Question

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of X.

Answer 1

`M_X(t)=\mathbb E(e^(Xt))`

`M_X(t)=\mathbb E\left(1+Xt+(t^2)/(2!)X^2+(t^3)/(3!)X^3+\cdots\right)`

`M_X(t)=\mathbb E(1)+t\mathbb E(X)+(t^2)/(2!)\mathbb E(X^2)+(t^3)/(3!)\mathbb E(X^3)+\cdots`

`M_X(t)=1+t+t^2+t^3+\cdots`

`M_X(t)=\displaystyle\sum_(k\ge0)t^k=\frac1{1-t}`

provided that
`|t|<1`.

Similarly,


`\varphi_X(t)=\mathbb E(e^(iXt))`

`\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots`

`\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)`

`\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)`

`\varphi_X(t)=(1+it)/(1+t^2)=\frac1{1-it}`

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute


`F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty(e^(itx)\varphi_X(-t)-e^(-itx)\varphi_X(t))/(it)\,\mathrm dt`

The integral can be rewritten as


`\displaystyle\int_0^\infty(2i\sin(tx)-2it\cos(tx))/(it(1+t^2))\,\mathrm dt`

so that


`F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty(\sin(tx)-t\cos(tx))/(t(1+t^2))\,\mathrm dt`

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to
`x`, which gives


`\displaystyle\mathcal L_s\left\{\int_0^\infty(\sin(tx)-t\cos(tx))/(t(1+t^2))\,\mathrm dt\right\}=\int_0^\infty(1-s)/((1+t^2)(s^2+t^2))\,\mathrm dt`

`=\displaystyle(\pi(1-s))/(2s(1+s))`

and taking the inverse transform returns


`F_X(x)=\frac12+\frac1\pi\left(\frac\pi2-\pi e^(-x)\right)=1-e^(-x)`

which describes an exponential distribution with parameter
`\lambda=1`.