In the proof of the delta method related with the convergence in distribution, I couldn't understand the statement below.

When $ \sqrt{n} (X_n - \mu) \rightarrow ^D N(0, \sigma^2 ) $ ,

\begin{equation} f(X_n) = f(\mu) + f'(\mu)(X_n -\mu ) + o_p (|X_n - \mu |), \end{equation}

where $a_n = o_p(b_n)$ means that $ \frac{a_n}{b_n} \rightarrow^p 0$ as $n \rightarrow \infty $.

I think I have to show this statement is true.

\begin{equation} g(X_n) \equiv \frac{f(X_n) - f(\mu)}{X_n -\mu} - f'(\mu) \rightarrow^p 0 \end{equation}

In some textbooks and pdf files, they define a new countuous function h(x) such that

$$ h(x) = \begin{cases} g(x) & \text{if x $\ne$ 0}\\ 0 & \text{if x $\eqcirc$ 0}\ \end{cases} $$

Then, by continuous mapping theorem, $ h(X_n) \rightarrow ^p 0$. So, they say $g(X_n) \rightarrow ^p 0$.

But, because $ h(X_n) \ne g(X_n)$ (at x=0 ), I couldn't convince myself that $ h(X_n) \rightarrow ^p 0$ => $g(X_n) \rightarrow ^p 0$ .

Is there no problem of insisting that $ h(X_n) \rightarrow ^p 0$ => $g(X_n) \rightarrow ^p 0$ ?

Thanks for reading my question in advance.

  • $\begingroup$ Note that for $h(x)$ to be continuous, surely you need $g(0)=0$. $\endgroup$ Sep 15, 2013 at 9:31
  • $\begingroup$ Thanks for replying./ I thought $lim_{x \rightarrow 0} g(x)=0$, so, h(x) is continuous at 0 by the contruction of h(x). But, I think g(0) is not defined. $\endgroup$
    – wikizero
    Sep 15, 2013 at 10:09
  • $\begingroup$ It would rather be $x \to a$ but what is $a$ ? $a= \mu$ ? $\endgroup$ Sep 15, 2013 at 10:43
  • $\begingroup$ wikizero, when you're talking to someone in particular, put the "@" symbol just before his/her pseudo and then he/she will be notified (for example probabilityislogic is notified if I do @probabilityislogic) $\endgroup$ Sep 15, 2013 at 10:45
  • $\begingroup$ @Stéphane Laurent Thanks for replying. I had a mistake and I didn't recognize it. $a=\mu$. I will correct it. $\endgroup$
    – wikizero
    Sep 15, 2013 at 11:28

1 Answer 1


First of all I think $a = \mu$ in the statement of your question. Moreover you have to use the fact that $X_n \overset{P}{\to} \mu$ (this fact can be derived from $\sqrt{n} (X_n - \mu) \overset{D}{\to} N(0, \sigma^2 )$ but I think you have derived it before).

The function $$g(x) \equiv \frac{f(x) - f(\mu)}{x -\mu} - f'(\mu)$$ is not defined at $x=\mu$ but it admits a continuous extension at $x=\mu$. To show that, one has to check that the function $$ h(x) = \begin{cases} g(x) & \text{if } x \ne \mu\\ 0 & \text{if } x = \mu\ \end{cases}$$ is continuous at $x=\mu$. This function $h$ is the continuous extension of $g$ at $x=\mu$. By the continuous mapping theorem and knowing that $X_n \overset{P}{\to} \mu$, one has then $h(X_n) \overset{P}{\to} h(\mu)=0$. But since $h$ is the continuous extension of $g$ at $x=\mu$, then this limit also holds for $g$, that is, one has $g(X_n) \overset{P}{\to} 0$. This latter fact is well known in elementary real analysis for classical limits, but it also holds true for limits in probability (proving it is a good elementary exercise about convergence in probability).


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