Show that $\frac{d}{dt}\Phi_{X_{1}}(0)=a$, then $\Phi_{X_{1}}(t)=1+(ia+o(1))t$ as $t\longrightarrow 0$.

Question

I am working on this exercise

Let $S_{n}:=\sum_{k=1}^{n}X_{k}$ where the i.i.d random variables $\{X_{k}\}$ have each the characteristic function $\Phi_{X}(\cdot).$ Prove that if $\frac{d}{d\theta}\Phi_{X}(0)=z\in\mathbb{C}$, then $z=ia$ for some $a\in\mathbb{R}$ and $S_{n}/n\longrightarrow_{p}a$ as $n\longrightarrow\infty.$

I found a solution of a similar question that directly assumes $\frac{d}{dt}\Phi_{X_{1}}(0)=ia,$ but I don't really understand it computation of finding $\Phi_{X_{1}}(t)$.

It argues that:

since $\frac{d}{dt}\Phi_{X_{1}}(0)=ia$, then $\Phi_{X_{1}}(t)=1+(ia+o(1))t$ as $t\longrightarrow 0$.

How did the solution get this?

I undertand that by hypothesis we have $$\lim_{t\longrightarrow 0}\dfrac{\Phi(t)-\Phi(0)}{\Phi(t)}=\lim_{t\longrightarrow 0}\dfrac{\Phi(t)-1}{\Phi(t)}=ia,$$ but does this imply $$\Phi(t)=\dfrac{1}{1-ia},\ \text{as}\ t\longrightarrow 0?$$

Also, how could I show that if $\frac{d}{dt}\Phi_{X_{1}}(0)=b+ia\in\mathbb{C}$, then $b=0$?

Thank you!

Edit 1:

Okay, I figure out my second question, how to show $z=ia$. The proof turns out really cute.

If $\phi_{0}(t)$ and $\phi_{1}(t)$ are the real and imaginary parts of $\Phi_{X_{1}}(t)$, we then can write $$\Phi_{X_{1}}(t)=\phi_{0}(t)+i\phi_{1}(t),$$ where $$\phi_{0}(t)=\int_{-\infty}^{\infty}\cos txdF(x)\ \text{and}\ \phi_{1}(t)=\int_{-\infty}^{\infty}\sin txdF(x).$$

Observe that $\phi_{0}(t)$ is an even function of $t$ and $\phi_{1}(t)$ is an odd function of $t$. Therefore, a derivative $\phi_{0}(t)$ of odd order which exists at $t=0$ must be zero there, and the same it true of an even derivative of $\phi_{1}(t)$.

It then follows immediately that in this exercise we must have $z=ia$ for some $a\in\mathbb{R}$, since we are taking the first order and by hypothesis the first order derivative exists at $0$.

I am still confused about why the quoted argument is true.

Edit 2:

Okay, I figure it out. It uses Taylor series around $0$, we know that $\Phi(0)=1$, and thus using Taylor series we have $$\Phi(t)=1+iat+t\mathcal{O}(t),$$ it could be $\mathcal{O}(t^{2})$, but we can factor out a $t$ and the remaining term is much smaller than $t$ as $t\longrightarrow 0$.

I will leave the post open for a while in case there are any other opinions.

Answer 1

I answer my post here to give a complete solution of the exercise:

Firstly, if $\phi_{0}(t)$ and $\phi_{1}(t)$ are the real and imaginary parts of $\Phi_{X_{1}}(t)$, we then can write $$\Phi_{X_{1}}(t)=\phi_{0}(t)+i\phi_{1}(t),$$ where $$\phi_{0}(t)=\int_{-\infty}^{\infty}\cos txdF(x)\ \text{and}\ \phi_{1}(t)=\int_{-\infty}^{\infty}\sin txdF(x).$$

Observe that $\phi_{0}(t)$ is an even function of $t$ and $\phi_{1}(t)$ is an odd function of $t$. Therefore, a derivative $\phi_{0}(t)$ of odd order which exists at $t=0$ must be zero there, and the same it true of an even derivative of $\phi_{1}(t)$.

It then follows immediately that in this exercise we must have $z=ia$ for some $a\in\mathbb{R}$, since we are taking the first order and by hypothesis the first order derivative exists at $0$.

Now, if we denote characteristic function of $X_{i}$ by $\phi$, then the characteristic function of $\frac{1}{n}\sum_{i=1}^{n}X_{i}$ is given by $\psi_{n}(t)=[\phi(\frac{t}{n})]^{n}$. By hypothesis we know that $\phi(t)$ is differentiable at $t=0$ with a derivative equal to $ia$. Therefore, by Taylor expansion, we have $$\phi\Big(\dfrac{t}{n}\Big)=1+\dfrac{iat}{n}+o\Big(\dfrac{1}{n}\Big).$$

Whenever $na_{n}\longrightarrow z$, follows that $(1+a_{n})^{n}\longrightarrow e^{z}$.

Therefore, we have $$\lim_{n\longrightarrow\infty}\psi_{n}(t)=\exp(iat),$$ where the limit is the characteristic function of a pointmass at $a$, and is continuous at $0$. Thus it follows from the continuity theorem that $S_{n}/n\Rightarrow a$.

Now, let $\epsilon>0$, since $S_{n}/n\Rightarrow a$, then by Portmanteau Theorem with the closed set $K=\mathbb{R}\setminus(a-\epsilon, a+\epsilon)$ we have $$\limsup_{n\longrightarrow\infty}\mathbb{P}(|S_{n}/n-a|\geq\epsilon)=\limsup_{n\longrightarrow\infty}\mathbb{P}(S_{n}/n\in K)\leq\mathbb{P}(a\in K)=0.$$

Thus, $S_{n}/n\longrightarrow_{p} a$ as desired.