# convergence in probability of function of random variables

Suppose that $X_1, X_2, \ldots, X_n$ be a sequence of i.i.d random variables. If we have $E(|X_1|^k) <\infty$ for some $k>0$ and $f(x)$ is a bounded continuous function on $\mathbb{R}$. Is the following theorem true? $$\frac 1n\sum_{i=1}^n f(X_i)X_i^k \overset{\mathbb{P}}{\longrightarrow} \mu$$

where $\mu = E\left[f(X_1)X_1^k\right]$. Thanks for any comment or suggestion.

• I guess you mean $\mu=\mathbb{E}(f(X_1) X_1^k)$?
– saz
May 30 '14 at 6:19
• @ saz: Yes, it's what I mean. Thank you, I corrected. May 30 '14 at 6:43
• You are summing integrable i.i.d. random variables, so we can even expect a.s. convergence by SLLN. May 30 '14 at 6:56

Hint: If $(X_n)_{n \in \mathbb{N}}$ is a sequence of iid random variables and $g: \mathbb{R} \to \mathbb{R}$ a measurable function, then $(g(X_n))_{n \in \mathbb{N}}$ is a sequence of iid random variables. If $g(X_1) \in L^1$, then by the strong law of large numbers
$$\frac{1}{n} \sum_{i=1}^n g(X_i) \stackrel{\text{a.s.}}{\to} \mathbb{E}g(X_1).$$