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.
Best Answer
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 $$\int_0^\infty \lambda^2 x e^{(t-\lambda)x}\,dx.$$