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I have a PDF:

$$f_y(y) = \frac{\lambda^n}{\Gamma(n)} (y-n\tau)^{n-1}e^{-\lambda(y-n\tau)}$$

I want to find the moment generating function for it: (I believe I made a mistake somewhere?)

$$\begin{aligned} M(t) = E[e^{Yt}] &= \int_0^\infty e^{yt} \frac{\lambda^n}{\Gamma(n)} (y-n\tau)^{n-1}e^{-\lambda(y-n\tau)} \; dy \\ &= \frac{\lambda^n e^{\lambda n \tau}}{\Gamma(n)} \int_0^\infty e^{-y(\lambda-t)(y-n\tau)^{n-1} \; dy} \\ & \text{... let } u = y(\lambda-t) \\ &= \frac{\lambda^n e^{\lambda n \tau}}{\Gamma(n)} \int_0^\infty e^{-u} (\frac{u}{\lambda-t}-n\tau)^{n-1} \frac{1}{\lambda-t} \, du \end{aligned}$$

Usually after the substitution I would get something that I know how to integrate like a PDF of some distribution or in this case perhaps a gamma function? But in this case it looks complex? Maybe I made a wrong substitution?

Since the PDF is that of a gamma distribution, I expected the MGF of a gamma distribution which I belive should be $$\frac{\lambda}{\lambda-t}n$$ in this case?

UPDATE

Ok probably 1 of my mistakes is after changing variables, my limits of integration should change ... but I am not sure how to proceed after:

after change of variables

$$\begin{aligned} &= \frac{\lambda^n e^{\lambda n \tau}}{\Gamma(n)} \int_{n\tau (\lambda - t)}^\infty e^{-u} (\frac{u}{\lambda-t} - n\tau)^{n-1} \frac{1}{\lambda-t} \, du \\ &= \frac{\lambda^n e^{\lambda n \tau}}{\Gamma(n) (\lambda - t)^n} \int_{n\tau (\lambda - t)}^\infty e^{-u} (u-n\tau (\lambda-t))^{n-1} \, du \end{aligned}$$

The integral looks like a gamma function? But how do I handle $u-n\tau (\lambda-t)$? I got a feeling it have something to do with the limits of integration, but I am not sure what ...

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1 Answer 1

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You wanted to shift the gamma distribution, but you didn't shift it. Your integral should start from $n\tau$.

\begin{aligned} M(t) = \int_{n\tau}^\infty e^{yt} \frac{\lambda^n}{\Gamma(n)} (y-n\tau)^{n-1}e^{-\lambda(y-n\tau)} \; dy. \end{aligned} Now apply the change of variables $u = y-n\tau$ to get \begin{aligned} M(t) = e^{n\tau t}\int_{0}^\infty e^{ut} \frac{\lambda^n}{\Gamma(n)} u^{n-1}e^{-\lambda u} \; du. = e^{n\tau t}\times\mbox{MGF of unshifted Gamma}(t) \end{aligned} Meanwhile, we have also shown that the MGF of the shifted PDF is just $e^{t\,\mathtt{SHIFTAMOUNT}}$ times the unshifted MGF, and all this calculation was actually not necessary :)

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  • $\begingroup$ Hmm, sorry but I am wondering even if I have $\int_{n\tau}^\infty$ how do I integrate that complex expression? $\endgroup$
    – Jiew Meng
    Nov 6, 2013 at 5:42
  • $\begingroup$ If you start from $n\tau$, it will be straightforward to integrate (after a change of variables) - You won't get that complicated integral. $\endgroup$
    – Lord Soth
    Nov 6, 2013 at 5:43
  • $\begingroup$ ok I updated the OP with what I have got after your help (many thanks btw!), but I am still alittle stuck ... $\endgroup$
    – Jiew Meng
    Nov 6, 2013 at 6:47
  • $\begingroup$ @JiewMeng Edited. $\endgroup$
    – Lord Soth
    Nov 6, 2013 at 7:19

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