My questions come from a pdf file and I have no idea how to put the symbols on here, so I took a screenshot and attached my questions as a picture. Can somebody please, please help me out with numbers 4 and 5? I've been trying but I never know what to do with proofs.

2. Originally Posted by babbagandu
My questions come from a pdf file and I have no idea how to put the symbols on here, so I took a screenshot and attached my questions as a picture. Can somebody please, please help me out with numbers 4 and 5? I've been trying but I never know what to do with proofs.
Help make you cry?
if you insist
This is not a real proof, it's a calculus calculation
And you can find this online too.
By definition

$\displaystyle E(e^{Xt})={1\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}e^{xt} x^{\alpha-1}e^{-x/\beta}dx$

NOW combine the exponents of e in the integrand and figure out your NEW beta and you are done.

3. Originally Posted by matheagle
Help make you cry?
if you insist
This is not a real proof, it's a calculus calculation
And you can find this online too.
Oh ok I think I see now. Do you think you could help me out on # 5? I really appreciate your help

Oh and I am still a little stuck on 4. What do you mean by calculate the new beta?

4. since densities integtrate to one....

$\displaystyle E(e^{Xt})={1\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}e^{xt} x^{\alpha-1}e^{-x/\beta}dx$

$\displaystyle ={1\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}x^{\al pha-1}e^{-x(-t+1/\beta)}dx$

this is infinite unless $\displaystyle -t+1/\beta >0$

NOW let $\displaystyle 1/\beta_*=-t+1/\beta$

$\displaystyle ={\beta_*^{\alpha}\over \Gamma(\alpha)\beta_*^{\alpha}\beta^{\alpha}}\int_ 0^{\infty}x^{\alpha-1}e^{-x(-t+1/\beta)}dx$

$\displaystyle = {\beta_*^{\alpha}\over \beta^{\alpha}} \int_0^{\infty}{1\over \Gamma(\alpha)\beta_*^{\alpha}} x^{\alpha-1}e^{-x(-t+1/\beta)}dx$

$\displaystyle ={\beta_*^{\alpha}\over \beta^{\alpha}}=\Biggl({\beta_*\over \beta}\Biggr)^{\alpha}$

using $\displaystyle 1/\beta_*=-t+1/\beta$

we have $\displaystyle \beta/\beta_*=-t\beta +1$

or $\displaystyle \beta_*/\beta={1\over 1-t\beta}$

raise that to the $\displaystyle \alpha$ and you're done