The Relationship Between Moments and Cumulants

The latest assignment in my biomedical signal processing class, followed by my modest attempt at the proof:

Show that
$$\begin{eqnarray*}   c_{1} &=& m_{1}=\mu \\   c_{2} &=& m_{2} - m_{1}^{2}=\sigma^{2}. \end{eqnarray*}$$
That is, show that the first and second cumulants of a distribution are equal to the mean ($\mu$) and variance ($\sigma^{2}$), respectively.

Proof:
The $n$th moment of a continuous distribution, $m_{n}$ is defined as:
$$\begin{eqnarray*}   m_{n} &=& E[x^{n}]\\   &=& \int_{-\infty}^{\infty}x^{n}p\left(x\right)\,dx. \end{eqnarray*}$$
A function that generates any moment is called a moment-generating function and is defined as  $$E[e^{tX}].$$  where $X$ is a random variable and $E[f\left(X\right)]$ is the expected value of a function, $f$, of a random variable $X$.

If we expand the definition of $E[e^{tX}]$ we can see that
$$\begin{eqnarray*}   E[e^{tX}] &=&   \int_{-\infty}^{\infty}p\left(X\right)\left(\sum_{n=0}^{\infty}\frac{t^{n}X^{n}}{n!}\right)\,\,\mathrm{d}x\\   &=& \int_{-\infty}^{\infty}p\left(X\right)\left(1 + tX + \frac{t^{2}X^{2}}{2!} + \cdots +     \frac{t^{n}X^{n}}{n!}\right)\,\,\mathrm{d}x\\   &=& 1 + tm_{1} + \frac{t^{2}m_{2}}{2!} + \cdots + \frac{t^{n}m_{n}}{n!}. \end{eqnarray*}$$

If we evaluate the $n$th derivative of $E[e^{tX}]$ with respect to $t$ at $t=0$, that is, $$\left(\frac{\mathrm{d}^{n}}{\mathrm{d}t^{n}}E[e^{tX}]\right)_{t=0} = \frac{\mathrm{d}^{n}}{\mathrm{d}t^{n}}\left(m_{0} + tm_{1} +   \frac{t^{2}m_{2}}{2!} + \cdots + \frac{t^{n}m_{n}}{n!}\right)$$ we get the $n$th moment of a probability distribution.

The $n$th cumulant, $c_{n}$, of a continuous distribution is defined as:
$$\ln\left(E[e^{tX}]\right).$$
Since the power series definition of the natural logarithm of $x$ is
$$\ln\left(x\right) = \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}\left(x - 1\right)^{n},$$
we can rewrite the definition of the $n$th cumulant, $c_{n}$, as
$$\begin{eqnarray*}   \ln\left(E[e^{tX}]\right) &=&   \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}\left(tm_{1} +     \frac{t^{2}m_{2}}{2!} + \cdots\right)^{n} \end{eqnarray*}$$
Expanding the terms from the previous equation we see that,
$$\begin{eqnarray*}   \ln\left(E[e^{tX}]\right) &=& \left(tm_{1} +     \frac{t^{2}m_{2}}{2!} + \cdots\right) - \frac{1}{2}\left(tm_{1} +     \frac{t^{2}m_{2}}{2!} + \cdots\right)^{2} + \cdots\\ &=&\left(tm_{1} +     \frac{t^{2}m_{2}}{2!} + \cdots\right)-\frac{1}{2}\left(t^{2}m_{1}^{2}+\frac{t^{3}m_{1}m_{2}}{2}+\cdots\right) + \cdots. \end{eqnarray*}$$
Now do a little algebra on the second term of $\ln\left(E[e^{tX}]\right)$ to get
$$\left(tm_{1}+\frac{t^{2}m_{2}}{2!}+\cdots\right)-\frac{t^{2}m_{1}^{2}}{2}-\frac{t^{3}m_{1}m_{2}}{4}+\cdots$$

The first derivative, with respect to $t$, evaluated at 0 gives $m_{1}$. The second derivative evaluated at 0 gives

$$\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\left(\frac{t^{2}m_{2}}{2!}\right) =m_{2}.$$
Then we see that $m_{2} - m_{1}^{2}$ gives the variance.

Thus it is shown that the first two cumulants of a distribution are the mean and variance, respectively.