Let $\varepsilon_1, \dots, \varepsilon_n$ be independent random variables with $E(\varepsilon_i) = 0$. Let $f: [0,1] \to \mathbb{R}$ be a Lipschitz function with constant $K > 0$, i.e., $$|f(x) - f(y)| \leq K |x - y| \quad \forall x, y \in [0,1].$$ Define $$Y_i = f\!\left(\frac{i}{n}\right) + \varepsilon_i, \quad i = 1, \dots, n.$$ Consider the estimator $$\hat{\sigma}_n = \frac{1}{2(n-1)} \sum_{i=1}^{n-1} (Y_{i+1} - Y_i)^2.$$ Prove that $$\lim_{n \to \infty} E(\hat{\sigma}_n) = \operatorname{var}(\varepsilon_1), \qquad \lim_{n \to \infty} \operatorname{var}(\hat{\sigma}_n) = 0.$$ From this, deduce $$\lim_{n \to \infty} E\!\left[ \left( \hat{\sigma}_n - \operatorname{var}(\varepsilon_1) \right)^2 \right] = 0.$$ I am getting stuck with the part $\lim_{n \to \infty} \operatorname{var}(\hat{\sigma}_n) = 0$, hope anyone help with any ideas . I very appreciate that.
1 Answer
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2 Firstly, I think you need some stronger assumptions to prove the required results, i.e. the $\varepsilon_i$ being i.i.d. with finite fourth moment?
Now, suppose $A_n$ and $B_n$ are two sequences of random variables with $\lim_{n\to\infty}\mathrm{var}(A_n) = 0$, $\lim_{n\to\infty}\mathrm{var}(B_n) = 0$. Then it follows that also $\lim_{n\to\infty}\mathrm{var}(A_n + B_n) = 0$. Prove this! Then split $\hat\sigma_n$ into a sum of simpler sequences and show that each of those sequences has variance that vanishes in the limit.
- $\begingroup$ thanks for your answer. In this case, we don't have the condition relate finite fourth moment, so I think we have to prove it $\endgroup$problematic– problematic2025-11-02 00:57:20 +00:00Commented Nov 2 at 0:57
- $\begingroup$ @Wei Here the number of terms in the sum is not fixed.So the convergence of $\mathrm{var}((Y_{i+1}-Y_i)^2/(2(n-1)))$ is not sufficient. $\endgroup$Christophe Boilley– Christophe Boilley2025-11-02 10:04:12 +00:00Commented Nov 2 at 10:04