3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Clarinetist Jan 21 '16 at 0:11
5
+50
$\begingroup$

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 $t=0$.
$\endgroup$
  • $\begingroup$ 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 $|x e^{x(t_0+h)}|$? $\endgroup$ – angryavian Jan 22 '16 at 6:47
  • $\begingroup$ @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 $x<0$. $\endgroup$ – saz Jan 22 '16 at 7:23
  • $\begingroup$ I don't think (1) is correct: the series should be $\sum_{k=0}^\infty \frac{(hx)^k}{(k+1)!}$. $\endgroup$ – angryavian Jan 25 '16 at 19:49
  • $\begingroup$ @angryavian That's exactly what's written there ... or what am I missing? $\endgroup$ – saz Jan 25 '16 at 19:51
  • $\begingroup$ 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! $\endgroup$ – angryavian Jan 25 '16 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.