Expansion of Characteristic function

Question

I am having troubles in understanding these passages. I am following a Probability course and I took those notes but I did not quite understand the passages.

Be $X$ a random variable, real with $\mathbb{E}[\vert X\vert^k] < +\infty$ and $k \leq 0$. Then I have $$\mathbb{E}[e^{iuX}] = \sum_{j = 0}^{k-1} \frac{(iu)^k}{k!} \mathbb{E}[X^j] + \frac{(iu)^k}{k}\delta(u)$$

with $\delta(u) \leq 3\mathbb{E}[\vert X \vert^k]$ and $\lim_{u\to 0} \delta(u) = 0$.

Proof

$$e^{iuX} = \sum_{j = 0}^{k-1} \frac{(iu)^j}{j!}X^k + \frac{(iu)^k}{k!}X^k\left(\cos(u \mu_1 x) + i\sin(u \mu_2 x)\right)$$

Then he said something about the measures $\mu_1$ and $\mu_2$ like they are random, then continuing

$$= \sum_{j = 0}^{k-1} \frac{(iu)^j}{j!}X^k + \frac{(iu)^k}{k!}X^k + \frac{(iu)^k}{k!}X^k(\cos(u \mu_1 x-1) + i\sin(u \mu_2 x))$$

$$ = \sum_{j = 0}^{k} \frac{(iu)^j}{j!}X^k + \frac{(iu)^k}{k!} \delta_1(u)$$

Then continuing, we perform $\mathbb{E}$ on both sides, concluding by taking $\delta(u) = \mathbb{E}[\delta_1(u)]$.

As you can see, it's quite a mess.

So I'm asking: does anybody of you know what all this is about? Can someone please explain me this better? For example, where do the measures $\mu$ come from, and all those shady passages, which seems like wrong to me...

I realy don't understand, and unfortunately my professor is one of those funny subjects that do not replies to emails.

Thank you!

Answer 1

Quite chaotic, but I did something similar when I studied Measure Theory.

Theorem

Be $X$ a random variable with $P(\vert X\vert ^k) < +\infty$. Then $$\hat{\mu}_X(u) = P(e^{iuX}) \equiv \mathbb{E}[e^{iuX}] = \sum_{j = 0}^k \frac{(iu)^j}{j!} P(X^j) + \frac{(iu)^k}{k!}\delta(u)$$

with $\delta(u) \leq 3 P(\vert X\vert^k)$ and $\lim_{u\to 0} \delta(u) = 0$

Proof

$$e^{iuX} = \sum_{j = 0}^{k-1} \frac{(iu)^j}{j!}X^j + \frac{(iu)^h}{h!}X^h(\cos(\mu_1 x) + i\sin(\mu_2 x))$$

$$ = \sum_{j = 0}^k \frac{(iu)^j}{j!} X^j + \frac{(iu)^h}{h!}X^h(\cos(\mu_1 x) - 1 + i\sin(\mu_2 x))$$

Hence

$$\mathbb{E}[e^{iuX} = \sum_{j = 0}^k \mathbb{E}[X^j] + \underbrace{\frac{(iu)^h}{h!} \mathbb{E}\left[X^k(\cos(\mu_1 x) - 1 + i\sin(\mu_2 x))\right]}_{\delta(u)}$$

Now it's easy to see that $\lim_{u\to 0} \delta(u) = 0$ as well as it's easy to check that

$$X^k(\cos(\mu_1 x) - 1 + i\sin(\mu_2 x)) \leq 3 P(\vert X\vert^k)$$

Hope this could help!

Answer 2

1. The number $\mu_1$ is a number from the Taylor expansion of real part of $e^{iuX}$ with Lagrange remainder. The number $\mu_2$ is a number from the Taylor expansion of imaginary part of $e^{iuX}$ with Lagrange remainder. As $X$ is r.v. hence $\mu_1$ and $\mu_2$ are r.v. (in fact there's a little problem with measurability of $\mu_i$ but it's better to forget about it)

2. You wrote $$= \sum_{j = 0}^{k-1} \frac{(iu)^j}{j!}X^k + \frac{(iu)^k}{k!}X^k + \frac{(iu)^k}{k!}X^k(\cos(u \mu_1 x-1) + i\sin(u \mu_2 x)).$$ There's a misprint. After fixing it we get $$= \sum_{j = 0}^{k-1} \frac{(iu)^j}{j!}X^k + \frac{(iu)^k}{k!}X^k + \frac{(iu)^k}{k!}X^k(\cos(u \mu_1 x)-1 + i\sin(u \mu_2 x))$$ which is equal to $$= \sum_{j = 0}^{k} \frac{(iu)^j}{j!}X^k + \frac{u^k}{k!}X^k(\cos(u \mu_1 x)-1 + i\sin(u \mu_2 x))i^k.$$ Note that $$| (\cos(u \mu_1 x-1) + i\sin(u \mu_2 x))i^k | = | \cos(u \mu_1 x)-1 + i\sin(u \mu_2 x) | \le $$ $$ \le | \cos(u \mu_1 x) | + |-1|+ |i\sin(u \mu_2 x) | \le 1+1+1=3.$$