the gamma distribution

[logistic_guy]

here is the question

Show that the \(\displaystyle r\)th moment about the origin of the gamma distribution is \(\displaystyle \mu'_r = \frac{\beta^r\Gamma(\alpha + r)}{\Gamma(\alpha)}\).


my attemb
i think \(\displaystyle \alpha\) and \(\displaystyle \beta\) are the parameter of the gamma distribution
first moment
\(\displaystyle \mu'_1 = \frac{\beta^1\Gamma(\alpha + 1)}{\Gamma(\alpha)}\)
second moment
\(\displaystyle \mu'_2 = \frac{\beta^2\Gamma(\alpha + 2)}{\Gamma(\alpha)}\)
third moment
\(\displaystyle \mu'_3 = \frac{\beta^3\Gamma(\alpha + 3)}{\Gamma(\alpha)}\)
so i think there is infinite moments. do my proof correct?

 

[fresh_42]

You have to start with the definition in order to show the equation. How are the [imath] \mu'_r [/imath] defined?

 

[logistic_guy]

You have to start with the definition in order to show the equation. How are the [imath] \mu'_r [/imath] defined?

thank

it don't tell me what \(\displaystyle \mu_r\) if i know it i'll take the derivative to get \(\displaystyle \mu'_r\)

 

[fresh_42]

thank

it don't tell me what \(\displaystyle \mu_r\) if i know it i'll take the derivative to get \(\displaystyle \mu'_r\)

Maybe you have the moment-generating function. The way from the distribution alone to that function is probably a bit longer.

 

[fresh_42]

Wikipedia says that [imath]\mu_r(x)=\left(\dfrac{b}{b-x}\right)^r [/imath] is the moment-generating function. This means
[math] \mu_r'(x)=\dfrac{rb^r}{(b-x)^{r+1}}=\dfrac{r}{b-x}\,\mu_r(x) [/math]but I have difficulties to translate this into [imath] \beta,\alpha. [/imath] At least, it looks like the functional equation of the Gamma-function:
[math] \Gamma(x+1)=x\cdot \Gamma(x). [/math]

 

[logistic_guy]

you're using \(\displaystyle x\) inside gamma function. in the question it's using \(\displaystyle \alpha\)

do you think if we change \(\displaystyle x\) to \(\displaystyle \alpha\) in \(\displaystyle \mu_r(x)\) it'll work?

 

[fresh_42]

you're using \(\displaystyle x\) inside gamma function. in the question it's using \(\displaystyle \alpha\)

do you think if we change \(\displaystyle x\) to \(\displaystyle \alpha\) in \(\displaystyle \mu_r(x)\) it'll work?

Maybe. I'm really not sure what all the variable and function names are supposed to represent.

As far as I could see, the moment-generating function [imath] \mu(x) [/imath] has the Taylor series
[math] \mu(x)=\sum_{k=0}^\infty \dfrac{x^k}{k!}\mu_k(x) \quad (*)[/math]with the [imath] k [/imath]-th moments [imath] \mu_k(x)=E(X^k). [/imath]

Let's see where we get at if we take the moment-generating function from Wikipedia:
[math] \mu(x)=\left(\dfrac{b}{b-x}\right)^p\quad (**). [/math]The parameters [imath] b,p [/imath] are defined by the density function
[math] f(x)=\begin{cases} \dfrac{b^p}{\Gamma(p)}\,x^{p-1}e^{-bx}&\text{ if }x>0\\0&\text{ if }x\leq 0 \end{cases} [/math]of the Gamma-distribution [imath] \mathcal{G}(p,b) [/imath] that are necessary to get the correct scaling factors since the integral over the density function has to equal [imath] 1. [/imath]

My suspicion is that [imath] \beta=b, \alpha=p, r=k [/imath] to match the various parameter names. Perhaps [imath] \beta=p, \alpha=b, r=k. [/imath] However, I cannot be sure at this point of the consideration. The next step would be to develop the Taylor series of [imath] \mu(x) [/imath] as defined in [imath] (**) [/imath] and compare the coefficients with the terms in [imath] (*). [/imath] I admit, I'm a bit lazy to perform all the differentiations, so I asked WA (https://www.wolframalpha.com/input?i=Taylor+expansion+of+(b/(b-x))^p) to do it for me and got
[math]\mu(x)= \left(\dfrac{b}{b-x}\right)^p = \sum_{k=0}^\infty \dfrac{p^k}{k!}\left[ \log \left( \dfrac{b}{b-x} \right)\right]^k [/math]Comparing the coefficients yields
[math]\mu_k(x)=\left(\dfrac{p}{x}\right)^k \left[ \log \left( \dfrac{b}{b-x} \right)\right]^k= \left[\dfrac{p}{x}\log\left( \dfrac{b}{b-x} \right)\right]^k.[/math]This leaves us with the task to compare
[math]\mu'_k(x)= \dfrac{d}{dx}\left[\dfrac{p}{x}\log\left( \dfrac{b}{b-x} \right)\right]^k\text{ with }\beta^k \dfrac{\Gamma(\alpha+k)}{\Gamma(\alpha)}, [/math]assuming [imath] r=k. [/imath]

