Mode of lognormal distribution

Question

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

Answer 1

$X$ 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.

Answer 2

The mode of a given random variable is a quantity that depends on the distribution of $X$, not only $X$, which is a measureable function from the sample space $\Omega$ to $\mathbb R$. For example, the quantity $\sup(X)=\sup_{\omega \in \Omega}X(w)$ depends only on $X$, so the method in the OP works, that is, $$\sup \left (e^X \right)=e^{\sup(X)}.$$

Thus, to check when the method given in the OP works for the mode, we need to focus on the original definition of mode, which is a distribution-dependent concept.

---

For discrete random variables, the method in the OP works because we have $$\mathbb P(X=m) \ge \mathbb P(X=x), x\in S_X \Leftrightarrow \mathbb P(e^X=e^m) \ge \mathbb P(e^X=e^x), e^x\in S_{e^X}.$$

In other words, $$\color{blue}{p_{e^X}(t)}=\mathbb P(e^X=t) =\mathbb P(X=\log t)\color{blue}{=p_{X}(\log t)}, t\in S_X;$$

hence, $p_{e^X}(t)$ preserves the order based on which the mode of $X$ is obtained, and if the point $m$ maximizes $p_{X}(x)$, $e^m$ maximizes $p_{e^X}(t).$

---

For continuous random variables, the method does not work because we cannot prove the following

$$ f_X(m) \ge f_X(x), x\in S_X \Leftrightarrow f_{e^X}(e^m) \ge f_{e^X}(e^x), e^x\in S_{e^X}.$$

The reason is that $f_{e^X}(t)$ is not $f_{X}(\log t)$, and is given by

$$\color{blue}{f_{e^X}(t)=\frac{1}{t} \times f_{X}(\log t)}, t>0$$

which does not preserve the order because of the term $\frac{1}{t}$, which is a decreasing function in $t$. Hence, if the point $m$ maximizes $f_{X}(x)$, $e^m$ does not necessarily maximize $f_{e^X}(t).$