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 $\mu'_r$ defined?

 

[logistic_guy]

You have to start with the definition in order to show the equation. How are the $\mu'_r$ 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 $\mu_r(x)=\left(\dfrac{b}{b-x}\right)^r$ is the moment-generating function. This means
$$\mu_r'(x)=\dfrac{rb^r}{(b-x)^{r+1}}=\dfrac{r}{b-x}\,\mu_r(x)$$but I have difficulties to translate this into $\beta,\alpha.$ At least, it looks like the functional equation of the Gamma-function:
$$\Gamma(x+1)=x\cdot \Gamma(x).$$

 

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

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

My suspicion is that $\beta=b, \alpha=p, r=k$ to match the various parameter names. Perhaps $\beta=p, \alpha=b, r=k.$ However, I cannot be sure at this point of the consideration. The next step would be to develop the Taylor series of $\mu(x)$ 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
$$\mu(x)= \left(\dfrac{b}{b-x}\right)^p = \sum_{k=0}^\infty \dfrac{p^k}{k!}\left[ \log \left( \dfrac{b}{b-x} \right)\right]^k$$Comparing the coefficients yields
$$\mu_k(x)=\left(\dfrac{p}{x}\right)^k \left[ \log \left( \dfrac{b}{b-x} \right)\right]^k= \left[\dfrac{p}{x}\log\left( \dfrac{b}{b-x} \right)\right]^k.$$This leaves us with the task to compare
$$\mu'_k(x)= \dfrac{d}{dx}\left[\dfrac{p}{x}\log\left( \dfrac{b}{b-x} \right)\right]^k\text{ with }\beta^k \dfrac{\Gamma(\alpha+k)}{\Gamma(\alpha)},$$assuming $r=k.$

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

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

My suspicion is that $\beta=b, \alpha=p, r=k$ to match the various parameter names. Perhaps $\beta=p, \alpha=b, r=k.$ However, I cannot be sure at this point of the consideration. The next step would be to develop the Taylor series of $\mu(x)$ 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
$$\mu(x)= \left(\dfrac{b}{b-x}\right)^p = \sum_{k=0}^\infty \dfrac{p^k}{k!}\left[ \log \left( \dfrac{b}{b-x} \right)\right]^k$$Comparing the coefficients yields
$$\mu_k(x)=\left(\dfrac{p}{x}\right)^k \left[ \log \left( \dfrac{b}{b-x} \right)\right]^k= \left[\dfrac{p}{x}\log\left( \dfrac{b}{b-x} \right)\right]^k.$$This leaves us with the task to compare
$$\mu'_k(x)= \dfrac{d}{dx}\left[\dfrac{p}{x}\log\left( \dfrac{b}{b-x} \right)\right]^k\text{ with }\beta^k \dfrac{\Gamma(\alpha+k)}{\Gamma(\alpha)},$$assuming $r=k.$

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

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)
$$\mu_r=E(X^r)=\dfrac{(\alpha +r-1)(\alpha +r-2)\cdots \alpha}{\beta^r}=\dfrac{1}{\beta^r}\dfrac{\Gamma(\alpha+r)}{\Gamma(\alpha)}$$So all that remains to note is, that your parameter $\beta^{-1}$ is the parameter $\beta$ 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)
$$\mu_r=E(X^r)=\dfrac{(\alpha +r-1)(\alpha +r-2)\cdots \alpha}{\beta^r}=\dfrac{1}{\beta^r}\dfrac{\Gamma(\alpha+r)}{\Gamma(\alpha)}$$So all that remains to note is, that your parameter $\beta^{-1}$ is the parameter $\beta$ 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 $\beta$ would require the definition from your side. It's a positive scaling factor anyway so whether we use $\beta$ or $1/\beta$ 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.

 

[logistic_guy]

I can explain it following the paper, but understanding your choice of $\beta$ would require the definition from your side. It's a positive scaling factor anyway so whether we use $\beta$ or $1/\beta$ 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

 

[mario99]

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

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

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

$\displaystyle \mu'_1 = E(X) = \frac{\beta\Gamma(\alpha + 1)}{\Gamma(\alpha)} = \frac{\alpha\beta\Gamma(\alpha)}{\Gamma{\alpha}} = \alpha\beta$

Now let us visit the Wikipedia page.

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 $\displaystyle\frac{\alpha}{\beta}$ because it matches the parameters of the OP. In fact, he should have chosen the left one with mean $k\theta$ because it matches the mean of the OP $\left(\alpha\beta\right)$. It only uses different parameters (variables).

$k \rightarrow \alpha$
$\theta \rightarrow \beta$

Therefore, the rth moment is:

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

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

$M'_X(0) = E(X)$
$M''_X(0) = E(X^2)$
$M'''_X(0) = E(X^3)$

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

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