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Let $X_n$ and $Y_n$ be random variables such that $X_n-Y_n\overset{p}{\longrightarrow}0$ as $n\rightarrow\infty$. Let $f(.)$ is a differentiable function. Is the following correct?

$f(X_n) = f(Y_n) + f'(Y_n)(X_n-Y_n) + o_p(X_n-Y_n)$ as $n\rightarrow\infty$

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To gain some intuition about the issues, let's consider a sequence of bivariate random variables $(X_n,Y_n)$ that have simple distributions and a function $f$ that is easy to calculate with. We want to give your supposition a hard time in the sense of contemplating functions $f$ for which the Taylor approximation $f(z+a) \approx f(z) + af^\prime(z)$ is particularly poor. If it can withstand such examination then we will proceed to developing a proof; otherwise, we will have constructed a counterexample.

There are many ways to perform such an investigation. For instance, we could examine functions $f$ that, although differentiable, are not continuously differentiable. But perhaps the simplest is to examine functions whose derivative can become arbitrarily large, and then let the sequence $(X_n,Y_n)$ approach points where the derivative "blows up."

Consider, then, the function

$$f:(0,\infty)\to \mathbb{R};\ f(x) = \frac{1}{x},$$

chosen to be easy to compute with but to have a derivative

$$f^\prime(x) = -\frac{1}{x^2}$$

that diverges as $x\to 0.$

Let $(X_n,Y_n)$ be a sequence of random discrete random variables converging towards each other and towards $0;$ specifically, suppose

$$\eqalign{ \Pr\left((X_n,Y_n) = \left(\frac{2}{n}, \frac{1}{n}\right)\right) &= \frac{1}{2}\\ \Pr\left((X_n,Y_n) = \left(\frac{3}{n}, \frac{2}{n}\right)\right) &= \frac{1}{2}. }$$

Equivalently, $nY_n-1$ has a Bernoulli$(1/2)$ distribution and $X_n = Y_n + 1/n.$ Their difference converges to zero in any sense you care to use: In particular, since $X_n-Y_n=1/n\to 0$ everywhere, $X_n-Y_n \overset{p}{\longrightarrow} 0.$

Let's compute the difference $f(X_n) - \left[ f(Y_n) + f^\prime(Y_n)(X_n-Y_n) \right].$ This is a quantity you hope will be growing smaller in some sense: you want to be able to assert it is "$o_p(X_n-Y_n).$"

There are two possible values, each with probability $1/2.$ I will show the computations with a table, one row for each possibility:

$$\begin{array}{cc|cccc} X_n & Y_n & f(X_n) & f(Y_n) & f^\prime(Y_n) & f(X_n) - \left[ f(Y_n) + f^\prime(Y_n)(X_n-Y_n) \right] \\ \hline \frac{2}{n} & \frac{1}{n} & \frac{n}{2} & n & -n^2 & \frac{n}{2} - [n - n^2\frac{1}{n}] = \frac{n}{2} \\ \frac{3}{n} & \frac{2}{n} & \frac{n}{3} & \frac{n}{2} & -\frac{n^2}{4} & \frac{n}{3} - [\frac{n}{2} - \frac{n^2}{4}\frac{1}{n}] = \frac{n}{12} \end{array}$$

Because the values $n/2$ and $n/12$ diverge everywhere, there is no possible definition of "$o_p$" that could make the statement true.


The insight provided by this simple example, and variations on it (in which $X_n$ and $Y_n$ may be independent or $f$ may be defined everywhere, for instance) is you cannot allow $X_n$ and $Y_n$ to have any appreciable probability of attaining values where the Taylor approximation becomes arbitrarily poor.

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