That's not easy to see and still possibly wrong since Wikipedia isn't the most reliable source plus I may have made mistakes. However, I will change the editor to do some calculations off-line.

 

[logistic_guy]

Maybe. I'm really not sure what all the variable and function names are supposed to represent.

As far as I could see, the moment-generating function [imath] \mu(x) [/imath] has the Taylor series
[math] \mu(x)=\sum_{k=0}^\infty \dfrac{x^k}{k!}\mu_k(x) \quad (*)[/math]with the [imath] k [/imath]-th moments [imath] \mu_k(x)=E(X^k). [/imath]

Let's see where we get at if we take the moment-generating function from Wikipedia:
[math] \mu(x)=\left(\dfrac{b}{b-x}\right)^p\quad (**). [/math]The parameters [imath] b,p [/imath] are defined by the density function
[math] f(x)=\begin{cases} \dfrac{b^p}{\Gamma(p)}\,x^{p-1}e^{-bx}&\text{ if }x>0\\0&\text{ if }x\leq 0 \end{cases} [/math]of the Gamma-distribution [imath] \mathcal{G}(p,b) [/imath] that are necessary to get the correct scaling factors since the integral over the density function has to equal [imath] 1. [/imath]

My suspicion is that [imath] \beta=b, \alpha=p, r=k [/imath] to match the various parameter names. Perhaps [imath] \beta=p, \alpha=b, r=k. [/imath] However, I cannot be sure at this point of the consideration. The next step would be to develop the Taylor series of [imath] \mu(x) [/imath] as defined in [imath] (**) [/imath] and compare the coefficients with the terms in [imath] (*). [/imath] I admit, I'm a bit lazy to perform all the differentiations, so I asked WA (https://www.wolframalpha.com/input?i=Taylor+expansion+of+(b/(b-x))^p) to do it for me and got
[math]\mu(x)= \left(\dfrac{b}{b-x}\right)^p = \sum_{k=0}^\infty \dfrac{p^k}{k!}\left[ \log \left( \dfrac{b}{b-x} \right)\right]^k [/math]Comparing the coefficients yields
[math]\mu_k(x)=\left(\dfrac{p}{x}\right)^k \left[ \log \left( \dfrac{b}{b-x} \right)\right]^k= \left[\dfrac{p}{x}\log\left( \dfrac{b}{b-x} \right)\right]^k.[/math]This leaves us with the task to compare
[math]\mu'_k(x)= \dfrac{d}{dx}\left[\dfrac{p}{x}\log\left( \dfrac{b}{b-x} \right)\right]^k\text{ with }\beta^k \dfrac{\Gamma(\alpha+k)}{\Gamma(\alpha)}, [/math]assuming [imath] r=k. [/imath]

That's not easy to see and still possibly wrong since Wikipedia isn't the most reliable source plus I may have made mistakes. However, I will change the editor to do some calculations off-line.

i'm lost

what am suppose to do?

 

[fresh_42]

i'm lost

what am suppose to do?

Me, too. This derivative is a mess. We need a different approach. Are you sure you don't have more information about [imath] \mu_r' [/imath] since I was only guessing that those were the coefficients of the Taylor series of the moment-generating function?

I switched to the English Wikipedia which uses the parameter names [imath] \alpha,\beta [/imath] and the density function
[math] f(x)=\dfrac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha -1}e^{-\beta x} [/math] with mean [imath] \alpha/\beta [/imath] and moment-generating function (MFG)
[imath] t\longmapsto \left(1-\dfrac{t}{\beta}\right)^{-\alpha}\, , \,t<\beta [/imath]

Let me see whether this makes more sense.

 

[fresh_42]

I have found the proof:

https://ocw.mit.edu/courses/18-443-statistics-for-applications-fall-2006/924817d33e1ccb6f3a6b944c985d0cdb_lecture6.pdf

At least I haven't told nonsense. The idea is to calculate (on the second page with page number 36)
[math] \mu_r=E(X^r)=\dfrac{(\alpha +r-1)(\alpha +r-2)\cdots \alpha}{\beta^r}=\dfrac{1}{\beta^r}\dfrac{\Gamma(\alpha+r)}{\Gamma(\alpha)} [/math]So all that remains to note is, that your parameter [imath] \beta^{-1} [/imath] is the parameter [imath] \beta [/imath] in the paper.

