Problem 6.7.2 of Resnick's A Probability Path says: Let $\{X_{n}\}$ be iid, with $EX_{n}=\mu$, $\mathrm{Var}(X_{n})=\sigma^2$. Set $\overline{X}=\sum_{i=1}^{n} X_{i}/n$. Show: $$ \frac{1}{n}\sum_{i=1}^{n}(X_{i}-\overline{X})^2\to \sigma^{2} $$ where the $\to$ means convergence in probability. There is no hypothesis on higher moments of $X_{n}$, so I believe that one cannot apply Chebyshev's theorem. Furthermore this precedes any discussion of characteristic functions or the strong law of large numbers. How does one get this result without such tools?

[This is probably the same question as this one : Convergence in probability of sample variance but the answers there apparently invoke the Strong Law which isn't proved until Chapter 7 of Resnick.]

  • $\begingroup$ See this question: math.stackexchange.com/q/243348 $\endgroup$ – saz Sep 13 '16 at 5:49
  • $\begingroup$ The answer to the question you cite also uses the strong law. The problem I'm asking about is posed at a point in the book before that's available; all you've got are generalities about the relationship between various types of convergence, plus Chebyshev's inequality (at least as far as I can figure out). $\endgroup$ – Jeremy Teitelbaum Sep 13 '16 at 10:45

Tools: Borel-Cantelli lemma, Tschebysheff inequality, dominated convergence theorem

The first step is to show the weak law of large numbers.

Theorem 1 Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of independent identically distributed random variables and suppose that $\mathbb{E}(|Y_1|)<\infty$. Then $$S_n := \frac{1}{n} \sum_{i=1}^n Y_i$$ converges in probability to $\mu := \mathbb{E}(Y_1)$ as $n \to \infty$.

For the proof we use the following auxiliary statement.

Lemma 1 Let $(Y_n)_{n \in \mathbb{N}}$ and $(Z_n)_{n \in \mathbb{N}}$ be two sequences of random variables. If $\sum_{n \geq 1} \mathbb{P}(Y_n \neq Z_n)<\infty$, then $\sum_{n \geq 1} (Y_n-Z_n)$ converges almost surely.

Proof of Lemma 1: It follows from the Borel Cantelli lemma that $\mathbb{P}(Z_n \neq Y_n$ infinitely often$)=0$, and therefore there exists a null set $N$ such that for any $\omega \notin N$, we have $Z_n(\omega) = Y_n(\omega)$ for $n$ sufficiently large. Obviously, this implies the (almost sure) convergence of the series.

Proof of Theorem 1: Recall that $\mathbb{E}(|Y_1|)$ implies that $\sum_{n \geq 1} \mathbb{P}(|Y_1|>n)<\infty$. If we define truncated random variables $Z_n := Y_n 1_{\{|Y_n| \leq n\}}$, then $$\sum_{n \geq 1} \mathbb{P}(Z_n \neq Y_n) = \sum_{n \geq 1} \mathbb{P}(|Y_n|>n) = \sum_{n \geq 1} \mathbb{P}(|Y_1|>n)<\infty.$$ Applying Lemma 1, we find that $S_n = n^{-1} \sum_{i=1}^n Y_i$ converges in probability to $\mu$ if (, and only if,) $T_n := n^{-1} \sum_{i=1}^n Z_i$ converges in probability to $\mu$. To show that $T_n \stackrel{\mathbb{P}}{\to} \mu$, we first calculate the variance of $T_n$. Since the random variables $Z_n$, $n \in \mathbb{N}$, are independent and bounded, we have

$$\text{var}(T_n) = \frac{1}{n^2} \sum_{i=1}^n \text{var}(Z_i) \leq \frac{1}{n^2} \sum_{i=1}^n \mathbb{E}(Z_i^2) \tag{1} $$

where we have used that

$$\text{var}(Z_i) = \mathbb{E}(Z_i^2)-[\mathbb{E}(Z_i)]^2 \leq \mathbb{E}(Z_i^2).$$

