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Let $X = (X_j)_{j=1}^\infty$ be a stochastic process (not necessarily independent) such that:  $$\frac{1}{n}\sum_{j=1}^nX_j^2 \longrightarrow c, \quad (n \to \infty,\,\,almost \,\, sure) $$ Now suppose that for each $n= 1,2,3,...$, $B_n = (B_{jn})_{j = 1}^\infty$ are iid such that $E[B_{jn}^2]= 1/n$ and $E[B_{jn}]=0$. Furthermore, $X = (X_j)_{j=1}^\infty$ and $B_n = (B_{jn})_{j = 1}^\infty$ are independent. Define $$Y_n = \sum_{j=1}^n X_jB_{jn}$$

I want to analyze the convergence of $E[Y_n^2]$. Note that $$E[Y_n^2 | X]= E[Y_n^2 | X_1, X_2, ...]= E\left[ \left(\sum_{j=1}^n X_jB_{jn}\right)^2 \Bigg| X_1, X_2, ...\right ] = E\left[ \sum_{j=1}^n X_j^2B_{jn}^2 + \sum_{j\neq k } 2X_j X_k B_{jn}B_{kn} \,\,\Bigg| X_1, X_2, ...\right ]$$ Given that $B_n = (B_{jn})_{j = 1}^\infty$ is iid, we have that $E[B_{jn}B_{kn}]= 0$. Therefore:

$$E[Y_n^2 | X]=  \sum_{j=1}^n X_j^2 E\left[ B_{jn}^2 \,\,\Bigg| X_1, X_2, ...\right ] = \sum_{j=1}^n X_j^2 E\left[ B_{jn}^2 \right ] = \frac{1}{n} \sum_{j=1}^n X_j^2 $$

Thus, we can conclude that  $$E[Y_n^2 | X]= E[Y_n^2 | X_1, X_2, ...] = \frac{1}{n} \sum_{j=1}^n X_j^2 \longrightarrow c, \quad (n \to \infty,\,\,almost \,\, sure) $$

My question then is:

Is it possible to show that $E[Y_n^2 ]$ converges to $c$, as $n \to \infty$? If yes, How to show?

user346624
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1 Answers1

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The first thing to determine is whether $\mathbb E\left[Y_n^2\right]$ exists. If all the random variable $X_j$, have a finite moment of order two, then $\mathbb E\left[Y_n^2\right]$ is finite and expanding the square and taking the expectation gives that $$ \mathbb E\left[Y_n^2\right]=\frac 1n\sum_{j=1}^n\mathbb E\left[X_j^2\right] =\mathbb E\left[\frac 1n\sum_{j=1}^nX_j^2\right] $$ If we assume that the sequence $\left(n^{-1}\sum_{j=1}^nX_j^2\right)_{n\geqslant 1}$ is uniformly integrable, then $\mathbb E\left[Y_n^2\right]\to c$ .

Davide Giraudo
  • 181,608