How to prove that $Y=\ln(X)$ approximately Normal when $X$ is a Normal random variable with $\mu\gg\sigma$

Question

I wanted to prove that PDF of $Y=\ln(X)$ tends to a Normal distribution with $\mathcal{N}(\ln(\mu_{x}),\sigma^{2}_{y})$ when $X\sim\mathcal{N}(\mu_{x},\sigma^{2}_{x})$. It is also important to note that $\mu_{x}\gg\sigma_{x}$. Important is that $Y$ tends to a Normal distribution with mean $\mu_{y}\simeq\ln(\mu_{x})$. Thanks!

Answer 1

Informal version: when $\mu_x\gg \sigma_x$, the logarithm is close to the linear function $t\mapsto \ln \mu_x+t/\mu_x$ on the interval $[\mu_x-5\sigma_x,\mu_x+5\sigma_x]$ where pretty much all of $X$ lives.

More formally: the pdf of $Y$ is $g(y)=e^y f(e^y) $ where $f$ is the pdf of $X$. Since $f(e^y)$ is extremely small outside of $$I= [\ln(\mu_x-5\sigma_x),\ln(\mu_x+5\sigma_x)]$$ we focus on $y$ in this interval. Using the Taylor approximation centered at $\ln \mu_x $ we obtain $$e^{y}= \mu_x + \mu_x(y-\ln \mu_x) +\theta(y)\, (y-\ln \mu_x)^2,\quad y\in I \tag2$$ where $|\theta(y)|\le \frac12 (\mu_x+5\sigma_x)$. Here $$|y-\ln \mu_x|\le \ln(1+5\sigma_x/\mu_x)\le 5\sigma_x/\mu_x$$ which is small. Thus, $$e^y f(e^y) = \mu_x\, f(\mu_x + \mu_x(y-\ln \mu_x))+O(y-\ln \mu_x) \tag3$$ where $O$ can be explicitly estimated as above. The function $\mu_x f(\mu_x + \mu_x(y-\ln \mu_x))$ describes the normal curve centered at $\ln\mu_x$ with variance $\sigma_x^2/\mu_x^2$.