lanedance]

ok, show us how you got those values

 

[BrownianMan]

E[Y^2] = (1/8)(0)+(3/8)(1)+(3/8)(4)+(1/8)(9) = 24/8 = 3

Var[Y^2] = [(1/8)(0^2)+(3/8)(1^2)+(3/8)(4^2)+(1/8)(9^2)] - 3^2 = 15/2

 

[lanedance]

method looks good to me

E[Y^2]

Var[Y^2] = E[(Y^2)^2] - (E[Y^2])^2

 

[BrownianMan]

Thanks!

Any ideas for the second question?