Characteristic function of a standardized sum

Question

Let ${X_n}$ be an iid binary random process with equal probability of $+1$ or $-1$ occurring at any time n.Now,if $Y_n$ is the standardized sum and equal to $\frac{1}{\sqrt{n}} \sum ^{n-1}_{k=0}X_{k}$,then please show that its characteristic function $M_{Y_{n}}(ju)=e^{nlogcos(\frac{u}{\sqrt{ n}})}$

There is my formula:

$f(k;p)=$ \begin{cases} 1/2, & \text{if $k=1$} \\ 1/2, & \text{if $k=-1$} \end{cases} so $f(x) = (\frac{1}{2})^{k}(\frac{1}{2})^{(1-k)}=\frac{1}{2}$

The characteristic function is $M_Y(t)= \sum^{\infty}_{-\infty}e^{jty} \times f(y)= \sum^{\infty}_{-\infty}e^{jty} \times \frac{1}{\sqrt{n}} \sum ^{n-1}_{k=0}X_{k}= \sum^{\infty}_{-\infty}e^{jty} \times \frac{1}{\sqrt{n}} \frac{1}{2} \times n=\sum^{\infty}_{-\infty}e^{jty} \sqrt{n}\frac{1}{2}=\sqrt{n}\frac{1}{2}\sum^{\infty}_{-\infty}e^{jty}$

now i stuck here, i don't know how do i calculate the answer($e^{nlogcos(\frac{u}{\sqrt{ n}})}$) from here.Can anyone told me my mistake?

Answer 1

The most important part of the hypothesis is independence. $Ee^{itY_n}=Ee^{i(t/\sqrt n) \sum X_i} =(Ee^{(it/\sqrt n) X_1} )^{n}$ by independence. Now $Ee^{(it/\sqrt n) X_1}=\frac {e^{(it/\sqrt n)}+e^{(-it/\sqrt n)}} 2 =\cos ( t/\sqrt n)$. Hence the characteristic function is $(\cos ( t/\sqrt n))^{n}$. Thus is same as $e^{n \log \cos(t/ \sqrt n})$.

Answer 2

Let $U_{k,n}=\frac{X_k}{\sqrt{n}}$. The char. function of $U_{k,n}$ is $cos(\frac{t}{\sqrt{n}})$. The char. function of $Y_n$ is $cos(\frac{t}{\sqrt{n}})^n$. Your expression $e^{nlogcos(\frac{u}{\sqrt{n}})}$ is identical (with t instead of u).

I don't know where you got the formula for char. function that you used. By definition it is $E(e^{itX})$ for a random variable X.