Find $\lim\limits_{n \to \infty} n^{\frac 3 4} \Bbb E \left [\left (Y_n - \Bbb E \left [Y_n \right ] \right )^2 \right ].$

Question

Let $X_i,\ i \geq 1$ be i.i.d. discrete random variables with mean $\mu$ and variance $\sigma^2.$ Let $k \gt 1.$ Define the sequence \begin{align*} Y_n : & = \dfrac {X_1 X_2 \cdots X_k + X_2 X_3 \cdots X_{k+1} + \cdots + X_{n-k+1} X_{n-k+2} \cdots X_n} {n}. \end{align*} Find $\lim\limits_{n \to \infty} n^{\frac 3 4} \Bbb E \left [\left (Y_n - \Bbb E \left [Y_n \right ] \right )^2 \right ].$

How do I find that? Is there any easy way to proceed?

Thanks in advance.

EDIT $:$ What I get is that $$ \Bbb {E}\ \left [ {Y_n}^2 \right ] = \dfrac {1} {n^2} \left [ \left (n - k + 1 \right ) \left ({\sigma}^2 + {\mu}^2 \right )^k + 2 \left [\sum\limits_{r = 0}^{k-2} \left (n-k-r \right ) \left (\mu^2 \right )^{r+1} \left ({\sigma}^2 + {\mu}^2 \right )^{k-r-1} + \dfrac {\left (n-2k+1 \right ) \left (n-2k+2 \right )} {2} \left ({\mu}^2 \right )^k \right ] \right ].$$ But then $$\begin{align*} \lim\limits_{n \to \infty} n^{\frac {3} {4}} \Bbb {Var}\ [Y_n] & = \lim\limits_{n \to \infty} \left [ \dfrac {\left (n-k+1 \right ) \left (n-k+2 \right)} {n^{\frac {5} {4}}} - \dfrac {\left (n-k+1\right)^2} {n^{\frac {5} {4}}} \right ] {\mu}^{2k} \\ & = \lim\limits_{n \to \infty} \dfrac {\left (n - k + 1 \right )} {n^{\frac {5} {4}}} {\mu}^{2k} = 0.\end{align*}$$ Am I right? Can anybody please check my computation whether it is correct or not.

Thanks in advance.

Source $:$ This question appeared in ISI PhD entrance test in Mathematics held in $20$th September this year (TEST CODE : MTB) in the afternoon session (Question No. $9$).

Answer 1

Let $Z_i≔X_i⋯X_{i+k-1}$.

As upper bounds we use:

$$\mathbb{V}(Z_i)=\mathbb{V}\left(X_i\cdots X_{i+k-1}\right)=\mathbb{E}\left(X_i^2\cdots X_{i+k-1}^2\right)-\mathbb{E}\left(X_i\cdots X_{i+k-1}\right)^2\le2{\max{\left(\max_{i\in\left[n\right]}{\left(X_i^2\right)},\max_{i\in\left[n\right]}{\left(\left|X_i\right|\right)}\right)}}^k=:v $$ and $$ |\mathbb{E}\left(Z_iZ_j\right)-\mathbb{E}\left(Z_i\right)\mathbb{E}\left(Z_j\right)| \le2{\max{\left(\max_{i\in\left[n\right]}{\left(X_i^2\right)},\max_{i\in\left[n\right]}{\left(\left|X_i\right|\right)}\right)}}^{2k}=:e $$

Using these we can now estimate the variance of $Y_n$:

$$\begin{align} \mathbb{V}\left(Y_n\right)&=\mathbb{V}\left(\frac{X_1X_2\cdots X_k+X_2X_3\cdots X_{k+1}+\cdots+X_{n-k+1}X_{n-k+2}\cdots X_n}{n}\right)\\&=\frac{1}{n^2}\mathbb{V}\left(Z_1+\ldots+Z_{n-k+1}\right)\\&=\frac{1}{n^2}\mathbb{V}\left(\sum_{i=1}^{n-k+1}Z_i\right)\\&=\frac{1}{n^2}\left(\sum_{i=1}^{n-k+1}\mathbb{V}\left(Z_i\right)\right)+\frac{2}{n^2}\left(\sum_{\begin{matrix}i,j=1\\i\neq j\\\end{matrix}}^{n-k+1}Cov\left(Z_i,Z_j\right)\right)\\&=\frac{1}{n^2}\left(\sum_{i=1}^{n-k+1}\mathbb{V}\left(Z_i\right)\right)+\frac{2}{n^2}\left(\sum_{\begin{matrix}i,j=1\\i\neq j\\\end{matrix}}^{n-k+1}{\mathbb{E}\left(Z_iZ_j\right)-\mathbb{E}\left(Z_i\right)\mathbb{E}\left(Z_j\right)}\right)\\&=\frac{1}{n^2}\left(\sum_{i=1}^{n-k+1}\mathbb{V}\left(Z_i\right)\right)+\frac{2}{n^2}\left(\sum_{\begin{matrix}i,j=1\\i\neq j\\\left|i-j\right|\le k\\\end{matrix}}^{n-k+1}{\mathbb{E}\left(Z_iZ_j\right)-\mathbb{E}\left(Z_i\right)\mathbb{E}\left(Z_j\right)}\right) \\&\le\frac{1}{n^2}\left(\sum_{i=1}^{n-k+1}{\ v}\right)+\frac{2}{n^2}\left(\sum_{\begin{matrix}i,j=1\\i\neq j\\\left|i-j\right|\le k\\\end{matrix}}^{n-k+1} e\right)\\&=\frac{1}{n^2}\left(\left(n-k+1\right)v\right)+\frac{2}{n^2}\left(\left(n-k+1\right)ke\right) \end{align}$$

Using this estimate, we can then conclude that the limit converges to $0$.

Answer 2

It is nice to start with a small problem and then generalize it, so I start with the variance for $k=1$.

First we have that $var(x_jx_{j+1})=\sigma^4$ and $cov(x_jx_{j+1})=\mu^2\sigma^2$ hence $$var(Y_n)=\frac1{n^2}\left((n-1)\sigma^4+2(n-2)\sigma^2\mu^2\right)$$ implying that for this simple case it holds that $\lim\limits_{n \to \infty} n^{\frac {3} {4}} \text {Var}\ [Y_n] = 0$. For $k=2$ we obtain $$var(Y_n)=\frac1{n^2}\left((n-2)\sigma^6+2(n-2)\sigma^2\mu^2(\sigma^2+2\mu^2)+2(n-3)\mu^4\sigma^2\right)$$ for which we have again that $\lim\limits_{n \to \infty} n^{\frac {3} {4}} \text {Var}\ [Y_n] = 0$. These too cases indicate that as long as $k$ is fixed and $n \to \infty$ the limit you are after will be zero.