The Stochastic Order of $e^{X_n}$

Question

Suppose a sequence of random variables $X_n = o_p(r_n)$ where $r_n \rightarrow 0$. Then is $Y_n = e^{X_n} - 1 = o_p(r_n)$? This seems like it should be true using the expansion $$e^{z} = \sum^\infty_{k=0} \frac{z^k}{k!},$$ which gives $$Y_n = e^{X_n} - 1 = X_{n} + \sum^\infty_{k=2} \frac{X^k_n}{k!},$$ where the leading term $X_n$ is $o_p(r_n)$. However does not seem so straightforward to prove because of the infinite summation.

Answer 1

Let $f:x\mapsto \begin{cases} \frac{e^x-1}x &\text{if } x\neq 0 \\ 1 &\text{if } x=0 \end{cases}$ so that $f$ is continuous.

Since $X_n=o_P(r_n)$ and $r_n\to 0$, we have $X_n\xrightarrow{P}0$ and by the continuous mapping theorem, $f(X_n)\xrightarrow{P}1$, thus $$e^{X_n}-1 = X_n+o_P(X_n)=o_P(r_n).$$

Answer 2

A slightly more analytic way to think about it: when $x$ is nonnegative and small enough (say $0\le x\le 1$), we have $e^x\le 1+2x$. Thus for all $\epsilon>0$,

$$\mathsf{P}\left(e^{X_n}-1>\epsilon r_n\right)\le \mathsf{P}\left(X_n>1\right)+\mathsf{P}\left(2X_n>\epsilon r_n\right).$$

Since $X_n=o_P(r_n)$ and $r_n\to 0$, both terms go to zero as $n\to\infty$.