Find the characteristic function for all n.

Question

Let $(Y_n)_{n\in \mathbb{N}}$ be independent and identically distributed random variables. Let $A_n=\sum_{p=1}^nY_p\quad \forall n\in \mathbb{N}$.

Find $\varphi_{S_n/n}$ for all $n\in \mathbb{N}$ expressed in terms of $\varphi_{Y_1}$.

I've tried the following;

We have $Y_p=Y_1$ for all $p\in \mathbb{N}$ since the variables are iid. Thus we get $\varphi_{S_n/n}=\varphi_{n\cdot Y_1 /n}=\varphi_{Y_1}$ for all $n\in \mathbb{N}$

This seems very simple, and I feel like i missed something?

Answer 1

We have $Y_p=Y_1$ for all $p\in \mathbb{N}$ since the variables are iid.

This is not true. For example, if $Y_1$ takes the values $0$ and $1$ with probability $1/2$, then the event $\left\{ Y_2=Y_1\right\}$ has probability $1/2$.

A worse situation is when $Y_1$ is uniformly distributed on the unit interval (or more generally, when the distribution of $Y_1$ has a density): the event $\left\{ Y_2=Y_1\right\}$ has probability $0$.

What is true is that $Y_1$ and $Y_p$ have the same characteristic function.

Call $\varphi_n$ the characteristic function of $S_n$. Since $Y_{n+1}$ is independent of $S_{n-1}$, one gets $$\varphi_{n+1} (t)=\varphi_{n} (t) \varphi_{Y_{n+1}}(t) $$
and since $ X_{n+1}$ has the same distribution as $X_1$, we derive that $$\varphi_{n+1} (t)=\varphi_{n} (t) \varphi_{Y_{1}}(t) $$
hence $$\varphi_{n} (t)=\left(\varphi_{Y_{1}}(t) \right)^n $$ hence $$\varphi_{S_n/n} (t)=\left(\varphi_{Y_{1}}\left(t/n\right) \right)^n.$$

Answer 2

\begin{align} \varphi_{S_n/n}(t) &= E[\exp(it \frac{S_n}{n})] \\ &= E[\prod_{k = 1}^n \exp(itY_k)^{1/n}] \\ &= \prod_{k = 1}^n E[\exp(itY_k)^{1/n}] \tag{$(Y_n)_n$ independent} \\ &= \prod_{k = 1}^n E[\exp(itY_1)^{1/n}] \tag{$(Y_n)_n$ identically distributed} \\ &= \prod_{k = 1}^n E[\exp(it\frac{Y_1}{n})] \\ &= \varphi_{Y_1}(\frac{t}{n})^n \end{align}