Showing stability under scaling and additivity of distriubtions

[stukbv]

Homework Statement

I need to show that if X ~ r(a1,B) Y ~ r(a2,b) where r means gamma distribution then if X and Y are independent
i) X+Y ~ r(a1+a2,B)
ii) cX ~ r(a1,cB)

Homework Equations

The Attempt at a Solution

i) i use the mgfs of x and y and ended up with mgf(x+y) = (1/1-Bt)^(a1 + a2 )
I am told this is enough to prove i, is this correct or what could i say to make it better?

ii) i am a bit stuck on this one, i mean, would you just put c in front of all the x's in the pdf and then re-derive the mgf? or is there something extra?

thanks

 

[Ray Vickson]

Homework Statement

I need to show that if X ~ r(a1,B) Y ~ r(a2,b) where r means gamma distribution then if X and Y are independent
i) X+Y ~ r(a1+a2,B)
ii) cX ~ r(a1,cB)

Homework Equations

The Attempt at a Solution

i) i use the mgfs of x and y and ended up with mgf(x+y) = (1/1-Bt)^(a1 + a2 )
I am told this is enough to prove i, is this correct or what could i say to make it better?

ii) i am a bit stuck on this one, i mean, would you just put c in front of all the x's in the pdf and then re-derive the mgf? or is there something extra?

thanks

I suppose your "i" is supposed to be "I", rather than the square root of -1? Anyway, your result in (i) would be OK if you wrote it properly, with brackets to make things clear. That is, instead of writing (1/1-Bt)^(a1+a2)---which equals [1 - Bt]^(a1+a2)---you should write 1/(1-Bt)^(a1+a2), or (1 - Bt)^(-a1-a2) or (1-Bt)^{-(a1+a2)}. As to (ii): what is the problem? If f(x) is the density function of a random variable X, what is the density function of Y = c*X for a constant c? Alternatively, if F(x) is the (cumulative) distribution function of X, that is, P{X >= x} = F(x), then what is the cumulative distribution of Y = c*X? Then you can differentiate the distribution to get the density.

RGV

 

[stukbv]

Is this where I say X = Y/c so then i put into fx y/c in place of all x's and then multiply by 1/c to get fY?

 

[Ray Vickson]

If F(x) = P{X <= x} and Y = c*X (with c > 0 a constant) then P{Y <= y} = P{c*X <= y} = P{X <] y/c} = F(y/c). The density of Y is g(y) = (d/dy)F(y/c) = f(y/c)/c, where f(x) = probability density of X. Alternatively: g(y)*dy = P{y < Y < y+dy} = P{y < c*X < y+dy} = P{y/c < X < y/c + dy/c} = f(y/c)*dy/c, so g(y) = f(y/c)/c. So, the answer to your question is YES, but I much prefer to get it from first principles.

RGV

 

[stukbv]

Ok so now I have 1/c ( 1/(r(a)B^a) * (y/c)^a-1 .e^(-y/cB)
Is that right ?

 

[Ray Vickson]

I don't know. You have all the formulas you need.

RGV