Proofing Varience in Gamma Distribution

[Philip Wong]

Homework Statement

Fy(y) = 1/2 λ^3 y^2 e^(-λy)

Assume E(1/y) = λ/2
Find E(1/y^2) and prove Var(1/y) = λ^2 / 4

Homework Equations

E(y) = k/λ
Var (y) = k/ λ^2

The Attempt at a Solution

Hi guys,

From the equation given from gamma distribution, I understands clearly that to get Var(y) what I have to do is to square y. i.e. 1/y^2 = (λ/2)^2 = λ^2 / 4. Which is equivalent to E(1/y^2)

but I'm no clue on how to set up a proof to show this is correct. I needed help on this, as I have no idea at all on how to proof this.

 

[lanedance]

the problem doesn't ask fior var(y) it asks for var(1/y)

the expectation of any function of y, say g(y), can be written
$$E(g(y)) = \int g(y) f_Y(y)dy$$

so for E(1/y^2) this gives
$$E(\frac{1}{y^2}) = \int \frac{1}{y^2} f_Y(y)dy$$

 

[Philip Wong]

Ok I've retried to do E(X). Can you please check if I have done it correctly before I moved on trying to work out the variance.

E(1/y^2) = ∫1/y^2 fy(y) dy
= ∫ 1/4 * λ^2 * 1/2 * λ^3 * y^2 * e^-λy dy
=∫1/8 * λ^5 * y^2 * e^-λy dy
=1/8 * 1/3 * λ^5 * y^3 * e^-λy - 1/λ * λ^6 * y^2 * e^-λy
=1/8 * 1/3 * λ^5 * y^3 * e^-λy - λe^-λy * λ^5 * y^2
=1/24 * λ^5 * y^3 * e^-λy - λ^6 * y^2 * e^-λy
=1/24 * λ^5 * y^2 * e^-λy (y-λ)

Up to this point I knew I did something wrong. But I run over my calculation again, but doesn't seem to be able to pick out where I did wrong.

It would be nice if you can pick out where I did wrong. thanks

 

[lanedance]

what happened to the 1/y^2 in the 2nd line?

 

[Philip Wong]

ar! I see where I get wrong now. I misunderstood it as (EX)^2 and substitute in the its value as 1/y^2. I got it now. should be

∫1/y^2 * fx(x) dy
∫1/y^2 * 1/2 λ^3 * y^2 * e(-λy) dy
∫1/2 * λ^3 * e(-λy) dy

= 1/2 λ^3 - 1/λ e(-λy)

 

[lanedance]

bit hard to read or follow, but if you mean times (-1/gamma) you're on the right track