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

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In summary: Hm. Are they fixing the bug? Is my addition below correct?I assume that this is the CDF: FY(y)=(1/k)FX(y/k).
  • #1
Shackleford
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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



[itex] λ = \frac{1}{β}

[/itex]

[itex] f(x) = \frac{λ^{α}}{Γ(α)}x^{α-1}e^{-λx}

= {(\frac{1}{β})^{α}}{\frac{1}{Γ(α)}}{(\frac{y}{k})}^{α-1}e^{-λ{(\frac{y}{k})}}

= k{(\frac{1}{kβ})^{α}}{\frac{1}{Γ(α)}}{({y})}^{α-1}e^{-{(\frac{y}{βk})}}

[/itex]

What's with the extra k?
 
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  • #2
Shackleford said:

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



[itex] λ = \frac{1}{β}

[/itex]

[itex] f(x) = \frac{λ^{α}}{Γ(α)}x^{α-1}e^{-λx}

= {(\frac{1}{β})^{α}}{\frac{1}{Γ(α)}}{(\frac{y}{k})}^{α-1}e^{-λ{(\frac{y}{k})}}

= k{(\frac{1}{kβ})^{α}}{\frac{1}{Γ(α)}}{({y})}^{α-1}e^{-{(\frac{y}{βk})}}

[/itex]

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) ##.
Shackleford said:

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



[itex] λ = \frac{1}{β}

[/itex]

[itex] f(x) = \frac{λ^{α}}{Γ(α)}x^{α-1}e^{-λx}

= {(\frac{1}{β})^{α}}{\frac{1}{Γ(α)}}{(\frac{y}{k})}^{α-1}e^{-λ{(\frac{y}{k})}}

= k{(\frac{1}{kβ})^{α}}{\frac{1}{Γ(α)}}{({y})}^{α-1}e^{-{(\frac{y}{βk})}}

[/itex]

What's with the extra k?

First, tell us what YOU think.
 
  • #3
Ray Vickson said:
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?
 
Last edited:
  • #4
Shackleford said:
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.
 
  • #5
Ray Vickson said:
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 ∫ fX(x)dx = ∫ fX(y/k)dy/k = ∫ fY(y)dy.

I assume that this is the CDF: FY(y)=(1/k)FX(y/k).[/sub]
 
Last edited:

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

1. What is the Gamma distribution?

The Gamma distribution is a continuous probability distribution that is commonly used to model the waiting time between events, such as the time between phone calls at a call center. It is also used to model the size of insurance claims, rainfall amounts, and radioactive decay.

2. What does it mean for a random variable to follow a Gamma distribution?

When a random variable follows a Gamma distribution, it means that the probability of the variable taking on certain values can be described by the Gamma probability density function. This function takes into account two parameters, shape and scale, which determine the shape and location of the distribution curve.

3. What is the relationship between the Gamma distribution and the exponential distribution?

The exponential distribution is a special case of the Gamma distribution, when the shape parameter is equal to 1. This means that the waiting time for an event to occur follows an exponential distribution. The Gamma distribution is a more general distribution that can be used to model a wider range of scenarios.

4. How can I use the Gamma distribution in my research?

The Gamma distribution can be used as a statistical tool to analyze and model data in various fields, such as finance, engineering, and biology. It can help you understand the behavior of a random variable and make predictions about future events based on past data. You can also use the Gamma distribution to perform hypothesis testing and make inferences about population parameters.

5. How do I show that a variable follows a Gamma distribution?

To show that a variable follows a Gamma distribution, you can use statistical software or programming languages to perform a goodness-of-fit test. This test compares the observed data to the expected values from a Gamma distribution and provides a measure of how well the data fits the distribution. Additionally, you can also visually inspect a histogram or density plot of the data to see if it resembles the shape of a Gamma distribution curve.

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