 

[logistic_guy]

I have found the proof:

https://ocw.mit.edu/courses/18-443-statistics-for-applications-fall-2006/924817d33e1ccb6f3a6b944c985d0cdb_lecture6.pdf

At least I haven't told nonsense. The idea is to calculate (on the second page with page number 36)
[math] \mu_r=E(X^r)=\dfrac{(\alpha +r-1)(\alpha +r-2)\cdots \alpha}{\beta^r}=\dfrac{1}{\beta^r}\dfrac{\Gamma(\alpha+r)}{\Gamma(\alpha)} [/math]So all that remains to note is, that your parameter [imath] \beta^{-1} [/imath] is the parameter [imath] \beta [/imath] in the paper.

thank fresh_42 very much

why \(\displaystyle \beta^r\) down?

 

[fresh_42]

thank fresh_42 very much

why \(\displaystyle \beta^r\) down?

I can explain it following the paper, but understanding your choice of [imath] \beta [/imath] would require the definition from your side. It's a positive scaling factor anyway so whether we use [imath] \beta [/imath] or [imath] 1/\beta [/imath] makes no significant difference. The crucial point is how the density function is written. The paper uses the version I mentioned in post #9. I cannot know your definition since you didn't give one. The other possibility would be a mistake, but I haven't seen one in the paper.

 

[logistic_guy]

I can explain it following the paper, but understanding your choice of [imath] \beta [/imath] would require the definition from your side. It's a positive scaling factor anyway so whether we use [imath] \beta [/imath] or [imath] 1/\beta [/imath] makes no significant difference. The crucial point is how the density function is written. The paper uses the version I mentioned in post #9. I cannot know your definition since you didn't give one. The other possibility would be a mistake, but I haven't seen one in the paper.

thank

i think i understand the paper. i hope it's a misake

appreciate your help

 

[mario99]

I have found the proof:
[math] \mu_r=E(X^r)=\dfrac{(\alpha +r-1)(\alpha +r-2)\cdots \alpha}{\beta^r}=\dfrac{1}{\beta^r}\dfrac{\Gamma(\alpha+r)}{\Gamma(\alpha)} [/math]

I will show why professor fresh_42 got a slightly different result.

The first moment is the mean. Let us calculate it:

[imath]\displaystyle \mu'_1 = E(X) = \frac{\beta\Gamma(\alpha + 1)}{\Gamma(\alpha)} = \frac{\alpha\beta\Gamma(\alpha)}{\Gamma{\alpha}} = \alpha\beta[/imath]

Now let us visit the Wikipedia page.

Gamma distribution - Wikipedia

en.wikipedia.org

According to the Wiki, Gamma distribution has two PDFs. Which one to choose? Professor fresh_42 chose the right one with mean [imath]\displaystyle\frac{\alpha}{\beta}[/imath] because it matches the parameters of the OP. In fact, he should have chosen the left one with mean [imath]k\theta[/imath] because it matches the mean of the OP [imath]\left(\alpha\beta\right)[/imath]. It only uses different parameters (variables).

[imath]k \rightarrow \alpha[/imath]
[imath]\theta \rightarrow \beta[/imath]

Therefore, the rth moment is:

[imath]\displaystyle \mu'_r = E(X^r) = \int_{-\infty}^{\infty}x^rf(x) \ dx = \int_{0}^{\infty} x^r\frac{x^{\alpha - 1}e^{-\frac{x}{\beta}}}{\Gamma(\alpha)\beta^{\alpha}} \ dx = \frac{\beta^r\Gamma(\alpha + r)}{\Gamma(\alpha)}[/imath]

------
Another way to solve the problem is to find the moment generating function, [imath]M_X(t),[/imath] of Gamma distribution, and differentiate it three times or more if necessary at [imath]t = 0[/imath].

[imath]M'_X(0) = E(X)[/imath]
[imath]M''_X(0) = E(X^2)[/imath]
[imath]M'''_X(0) = E(X^3)[/imath]

This will give you a pattern that will lead you to derive [imath]\displaystyle \frac{\beta^r\Gamma(\alpha + r)}{\Gamma(\alpha)}[/imath].

Note: The notation [imath]\mu'_r[/imath] is very ambiguous if someone is not very careful.