# Let $Y_n := \frac{X_1\xi_1 + \cdots + X_n \xi_n}{\xi_1 + \cdots + \xi_n}$. Is $\operatorname{var} Y_n \to 0$ as $n \to \infty$?

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 $$\operatorname{var} Y_n \to 0 \quad \text{as} \quad n \to \infty$$ ?

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.