Let $Y$ be Log-Normal with parameters $\mu$ and $\sigma^{2}$. So $Y=e^{x}$ with $X \sim \mathcal{N}\left(\mu, \sigma^{2}\right)$. Which of the following statements is correct?
(A) The median of $Y$ is $e^{\mu}$ because the median of $X$ is $\mu$ and the exponential function is continuous and strictly increasing, so the event $Y \leq e^{\mu}$ is the same as the event $X \leq \mu .$
(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 mode of $Y$ is $\mu$ because the mode of $X$ is $\mu$ and the exponential function is continuous and strictly increasing, so maximizing the PDF of $X$ is equivalent to maximizing the PDF of $Y=e^{x}$.
How do I proceed with this? I know that the median condition is not true? But do we have some expression for mode?
Reference: This problem is from Chapter 6 of Introduction to Probability by Blitzstein and Hwang.