How Do You Calculate the Third Moment of a Gamma Distribution?

[Artusartos]

Let X be $$\Gamma(3.5,2)$$. Find $$E(X^3)$$.

We can't write E(X^3)=E(X)E(X)E(X), right? (since they are not independent)...

So I just tried to compute:

$$E(X^3) = int_0^{/infty} \frac{x^3}{\Gamma(3.5)2^{3.5}} x^{3.5-1} e^{-x/\beta} dx$$

But I'm having a problem, since $$\Gamma(3.5) = (3.5)!$$ and I don't know how to compute (3.5)!...

Thanks in advance

 

[Ray Vickson]

Let X be $$\Gamma(3.5,2)$$. Find $$E(X^3)$$.

We can't write E(X^3)=E(X)E(X)E(X), right? (since they are not independent)...

So I just tried to compute:

$$E(X^3) = int_0^{/infty} \frac{x^3}{\Gamma(3.5)2^{3.5}} x^{3.5-1} e^{-x/\beta} dx$$

But I'm having a problem, since $$\Gamma(3.5) = (3.5)!$$ and I don't know how to compute (3.5)!...

Thanks in advance

$x^3 x^{3.5 - 1} = x^{6.5 - 1},$ so after supplying the appropriate factors you essentially have $X^3 \Gamma(3.5,2) = c \Gamma(6.5,2)$ for some constant c.

 

[Artusartos]

$x^3 x^{3.5 - 1} = x^{6.5 - 1},$ so after supplying the appropriate factors you essentially have $X^3 \Gamma(3.5,2) = c \Gamma(6.5,2)$ for some constant c.

Thanks, but I don't understand how you got this equality...

 

[Ray Vickson]

Thanks, but I don't think I understand where this equality came from...$X^3 \Gamma(3.5,2) = c \Gamma(6.5,2)$

What I really meant (but it was harder to write) was: if f1(x) = probability density of Gamma(3.5,2) and f2(x) = probability density of Gamma(6.5,2), then we have x^3*f1(x) = c*f2(x) for some constant c. Now I will leave the rest up to you.

 

[Artusartos]

What I really meant (but it was harder to write) was: if f1(x) = probability density of Gamma(3.5,2) and f2(x) = probability density of Gamma(6.5,2), then we have x^3*f1(x) = c*f2(x) for some constant c. Now I will leave the rest up to you.

Ok, thanks... :)