Almost sure convergence of the sequence of partial averages of a random variable $T_n = X*\mathbb{1}_{X_n \geq X}$

conditional probabilityprobability theoryprobability-limit-theoremsrandom variables

Let $X_1,X_2,\dots$ be independent and identically distributed random variables according to the uniform distribution on the interval $[0, 100]$. Define the sequence of random variables $(T_n)_{n\in\mathbb{N}}$ by $T_n = X*\mathbb{1}_{X_n \geq X}$, where $X$ is a random constant chosen by the distribution of the r.v.s of the sequence $(X_n)_{n\in\mathbb{N}}$, making $X$ independent of all $X_n$ (and vice versa). The problem I'm trying to solve is that does the sequence of averages $S_n = \frac{1}{n}\sum_{i=1}^nT_i$ converge almost surely to a constant value? We immediately see that each $T_n$ is an integrable r.v., since each $T_n$ can be represented by a deterministic mapping of $X$ and $X_n$, $(X, X_n)\mapsto X*\mathbb{1}_{X_n\geq X}$, and $X \sim U([0, 100])$. However, the expectation of $T_n$ depends enterily upon the realized value of $X$, namely $\mathbb{E}(T_n\mid X = 0) = 0*\mathbb{P}(X_n \geq 0) = 0.$ and $\mathbb{E}(T_n\mid X = 100) = 100*\mathbb{P}(X_n \geq 100) = 100 * \frac{1}{100} = 1$. Since each $T_n$ is independent of one another by the independence of $X_ns$ and $X$, we have (to my knowledge) that sequence of partial averages of each of the conditioned r.v.s $L_n = T_n\mid X = a, a \in [0, 100]$ (I have no clue what is the accepted way of writing a conditioned r.v.) converge almost surely to the expectation of $L_n$. But then, what about the original sequence of averages of the unconditioned r.v. $T_n$?

Best Answer

Let $(X,X_1,X_2,...)$ be IID, $X\sim \mathcal{U}[0,m]$ $$S_n=\frac{1}{n}\sum_{k\leq n}X\mathbf{1}_{\{X_n\geq X\}}=X\cdot \frac{1}{n}\sum_{k\leq n}\mathbf{1}_{\{X_n\geq X\}}=XY_n$$ Conditional on $X$, we have that $(\mathbf{1}_{\{X_n\geq X\}})_{n \in \mathbb{N}}$ are IID with $E[\mathbf{1}_{\{X_n\geq X\}}|X]=P(X_1\geq X|X)$. Therefore (a.s.) $Y_n \to P(X_1\geq X|X)$ and $S_n \to XP(X_1\geq X|X)=X(1-X/m)$, so that $S_n$ converges a.s. to a random variable and not to a constant.

Related Question