Suppose $$y=e^{x}$$ where x is normal with mean mu and variance sigma. Then I see how to derive mode of f(y) (distribution of y), as we need to find the value y that makes $$f'(y)==0$$ However, why is mode not simply $$e^{\mu}$$?
y is a monotonic function of x, and so when x reaches its mode, then y should also reach its mode. The mode of x is its mean (mu) hence y's mode should be $$e^{\mu}$$ what mistake have I made?
Thanks
1 Answer
$X$ has lognormal distribution if $X=e^Z$ where Z has normal distribution, $Z \sim N(\mu, \sigma^2)$. However, the density of X is then given by:
$f(x)=\frac{1}{x\sqrt{2\pi \sigma^2}} e^{-\frac{1}{2 \sigma^2}\left(\ln(x)-\mu\right)^2}$
Differentiating the density with respect to $x$ we get
$-\frac{1}{x^2\sqrt{2\pi \sigma^2}} e^{-\frac{1}{2 \sigma^2}\left(\ln(x)-\mu\right)^2} - \frac{1}{x\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2 \sigma^2}\left(\ln(x)-\mu\right)^2} \frac{\ln(x)-\mu}{\sigma^2} \frac{1}{x} $
The mode is the value of x that maximizes the density. Thus, equating the above derivative to zero and simplifying, we get
$-1-\frac{\ln(x)-\mu}{\sigma^2}=0$
or
$x= e^{(\mu-\sigma^2)}$
To sum up, your mistake is that you have not used the correct density but the equation for the transformation of variables that gives us the random variable with log-normal distribution.
