the gamma distribution

logistic_guy

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here is the question

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


my attemb
i think α\displaystyle \alpha and β\displaystyle \beta are the parameter of the gamma distribution
first moment
μ1=β1Γ(α+1)Γ(α)\displaystyle \mu'_1 = \frac{\beta^1\Gamma(\alpha + 1)}{\Gamma(\alpha)}
second moment
μ2=β2Γ(α+2)Γ(α)\displaystyle \mu'_2 = \frac{\beta^2\Gamma(\alpha + 2)}{\Gamma(\alpha)}
third moment
μ3=β3Γ(α+3)Γ(α)\displaystyle \mu'_3 = \frac{\beta^3\Gamma(\alpha + 3)}{\Gamma(\alpha)}
so i think there is infinite moments. do my proof correct?
 
You have to start with the definition in order to show the equation. How are the μr \mu'_r defined?
 
You have to start with the definition in order to show the equation. How are the μr \mu'_r defined?
thank

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

it don't tell me what μr\displaystyle \mu_r if i know it i'll take the derivative to get μr\displaystyle \mu'_r
Maybe you have the moment-generating function. The way from the distribution alone to that function is probably a bit longer.
 
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Wikipedia says that μr(x)=(bbx)r\mu_r(x)=\left(\dfrac{b}{b-x}\right)^r is the moment-generating function. This means
μr(x)=rbr(bx)r+1=rbxμr(x) \mu_r'(x)=\dfrac{rb^r}{(b-x)^{r+1}}=\dfrac{r}{b-x}\,\mu_r(x) but I have difficulties to translate this into β,α. \beta,\alpha. At least, it looks like the functional equation of the Gamma-function:
Γ(x+1)=xΓ(x). \Gamma(x+1)=x\cdot \Gamma(x).
 
you're using x\displaystyle x inside gamma function. in the question it's using α\displaystyle \alpha

do you think if we change x\displaystyle x to α\displaystyle \alpha in μr(x)\displaystyle \mu_r(x) it'll work?
 
you're using x\displaystyle x inside gamma function. in the question it's using α\displaystyle \alpha

do you think if we change x\displaystyle x to α\displaystyle \alpha in μr(x)\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 μ(x) \mu(x) has the Taylor series
μ(x)=k=0xkk!μk(x)() \mu(x)=\sum_{k=0}^\infty \dfrac{x^k}{k!}\mu_k(x) \quad (*)with the k k -th moments μk(x)=E(Xk). \mu_k(x)=E(X^k).

Let's see where we get at if we take the moment-generating function from Wikipedia:
μ(x)=(bbx)p(). \mu(x)=\left(\dfrac{b}{b-x}\right)^p\quad (**). The parameters b,p b,p are defined by the density function
f(x)={bpΓ(p)xp1ebx if x>00 if x0 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} of the Gamma-distribution G(p,b) \mathcal{G}(p,b) that are necessary to get the correct scaling factors since the integral over the density function has to equal 1. 1.

My suspicion is that β=b,α=p,r=k \beta=b, \alpha=p, r=k to match the various parameter names. Perhaps β=p,α=b,r=k. \beta=p, \alpha=b, r=k. However, I cannot be sure at this point of the consideration. The next step would be to develop the Taylor series of μ(x) \mu(x) as defined in () (**) and compare the coefficients with the terms in (). (*). 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
μ(x)=(bbx)p=k=0pkk![log(bbx)]k\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 Comparing the coefficients yields
μk(x)=(px)k[log(bbx)]k=[pxlog(bbx)]k.\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.This leaves us with the task to compare
μk(x)=ddx[pxlog(bbx)]k with βkΓ(α+k)Γ(α),\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)}, assuming r=k. r=k.

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.
 
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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 μ(x) \mu(x) has the Taylor series
μ(x)=k=0xkk!μk(x)() \mu(x)=\sum_{k=0}^\infty \dfrac{x^k}{k!}\mu_k(x) \quad (*)with the k k -th moments μk(x)=E(Xk). \mu_k(x)=E(X^k).

