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the problem in question is: Let $\{X_n\mid n\ge 1\}$ and $Y$ be random variables on probability space $(\Omega, \mathcal F, \mathbb P)$. Suppose that the $X_n$ are independent, $E[X_n] = 0$ and $E[X_n^2] = 1$ for all $n$, and $Y$ is square-integrable. Show that $E[X_n Y] \to 0$ as $n\to \infty$.

My attempt at a solution:

First define $Z_n = E[X_n Y]X_n$. The $Z_n$ are independent, and $E[Z_n^2] = E^2[X_nY]E[X_n^2] = E^2[X_nY]$. By Cauchy Schwarz, this last probability is bounded above by $E[Y^2] = M <\infty$. In particular, the $Z_n$ are uncorrelated and have uniformly bounded second moment, and $E[Z_n] = 0$, so $E[\sum_{i=1}^n Z_n] = 0$. We can apply a SLLN theorem and get $$\frac{\sum_{i=1}^n Z_i}{n} \to 0$$ almost surely. I'm not sure where to go from here or if all this is exactly correct... the prof left a hint to consider $\sum_{i=1}^n E[X_i Y]X_i$

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  • $\begingroup$ You need to include some additional conditions/check the existing ones, because $E(X_nY)\to 0$ simply doesn't have to hold. $\endgroup$ – A.S. Nov 3 '15 at 18:47
  • $\begingroup$ @A.S. Ah, sorry I forgot to include that the $X_n$ are independent - obviously that's a big assumption to add. $\endgroup$ – user62748 Nov 3 '15 at 19:02
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It suffices to show that if $\left(n_j\right)_{j\geqslant 1}$ is an increasing sequence of real numbers such that the sequence $\left(\mathbb E\left[X_{n_j}Y\right]\right)_{j\geqslant 1}$ converges to some real number $l$, then $l=0$. Indeed, the sequence $\left(\mathbb E\left[X_{n}Y\right]\right)_{n\geqslant 1}$ is bounded, hence it is possible to extract a convergent subsequence. By what we assumed, the potential limit is $0$. Applying this reasoning to each subsequence, we can deduce that the sequence $\left(\mathbb E\left[X_{n}Y\right]\right)_{n\geqslant 1}$ converges to $0$.

Now, consider an increasing sequence of real numbers $\left(n_j\right)_{j\geqslant 1}$ and define $c_j:=\mathbb E\left[X_{n_j}Y\right]$.

Here are some hints to show the wanted claim.

  • The sequence $\left(1/n\sum_{j=1}^nc_j\right)_{n\geqslant 1}$ converges to $0$.

  • If $c_j\to l$, then $\left(1/n\sum_{j=1}^nc_j\right)_{n\geqslant 1}$ converges to $l$.

  • Conclude.
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