limiting distribution and continuous transformation

Question

Assume that $\sqrt{n}(b-\beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $\beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $\sqrt{n}(e^{b}-e^\beta)$.

I know the conversation property of convergence for continuous transformations, thus if I can rewrite $\sqrt{n}(e^{b}-e^\beta)$ in terms of $\sqrt{n}(b-\beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $\sqrt{n}(b-\beta)$. Does anybody have any idea about how to proceed?

edit: I guess the delta method suits well into this case, I will proceed using the delta method.

Answer 1

Yes, delta method applies here.

$$\sqrt{n}(X_n - \theta) \xrightarrow{D} N(0, \sigma^2)$$

then we have

$$\sqrt{n}(g(X_n) - g(\theta)) \xrightarrow{D} N(0, \sigma^2[g'(\theta)]^2)$$

if $g'(\theta)$ exists and non-zero.

Here, $\sigma^2 = S, \theta = \beta$, $g(x)=\exp(x)=g'(x)$.

Hence

$$\sqrt{n}(e^{b_n} - e^\beta) \xrightarrow{D} N(0, S\exp(2\beta)).$$