Now choose some sequence $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{N}$ such that $a_n \to \infty$ and $a_n/n \to 0$ as $n \to \infty$. Using that $|Z_i| \leq i$, we find

$$\begin{align*} \sum_{i=1}^n \mathbb{E}(Z_i^2) &= \sum_{i=1}^{a_n} \mathbb{E}(Z_i^2) + \sum_{i=a_{n}+1}^n \mathbb{E}(Z_i^2) \\ &\leq a_n \sum_{i=1}^{a_n} \mathbb{E}(|Z_i|) + \sum_{i=a_n+1}^n \mathbb{E}(Z_i^2 1_{|Z_i| \leq a_n}) + \sum_{i=a_n+1}^n \mathbb{E}(Z_i^2 1_{|Z_i| > a_n}) \\ &\leq a_n \sum_{i=1}^{a_n} \mathbb{E}(|Z_i|) + a_n \sum_{i=a_n+1}^n\mathbb{E}(|Z_i|) + n \sum_{i=a_n+1}^n \mathbb{E}(|Z_i| 1_{|Z_i|>a_n})\\ &\leq a_n \sum_{i=1}^n \mathbb{E}(|Y_i|) + n \sum_{i=a_{n}+1}^n \mathbb{E}(|Y_i| 1_{|Y_i|>a_n}) \\ &\leq a_n n \mathbb{E}(|Y_1|) +n^2 \mathbb{E}(|Y_1| 1_{|Y_1|>a_n}). \end{align*}$$

Since $Y_1 \in L^1$, $a_n/n \to 0$ and $a_n \to \infty$, the dominated convergence theorem gives

$$\lim_{n \to \infty} \frac{1}{n^2} \text{var}(T_n) \stackrel{(1)}{\leq} \lim_{n \to \infty} \frac{1}{n^2} \sum_{i=1}^n \mathbb{E}(Z_i^2) =0.$$

Applying Tschebysheff's inequality, it follows easily that $T_n$ converges in probability to $\mu$. This finishes the proof.

Theorem 2: Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of independent identically distributed random variables with $\mathbb{E}(X_n) = \mu$ and $\text{var}(X_n) = \sigma^2<\infty$. Set $\bar{X}_n = n^{-1} \sum_{i=1}^n X_i$ and $S_n := n^{-1} \sum_{i=1}^n (X_i-\bar{X}_n)^2$. Then $S_n \to \sigma^2$ in probability.

Proof: Since $$X_i-\mu+(\mu-\bar{X}))^2 = (X_i-\mu)^2 + 2 (X_i-\mu) \cdot (\mu-\bar{X})+(\mu-\bar{X})^2 $$

we have $$\begin{align} S_n= \frac{1}{n} \sum_{i=1}^n (X_i-\bar{X})^2 &= \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 + 2 (\mu-\bar{X}) \underbrace{\frac{1}{n} \cdot \sum_{i=1}^n (X_i-\mu)}_{\left(\frac{1}{n} \sum_{i=1}^n X_i\right)-\mu =(\bar{X}-\mu)} + (\mu-\bar{X})^2 \\ &= \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 - (\bar{X}-\mu)^2. \end{align}$$

By the weak law of large numbers, Theorem 1, we have

$$\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i \to \mu \quad \text{in probability} \\ \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 \to \mathbb{E} \left( (X_i-\mu)^2 \right)=\sigma^2 \quad \text{in probability}. $$

Hence $S_n \to \sigma^2$ in probability.

Remark: The idea for this proof is taken from Chung's book A Course in Probability Theory.

  • $\begingroup$ The version of the Weak Law (Theorem 1) you used is actually proved in Section 7.2 of Resnick (it's Khintchin's version). As you show it gives convergence of the mean without a hypothesis on the variance. With that result, the original question is much easier (as you show). This does still leave me wondering whether Resnick expected me to figure Khintchin's result out on my own. Considering the original question is problem 2 in the Chapter 6 exercises, I expected an easier proof! $\endgroup$ – Jeremy Teitelbaum Sep 13 '16 at 18:27
  • $\begingroup$ @JeremyTeitelbaum Well, I agree with you. However, in my experience (as a student) there are plenty of exericses in books which are either plainly wrong or much more complicated to prove than the author intended (... perhaps because he didn't write the proof up and missed some subtlety). Let's see whether someone else knows an easier proof... $\endgroup$ – saz Sep 13 '16 at 18:37
  • $\begingroup$ I strongly suspect that Resnick goofed here, and should either have added a hypothesis on the 4th moment so you can use Chebyshev's theorem, or else moved the exercise to Chapter 7! $\endgroup$ – Jeremy Teitelbaum Sep 13 '16 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.