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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

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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 $n$.

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  • $\begingroup$ Thank you for your response, I understand why the $X_k$ are not independent, however when expressing $\sum_{k=1}^n X_k$ using $\sum_{k=1}^n \xi_k$, what I get is $\sum_{k=1}^n X_k$ = $\sum_{k=1}^n\xi_k -3\xi_2 + \xi_{n+2} - 2\sum_{k=3}^{n+1}\xi_k$, and I do not clearly see how to use this result. Furthermore I do not know how to use the facts that $S_n$ grow like $-n\mu$ and $\mathbb{Var}[S_n]$ $\lt$ $11\sigma^2n$ $\endgroup$ – fidel castro May 31 '18 at 20:35
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    $\begingroup$ @fidel castro: collapse the two sums. You should get $-S_n$ plus a few boundary terms ($2\xi_1$, $- \xi_2$, $-2\xi_{n+1}$, $\xi_{n+2}$). When you divide by $\sqrt{n}$, the boundary terms converge in distribution to $0$, and you can use the CLT on $S_n/\sqrt{n}$. $\endgroup$ – D. Thomine May 31 '18 at 21:41
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    $\begingroup$ Also, the remark about the growth in $-n\mu$ and $Var(S_n)$ was absolute hand-waving. Don't focus on them for now; maybe their signification will be clearer once you manage to solve the exercise. $\endgroup$ – D. Thomine May 31 '18 at 21:44
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    $\begingroup$ Yes, that's almost it. Just beware of the CLT: you need to subtract the expectation before taking the limit. The CLT says that $(\sum_1^n \xi_k-\mu n)/\sqrt{n}$ converges in distribution to $N(0, \sigma^2)$, not that $(\sum_1^n \xi_k)/\sqrt{n}$ converges to $N(\mu, \sigma^2)$. $\endgroup$ – D. Thomine Jun 3 '18 at 20:45
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    $\begingroup$ Yes, that's it. There are a couple of issues of redaction, but you can probably get it right when writing the argument down. $\endgroup$ – D. Thomine Jun 3 '18 at 21:25

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