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In Blitzsteins book "Intro to probability" he asks the following question in chapter 6:

Let $Y$ be Log-Normal with parameter $\mu$ and $\sigma^2$. So $Y=e^X$ with $X\sim N(\mu, \sigma^2)$. Evaluate and explain whether or not each of the following arguments is correct

(a) ...

(b) Student B: "The mode of $Y$ is $e^\mu$ because the mode of $X$ is $\mu$, which corresponds to $e^\mu$ for $Y$ since $Y=e^X$."

(c) ...

The statement of student B is wrong. This can be seen by comparing it to the correct mode of a Log-Normal $y=e^{\mu-\sigma^2}$, see e.g. this link. However, this is not the answer to the above question, because the "correct" mode was not stated in the book.

What I got:

  • The mode of a distribution is defined by $$f_X(x=\mu) \ge f_X(x), \text{ for all $x$}$$ where $f_X(x)$ is the PDF of the random variable $X$.
  • Now express the PDF as the derivative of the CDF: $$f_X(x_0) = \frac{d}{dx} F_X(x)\big|_{x_0} = \frac{d}{dx} P(X \le x)\big|_{x_0}$$ where $F_X(x)$ is the CDF of the random variable $X$ and the derivative is evaluated at $x=x_0$.
  • Now change the argument by taking both sides to the $e$-th power. This yields $$f_X(x_0) = \frac{d}{dx} P(e^X \le e^x)\big|_{x_0}$$ and change the variable to $Y$ $$f_X(x_0) = \frac{d}{dy} P(Y \le y)\big|_{y_0} \cdot \left| \frac{dy}{dx}(y_0)\right|$$ where $y_0 = e^{x_0}$. Here $\frac{dy}{dx}(y_0) = \frac{dy}{dx}\big|_{y_0}$ is the derivative of $y$ with respect to $x$ evaluated at $y=y_0$.
  • Put this into the above inequality, we obtain $$\frac{d}{dy} P(Y \le y_0) \cdot \left| \frac{dy}{dx}({y_0})\right| \ge \frac{d}{dy} P(Y \le y) \cdot \left| \frac{dy}{dx}(y)\right|$$
  • Finally I use that $\frac{dy}{dx} = e^x = y$. Hence we obtain the inequality $$f_Y(y_0) \cdot y_0 \ge f_Y(y) \cdot y, \text{ where $y_0 = e^\mu$.}$$However, for $y\lt y_0$ this does not imply $f_Y(y_0) \ge f_Y(y)$. Hence, the mode of the Log-Normal does not have to be $e^\mu$.

Is this correct? Is there a simpler argument, which uses less calculus?

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