Let's see where we get at if we take the moment-generating function from Wikipedia:
μ(x)=(bbx)p(). \mu(x)=\left(\dfrac{b}{b-x}\right)^p\quad (**). The parameters b,p b,p are defined by the density function
f(x)={bpΓ(p)xp1ebx if x>00 if x0 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} of the Gamma-distribution G(p,b) \mathcal{G}(p,b) that are necessary to get the correct scaling factors since the integral over the density function has to equal 1. 1.

My suspicion is that β=b,α=p,r=k \beta=b, \alpha=p, r=k to match the various parameter names. Perhaps β=p,α=b,r=k. \beta=p, \alpha=b, r=k. However, I cannot be sure at this point of the consideration. The next step would be to develop the Taylor series of μ(x) \mu(x) as defined in () (**) and compare the coefficients with the terms in (). (*). 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
μ(x)=(bbx)p=k=0pkk![log(bbx)]k\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 Comparing the coefficients yields
μk(x)=(px)k[log(bbx)]k=[pxlog(bbx)]k.\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.This leaves us with the task to compare
μk(x)=ddx[pxlog(bbx)]k with βkΓ(α+k)Γ(α),\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)}, assuming r=k. r=k.

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?
 
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 μr \mu_r' 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 α,β \alpha,\beta and the density function
f(x)=βαΓ(α)xα1eβx f(x)=\dfrac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha -1}e^{-\beta x} with mean α/β \alpha/\beta and moment-generating function (MFG)
t(1tβ)α,t<β t\longmapsto \left(1-\dfrac{t}{\beta}\right)^{-\alpha}\, , \,t<\beta

Let me see whether this makes more sense.
 
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I have found the proof:

At least I haven't told nonsense. The idea is to calculate (on the second page with page number 36)
μr=E(Xr)=(α+r1)(α+r2)αβr=1βrΓ(α+r)Γ(α) \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)} So all that remains to note is, that your parameter β1 \beta^{-1} is the parameter β \beta in the paper.
 
I have found the proof:

At least I haven't told nonsense. The idea is to calculate (on the second page with page number 36)
μr=E(Xr)=(α+r1)(α+r2)αβr=1βrΓ(α+r)Γ(α) \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)} So all that remains to note is, that your parameter β1 \beta^{-1} is the parameter β \beta in the paper.
thank fresh_42 very much

why βr\displaystyle \beta^r down?
 
thank fresh_42 very much

why βr\displaystyle \beta^r down?
I can explain it following the paper, but understanding your choice of β \beta would require the definition from your side. It's a positive scaling factor anyway so whether we use β \beta or 1/β 1/\beta 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.
 
I can explain it following the paper, but understanding your choice of β \beta would require the definition from your side. It's a positive scaling factor anyway so whether we use β \beta or 1/β 1/\beta 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🙏
 
I have found the proof:
μr=E(Xr)=(α+r1)(α+r2)αβr=1βrΓ(α+r)Γ(α) \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)}
I will show why professor fresh_42 got a slightly different result.

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

μ1=E(X)=βΓ(α+1)Γ(α)=αβΓ(α)Γα=αβ\displaystyle \mu'_1 = E(X) = \frac{\beta\Gamma(\alpha + 1)}{\Gamma(\alpha)} = \frac{\alpha\beta\Gamma(\alpha)}{\Gamma{\alpha}} = \alpha\beta

Now let us visit the Wikipedia page.


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

kαk \rightarrow \alpha
θβ\theta \rightarrow \beta

Therefore, the rth moment is:

μr=E(Xr)=xrf(x) dx=0xrxα1exβΓ(α)βα dx=βrΓ(α+r)Γ(α)\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)}

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

MX(0)=E(X)M'_X(0) = E(X)
MX(0)=E(X2)M''_X(0) = E(X^2)
MX(0)=E(X3)M'''_X(0) = E(X^3)

This will give you a pattern that will lead you to derive βrΓ(α+r)Γ(α)\displaystyle \frac{\beta^r\Gamma(\alpha + r)}{\Gamma(\alpha)}.

Note: The notation μr\mu'_r is very ambiguous if someone is not very careful.
 
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