Proving Y ~ Gamma(α, scale = kβ) from X ~ Gamma(α, scale = β)

[Shackleford]

Homework Statement

Suppose X ~ Gamma(α, scale = β and that Y = kX with k > 0 a constant. Show that Y ~ Gamma(α, scale = kβ).

Homework Equations

Gamma distribution, etc.

The Attempt at a Solution

$λ = \frac{1}{β}<br />$

$f(x) = \frac{λ^{α}}{Γ(α)}x^{α-1}e^{-λx}<br /> <br /> = {(\frac{1}{β})^{α}}{\frac{1}{Γ(α)}}{(\frac{y}{k})}^{α-1}e^{-λ{(\frac{y}{k})}}<br /> <br /> = k{(\frac{1}{kβ})^{α}}{\frac{1}{Γ(α)}}{({y})}^{α-1}e^{-{(\frac{y}{βk})}}<br />$

What's with the extra k?

 

[Ray Vickson]

Homework Statement

Suppose X ~ Gamma(α, scale = β and that Y = kX with k > 0 a constant. Show that Y ~ Gamma(α, scale = kβ).

Homework Equations

Gamma distribution, etc.

The Attempt at a Solution

$λ = \frac{1}{β}<br />$

$f(x) = \frac{λ^{α}}{Γ(α)}x^{α-1}e^{-λx}<br /> <br /> = {(\frac{1}{β})^{α}}{\frac{1}{Γ(α)}}{(\frac{y}{k})}^{α-1}e^{-λ{(\frac{y}{k})}}<br /> <br /> = k{(\frac{1}{kβ})^{α}}{\frac{1}{Γ(α)}}{({y})}^{α-1}e^{-{(\frac{y}{βk})}}<br />$

What's with the extra k?

It comes from $dy = k dx$, so that $\int f_X(x) \, dx = \int f_X(y/x) \, dy/x$.
Thus, $f_Y(y) = (1/k) f_X(y/k)$.

Homework Statement

Suppose X ~ Gamma(α, scale = β and that Y = kX with k > 0 a constant. Show that Y ~ Gamma(α, scale = kβ).

Homework Equations

Gamma distribution, etc.

The Attempt at a Solution

$λ = \frac{1}{β}<br />$

$f(x) = \frac{λ^{α}}{Γ(α)}x^{α-1}e^{-λx}<br /> <br /> = {(\frac{1}{β})^{α}}{\frac{1}{Γ(α)}}{(\frac{y}{k})}^{α-1}e^{-λ{(\frac{y}{k})}}<br /> <br /> = k{(\frac{1}{kβ})^{α}}{\frac{1}{Γ(α)}}{({y})}^{α-1}e^{-{(\frac{y}{βk})}}<br />$

What's with the extra k?

First, tell us what YOU think.

 

[Shackleford]

It comes from $dy = k dx$, so that $\int f_X(x) \, dx = \int f_X(y/x) \, dy/x$.
Thus, $f_Y(y) = (1/k) f_X(y/k)$.First, tell us what YOU think.

Eh. It's related to the scale factor and property of the gamma function. In the first line, shouldn't it be y/k and dy/k?

 

[Ray Vickson]

Eh. It's related to the scale factor and property of the gamma function. In the first line, shouldn't it be y/k and dy/k?

Yes.

Unfortunately, when I was composing the response, I first gave that detailed answer, then changed my mind and went instead with the "non-answer" that would leave the work to you. I thought I had deleted the detailed answer, but the PF editor is tricky: if one is not careful, material one thinks has been deleted turns out to not have been. That is what happened here! I told you more than I wanted to.

 

[Shackleford]

Yes.

Unfortunately, when I was composing the response, I first gave that detailed answer, then changed my mind and went instead with the "non-answer" that would leave the work to you. I thought I had deleted the detailed answer, but the PF editor is tricky: if one is not careful, material one thinks has been deleted turns out to not have been. That is what happened here! I told you more than I wanted to.

Hm. Are they fixing the bug? Is my addition below correct?

dy=kdx, so that ∫ f_X(x)dx = ∫ f_X(y/k)dy/k = ∫ f_Y(y)dy.

I assume that this is the CDF: F_Y(y)=(1/k)F_X(y/k).[/sub]