Friday, 7 March 2014

Inversion of probability and moment generating functions

For discrete random variables working with the p.g.f. is easy. We have

Theorem 12.2The distribution of $X$ is uniquely determined by the p.g.f. $p(z)$. 

The proof is trivial because the distribution of $X$ is contained in coefficients of the power series $p(z)=p_0+p_1z+p_2z^2+\cdots$. It is also easy to tell if any given power series is a p.g.f. All we need is check is that the coefficients are nonnegative and sum to 1.

However, things are not so easy for moment generating functions. Given a power series $m(\theta)=1+m_1\theta  + m_2\frac{1}{2}\theta^2 +\cdots$
it is not easy to tell if there exists a r.v. that has moments $m_1,m_2,\dotsc$.
Also we had without proof the theorem analogous to Theorem 12.2,

Theorem 21.4 The moment generating function determines the distribution of $X$, provided $m(\theta)$ is finite for all $\theta$ in some interval containing the origin.

To prove Theorem 21.4 we would need to show how to recover the p.d.f. from the m.g.f.
In fact, it is easier to use the characteristic function. Let's take a small look at that.

Today we saw that the moment generating function of $N(0,1)$ is $m(\theta)=e^{\theta^2/2}$. This looks a lot like the p.d.f., which is $f(x)=e^{-x^2/2}/\sqrt{2\pi}$.

The characteristic function (c.f.) $\chi$ is obtained by replacing $\theta$ with $it$, $t$ real,

$\chi(t) = \int_{-\infty}^\infty f(x)e^{itx} dx$. This is $\chi(t)=e^{-t^2/2}$ for $N(0,1)$.

We can compute nearly the same integral of $\chi$ to recover $f$:

$\int_{-\infty}^\infty \chi(t)e^{-itx} dt=2\pi e^{-x^2/2}=2\pi f(x)$.

In general, given the c.f. we can recover the p.d.f. using the inversion formula

$f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty \chi(t)e^{-itx} dt$.

The m.g.f. exists only if all moments are finite. However, the c.f. exists even when moments do not exist, since $e^{itX}$ is just some point on a unit circle in the complex plane, and hence nothing is going off to infinity. The Cauchy distribution is an example of a distribution that has no m.g.f, but does have a c.f., namely $e^{-|t|}$.