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Let $x_n\overset{p}{\to}c$ and $x_n\overset{d}{\to}N(0,\sigma^2)$ denote convergence in probability to a constant $c$ and convergence in distribution to a random normal variable (with some abuse of notation), respectively. Suppose that $\frac{1}{n}\sum_{i=1}^{n}y_i \overset{p}{\to} c$ and $x_n\overset{d}{\to}N(0,\sigma^2)$ as $n\to\infty$. What can be said about the convergence of the sequence of random variables $\frac{1}{n}\sum_{i=1}^nx_iy_i$?

This somehow confuses me. It is reminiscent of Slutsky's theorem, but Slutsky cannot be directly applied.

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  • $\begingroup$ Consider the case $y_n=1$ for all $n\ge 1$ and $x_n = (-1)^{n+1}Z$ where $Z \sim N(0,\sigma^2)$. Then $\frac{1}{n} \sum_{i=1}^n x_iy_i = Z \cdot 1(n\text{ is odd})$, which does not converge. $\endgroup$ Commented Nov 19 at 14:37
  • $\begingroup$ @raj: it is true that $\sum_{i=1}^n x_iy_i = Z \cdot 1(n\text{ is odd})$ but then dividing by $n$ gives a convergence to $0$. $\endgroup$ Commented Nov 19 at 14:59
  • $\begingroup$ @DavideGiraudo Yep my bad, thanks for the correction. Nice answer $\endgroup$ Commented Nov 19 at 15:39

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It is possible that $n^{-1}\sum_{i=1}^nx_iy_i$ does not converge in distribution. Let $(n_u)_{u\geqslant 1}$ be an increasing sequence of integers that will be specified later, let $Z$ be a random variable having standard normal distribution and let $y_i=1$. For $n_{2\ell}\leqslant i<n_{2\ell+1}$, $y_i:=(-1)^i Z$ and for $n_{2\ell+1}\leqslant i<n_{2\ell+2}$, let $y_i:=Z$.

Consequently, $$ \left\lvert \frac 1{n_{2\ell+1}}\sum_{i=1}^{n_{2\ell+1}}x_iy_i\right\rvert=\left\lvert \frac 1{n_{2\ell+1}}\sum_{k=1}^{\ell-1}(n_{2k+2}-n_{2k+1})Z\right\rvert \leqslant \ell \frac{n_{2\ell}}{n_{2\ell+1}}\lvert Z\rvert $$ and $$ \frac 1{n_{2\ell+2}}\sum_{i=1}^{n_{2\ell+2}}x_iy_i=\frac 1{n_{2\ell+2}}\sum_{k=1}^{\ell}(n_{2k+2}-n_{2k+1})Z =\frac{n_{2\ell+2}-n_{2\ell+1}}{n_{2\ell+2}}Z+\frac 1{n_{2\ell+2}}\sum_{k=1}^{\ell-1}(n_{2k+2}-n_{2k+1})Z. $$ If we assume that $\ell n_{2\ell}/n_{2\ell+1}\to 0$, then the subsequence $\left(\frac 1{n_{2\ell+1}}\sum_{i=1}^{n_{2\ell+1}}x_iy_i\right)$ converges to $0$ and if we furthermore assume that $n_{2\ell+1}/n_{2\ell+2}\to 0$, then the subsequence $\left(\frac 1{n_{2\ell+2}}\sum_{i=1}^{n_{2\ell+2}}x_iy_i\right)$ converges to $Z$ in distribution.

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