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I am trying to find the moment generating function of a random variable $X$, which has probability density function given by

$$f_{X}\left( x\right) =\dfrac {\lambda ^{2}x} {e^{\lambda x}}$$ Where $x>0$ and $λ>0$.

The standard approach to doing this i believe is to follow the following algorithm The moment generating function (mgf) of a continuous rv X, which is not necessarily non negative, is defined as

$$M_{x}\left( t\right) =E\left[ e^{tX}\right] = \int _{-\infty }^{\infty }e^{tx}f_{X}\left( x\right) dx$$

Since $x>0$ for the given probability density function this is the same as Laplace transform but with $t > 0$.

If this integral does not converge, which i believe is the case for the given probability density function, then we switch to using Characteristic functions( Fourier transform). Given as

$$\phi_{x}\left( t\right) =E\left[ e^{itX}\right] = \int _{-\infty }^{\infty }e^{itx}f_{X}\left( x\right) dx$$

When this integral converges we substitute $t=0$ and viola we have our mgf. As per my readings this integral is always meant to converge, but i believe this does not converge for the given probability density function. This is what i managed to compute

$$\int _{-\infty }^{\infty }e^{x(it-\lambda)}\lambda ^{2}xdx = \lambda ^{2}\left[ \dfrac {x+1} {e^{x\left( \lambda-ti\right) }\left( it-\lambda \right) }\right] _{-\infty }^{\infty }$$

I was hoping some one could point out what am i doing wrong in calculating this integral. I used Integration by parts treating $i$, $t$ and $\lambda$ as constants.
If it does indeed not converge then if there are any more steps one could take to get the mgf ?

Thanks in advance.

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If you "substitute $t=0$" you aren't left with a function of $t$ anymore. You "believe" the original integral doesn't converge; have you explored the possibility the region of convergence might depend on $\lambda$ and $t$'s relative size? –  anon Jul 31 '12 at 21:15
    
@anon That is a valid point though i am unsure if more information is available about $t$ or $\lambda$. Although with that said $t$ is a variable introduced by the transform so there maybe ways to retrieve more information. I'll try doing that –  Hardy Jul 31 '12 at 21:18
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Two other questions to ask yourself: did you compute the antiderivative of $xe^{ax}$ correctly, and why is the lower bound $-\infty$ in your integrals? –  anon Jul 31 '12 at 21:20
    
@anon Thanks i 'll try recomputing as per the answers. –  Hardy Jul 31 '12 at 21:25
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2 Answers

up vote 3 down vote accepted

You did not quite write the density function correctly. It is zero for $x\lt 0$, and $\dots$. So you will be integrating from $0$ to $\infty$, and it will be quite doable by integration by parts.

If $f_X(x)$ is our density function, then indeed we want $$\int_{-\infty}^\infty e^{tx}f_X(x)\,dx.$$ This is equal to $\int_{-\infty}^0 (0)e^{tx}\,dx\,$ (which is $0$) plus $$\int_0^\infty \lambda^2 x e^{(t-\lambda)x}\,dx.$$

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The integral for this mgf $M_X(t)$ converges if $\text{Re}(t) < \lambda$: $$M_X(t) = \int_0^\infty \lambda^2 x e^{(t-\lambda) x}\ dx = \dfrac{\lambda^2}{(\lambda-t)^2}$$ You can use integration by parts.

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Thanks for your help too. Sorry I could only pick one correct answer. –  Hardy Jul 31 '12 at 21:28
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