# Taylor expansion for sequences

Let $$\theta$$, $$\{\theta_n\}_{n=1}^{\infty}$$ and $$\{X_n\}_{n=1}^{\infty}$$ be a real number, a sequence of real numbers, and a sequence of random variables, respectively, satisfying that $$\theta_n\rightarrow \theta$$ and $$X_n\overset{p}{\longrightarrow}\theta$$. Let $$f(.)$$ be a differentiable function. By Taylor expansion, we have

$$f(\theta_n) = f(\theta) + f'(\theta)(\theta_n-\theta) + o(|\theta_n-\theta|).$$

However, is the following generalization also correct?

$$f(X_n) = f(\theta_n) + f'(\theta_n)(X_n-\theta_n) + o_p(\|X_n-\theta_n\|)$$

I am assuming that the generalization is rather $$f(X_n) = f(\theta) + f'(\theta)(X_n - \theta) + o_p(X_n - \theta).$$ In that case, notice that by assumption, there exists a function $$g$$ such that $$f(\theta_n) = f(\theta) + f'(\theta)(\theta_n - \theta) + (\theta_n - \theta)g(\theta_n - \theta)$$ with $$\lim_{x \to 0} g(x) = 0$$. Applying this to $$X_n$$, we get $$f(X_n) = f(\theta) + f'(\theta)(X_n - \theta) + (X_n - \theta)g(X_n - \theta)$$ and the proof will be complete if we are able to prove that $$g(X_n - \theta) \to 0$$ in probability. Now fix $$\epsilon >0$$, then there exists $$\delta >0$$ such that $$|g(x)| < \epsilon$$ for $$|x| < \delta$$. So $$\mathbb{P}(|g(X_n - \theta)| \geq \epsilon) \leq \mathbb{P}(|X_n - \theta| \geq \delta) \to 0$$ since $$X_n \to \delta$$ in probability. The inequality is due to the inclusion $$\{|g(X_n - \theta)| \geq \epsilon\} \subset \{|X_n - \theta| \geq \delta\}$$.
• If we subtract the the second equation from the first one in your comment, then we have $f(X_n) - f(\theta_n) = f'(\theta)(X_n-\theta_n) + o_p(X_n-\theta) + o(\theta_n-\theta)$ So is the following true? $o_p(X_n-\theta) + o(\theta_n-\theta) = o_p(X_n-\theta_n)$ If it is, then we have the one in the question, except that $f'(\theta)$ replaces $f'(\theta_n)$.. – Hercules Dec 30 '18 at 9:26