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Let $X, X_1, X_2,\ldots$ be an i.i.d. sequence of random variables with finite variance. Then $$ \operatorname{var} \left ( \frac{X_1 +\cdots + X_n}{n} \right ) = \frac{\operatorname{var} X}{n} \to 0 \quad \text{as} \quad n \to \infty. $$

Let $\xi,\xi_1, \xi_2,\ldots$ be another i.i.d. sequence of positive bounded random variables. Assume that $\xi,\xi_1, \xi_2,\ldots$ is independent of $X, X_1, X_2,\ldots$. Consider the weighted sum $$ Y_n := \frac{X_1\xi_1 + \cdots + X_n \xi_n}{\xi_1 + \cdots + \xi_n}. $$

Can we show that $$ \operatorname{var} Y_n \to 0 \quad \text{as} \quad n \to \infty $$ ?

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1 Answer 1

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By the law of total variance,

\begin{align*} \mathbf{Var}(Y_n) &= \mathbf{E}[\mathbf{Var}(Y_n\mid\xi_1,\xi_2,\ldots)] + \mathbf{Var}(\mathbf{E}[Y_n\mid\xi_1,\xi_2,\ldots]) \\[0.25em] &= \mathbf{E}\biggl[ \frac{\xi_1^2\mathbf{Var}(X_1) + \cdots + \xi_n^2\mathbf{Var}(X_n)}{\xi_1^2 + \cdots + \xi_n^2} \biggr] + \mathbf{Var}\biggl( \frac{\xi_1\mathbf{E}[X_1] + \cdots + \xi_n\mathbf{E}[X_n]}{\xi_1 + \cdots + \xi_n} \biggr). \\ &= \mathbf{Var}(X) \mathbf{E}\biggl[ \frac{\xi_1^2 + \cdots + \xi_n^2}{(\xi_1 + \cdots + \xi_n)^n} \biggr]. \end{align*}

Now we note that the random variable

$$ Z_n = \frac{\xi_1^2 + \cdots + \xi_n^2}{(\xi_1 + \cdots + \xi_n)^n} $$

is bounded by $1$ and $n Z_n \to \mathbf{E}[\xi^2]/\mathbf{E}[\xi]^2$ a.s. by the strong law of large numbers. So, $Z_n \to 0$ and therefore

$$ \mathbf{Var}(Y_n) = \mathbf{Var}(X)\mathbf{E}[Z_n] \to 0$$

by the dominated convergence theorem.

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