# Why use the exponential in moment-generating function?

When trying to understand the moment-generating function, I've stumbled upon this general function:

Mx(t) = E[e^tx], t ∈ R

I understand that you have X = (X1, ... Xn) a series of random variables and that you draw a vector through tX.

It may sounds naive, but why use the exponential of that series? Why not ln for instance? After all, both are monotonous, continuous, positive on R? (and are also closely connected, though ln is much slower than e).

Thank you very much for your insights!

Jordane

• For any sequence $a_0, a_1, a_2, \ldots$, the (ordinary) generating function is $g(x) = a_0 + a_1 x + a_2 x^2 + \cdots = \sum_n a_n x^n$. The idea is that this single function contains the complete information about the whole sequence, since the coefficient of $x^n$ is exactly the $n$-th term of the sequence. However, given a generating function in closed form, it's not always easy to find this coefficient for a general $n$. – M. Vinay Mar 22 at 17:13
• On the other hand, we can also define the exponential generating function of the sequence as $f(x) = \sum_n a_n \frac{x^n}{n!}$. Again, the $n$-th term is exactly the coefficient of $\frac{x^n}{n!}$ in this series. If we know the function in closed form, then one way to obtain this coefficient is to differentiate the function $n$ times and set $x = 0$. That is, $a_n = f^{(n)}(0)$ [you can verify this easily; or else recall the form of Maclaurin's series]. – M. Vinay Mar 22 at 17:14
• $$M_X(t) = \mathbb E[e^{tX}] = \mathbb E\left[\sum_n \frac{t^n}{n!} X^n\right] = \sum_n \mathbb E[X^n] \frac{x^n}{n!}$$ is the exponential generating function of the series of $n$-th (raw) moments $\mathbb E[X^n]$ of the random variable $X$. – M. Vinay Mar 22 at 17:16
• Thank you so much M. Vinay this makes total sense now! – Jordane Bdss Mar 25 at 19:31
• Glad to help. But I see I've made a stupid typo there. It should be $$M_X(t) = \sum_n \mathbb E[X^n] \dfrac{t^n}{n!}$$ (not $x^n$). – M. Vinay Mar 26 at 1:10