Let ($\xi_{k}$)$_{k \in \mathbb{N}}$ be a sequence of i.i.d random variables with expectation $\mu$ and variance $\sigma^2$ $\in (0, \infty)$. We define $X_k$ = $\xi_k$ - $3\xi_{k+1}$ + $\xi_{k+2}$, $S_n$ = $\sum_{k=1}^n X_k$, $k,n \in \mathbb{N}$
Compute for $x \in \mathbb{R}$ the limit
$\lim_{n \to \infty}$ $\mathbb{P}$$(\frac{S_n - n\mathbb{E}[X_1]}{\sqrt{n\mathbb{Var}[X_1]}}$ $\le$ $x$).
My thoughts on this:
First note that $\mathbb{E}[X_1]$ = -$\mu$,$\mathbb{Var}$[$X_1$] = $11\sigma^2$
Then with the central limit theorem, it holds
$\frac{S_{n}-n\mu}{\sigma\sqrt{n}}$ $\overset d \longrightarrow$ ${N}(0,1)$ $\Rightarrow$ $S_{n} - n\mu$ $\overset d \longrightarrow$ $N(0,n\sigma^2)$
$\Rightarrow$ $S_n$ $\overset d \longrightarrow$ $N(n\mu,n\sigma^2)$
It then follows that $S_n +n\mu$ $\overset d \longrightarrow$ $N(2n\mu,n\sigma^2)$ $\Rightarrow$ $\frac{S_n +n\mu}{n}$ $\overset d \longrightarrow$ $N(2\mu,\frac{\sigma^2}{n})$
I would now like to reach $\frac{S_n+n\mu}{\sqrt{11n\sigma^2}}$ $\overset d \longrightarrow$ $N(2\mu,\frac{1}{11})$, as I feel this is the correct result, but I am unable to get to the correct steps.
I am grateful for any tip and suggestion
Your intuition, in this case, is off the target. But that's a very interesting mistake you made (in a good sense!).
The problem is that the version of the CLT you want to use is valid for sequences of independent random variables. And the $X_k$ are not independent. For instance, if $\xi_3$ is large and positive, then $X_1$ and $X_3$ will be large and positive, but $X_2$ will be very large and negative. Then, when you sum them, a compensation occurs.
On average, the $X_k$ have expectation $-\mu$, so $S_n$ grow like $-\mu n$. However, these compensations make the variance of the sum smaller; that is, $Var(S_n) < 11 \sigma^2 n$.
If you want to get the correct result, try to express $\sum_{k=1}^n X_k$ using $\sum_{k=1}^n \xi_k$. There are some boundary effects in the sum, but the asymptotic behaviour should be clearly visible.
In addition, be more cautious in the way you divide by $n$ or $\sqrt{n}$. The CLT tell you that $(S_n-n\mu)/(\sigma \sqrt{n}) \to N(0,1)$, but the expression $S_n \to N(n \mu, n \sigma^2)$ has no meaning. It's as if I were saying that the sequence $(n+\sqrt{n})$ converge to $n$.