# Characteristic function and moment generating function: differentiating under the integral

In order to justify the interchange of the derivative and integral when differentiating a characteristic function, one can use the dominated convergence theorem:

$$\frac{d}{dt} \int e^{itx} P(dx) = \lim_{h \to 0} \frac{1}{h} \int (e^{ihx}-1) e^{itx} P(dx).$$

Since $|e^{ihx}-1| \le |hx|$, we have

$$\frac{1}{h} \int |e^{ihx}-1| P(dx) \le \int |x| P(dx),$$ so if we assume the random variable is in $L^1$, we may push the derivative under the integral. Similarly, if the random variable is in $L^k$, then we can push the $k$th derivative under the integral.

I am trying to find an analogous statement for moment generating functions, but I am having trouble generalizing the above argument. Under what conditions can we do this for MGFs? Any hints would be appreciated, but I would prefer an argument that uses dominated convergence rather than Leibniz's integral rule.

• I'm not a probability theory expert. However, the book I have (Athreya and Lahiri) says that this is justified since $M_{X}$ (MGF of $X$) "admits a power series expansion convergent in $|t| < \epsilon$, it is infinitely differentiable in $|t| < \epsilon$ and the derivatives of $M_{X}$ can be found by term-by-term differentiation of the power series" and then it cites Rudin's PMA, Ch. 9. I don't think this is going to be helpful, but this is what I found, for what it's worth. – Clarinetist Jan 21 '16 at 0:11

Denote by

$$M(t) := \int e^{tx} \, \mathbb{P}(dx)$$

the moment generating function of the measure $\mathbb{P}$.

Suppose that there exist $t_0 \in \mathbb{R}$ and $\epsilon>0$ such that $M(t)<\infty$ for all $t \in [t_0-\epsilon,t_0+\epsilon]$. Then

1. If $t_0>0$ and $\int_{(-\infty,0)} |x| \, \mathbb{P}(dx)<\infty$, then $M$ is differentiable at $t=t_0$.
2. If $t_0<0$ and $\int_{(0,\infty)} |x| \, \mathbb{P}(dx) <\infty$, then $M$ is differentiable at $t=t_0$.
3. If $t_0=0$, then $M$ is differentiable at $t = t_0 = 0$.

Proof:

1. Choose $\epsilon \in (0,1)$ sufficiently small such that $(t_0-\epsilon,t_0+\epsilon) \subseteq (0,\infty)$ and fix $h \in (-\epsilon/2,\epsilon/2)$. It follows from the mean value theorem that $$\left|\frac{e^{(t_0+h)x} -e^{t_0 x}}{h} \right| \leq |x| e^{\zeta x} \tag{1}$$ for some intermediate value $\zeta \in (t_0, t_0+h) \subseteq (0,\infty)$. If $x \geq 0$, we get $$\left|\frac{e^{(t_0+h)x} -e^{t_0 x}}{h} \right| \leq x e^{(t_0+h)x}.$$ Since $t_0+\epsilon>t_0+h>0$, we can choose $C>0$ (not depending on $h$, $x$) such that $$\left|\frac{e^{(t_0+h)x} -e^{t_0 x}}{h} \right| \leq C e^{(t_0+\epsilon)x} \tag{2}$$ for all $x \geq 0$. For $x \leq 0$, $(1)$ yields $$\left|\frac{e^{(t_0+h)x} -e^{t_0 x}}{h} \right| \leq |x|. \tag{3}$$ Combining $(2)$ and $(3)$, we get $$\left|\frac{e^{(t_0+h)x} -e^{t_0 x}}{h} \right| \leq w(x)$$ for $$w(x) := \begin{cases} C e^{x(t_0+\epsilon)}, & x \geq 0, \\ |x|, & x < 0 \end{cases}$$ Because of our assumptions, $w$ is an integrable dominating function. Applying the dominated convergence theorem proves the differentiability.
2. Apply statement 1 to the measure $\mathbb{Q}(B) := \mathbb{P}(-B)$.
3. Choose $h \in (0,\epsilon/2)$. Using $(1)$ for $t_0 = 0$, we get $$\left| \frac{e^{hx}-1}{h} \right| \leq |x| e^{\zeta x}$$ for some intermediate value $\zeta=\zeta(h) \in (0,h)$. Hence, $$\left| \frac{e^{hx}-1}{h} \right| \leq |x| (1_{\{x \leq 0\}} + e^{hx} 1_{\{x>0\}}).$$ Using that $|x| \leq C(e^{-\epsilon x}+e^{\epsilon x})$ for some constant $C$, we get $$\left| \frac{e^{hx}-1}{h} \right| \leq (C+1) e^{x \epsilon} + C e^{-x \epsilon}.$$ Consequently, we may again apply the dominated convergence theorem to interchange limit & integration. A very similar argumentation works for $h \in (-\epsilon/2,0)$. This gives the differentiability of $M$ at $t=0$.
• I think you mean $h \in (-\epsilon,\epsilon)$ rather than $h \in (t_0-\epsilon, t_0+\epsilon)$. Could you explain how you get the upper bound on $|x e^{x(t_0+h)}|$? – angryavian Jan 22 '16 at 6:47
• @angryavian You are right; actually it's even better to take $h \in (-\epsilon/2,\epsilon/2)$. If $h \in (-\epsilon/2,\epsilon/2)$, then $$x e^{x(h-\epsilon)} \leq x e^{-x \epsilon/2}$$ for all $x \geq 0$ and therefore $$x e^{x(t_0+h)} = x e^{x(h-\epsilon)} e^{x(t_0+\epsilon)} \leq C e^{x(t_0+\epsilon)}$$ for any $x \geq 0$; here $$C := \sup_{x \geq 0} (xe^{-x \epsilon/2})< \infty.$$ With a very similar reasoning, we also get a bound for $x<0$. – saz Jan 22 '16 at 7:23
• I don't think (1) is correct: the series should be $\sum_{k=0}^\infty \frac{(hx)^k}{(k+1)!}$. – angryavian Jan 25 '16 at 19:49
• @angryavian That's exactly what's written there ... or what am I missing? – saz Jan 25 '16 at 19:51
• The denominator is different. Going from the previous line to (1), you've shown that $\frac{e^{hx}-1}{h} = x e^{hx}$ which I don't think is true. Apologies if I'm overlooking something! – angryavian Jan 25 '16 at 19:57