Intuition for why mean of lognormal distribution depends on variance of normally distributed rv Let $X\sim\mathcal{N}(\mu,\sigma^2)$, which is a normal distribution. Then, $\text{exp}(X)\sim\text{Lognormal}(\mu,\sigma^2)$, and its mean is
$$
\mathbb{E}[\text{exp}(X)]=\text{exp}\left(\mu+\dfrac{\sigma^2}{2}\right)
$$
where the two parameters $\mu$ and $\sigma^2$ are carried over from the normal distribution.
I am looking for an intuitive explanation for why the mean of the log-normal distribution positively depends on $\sigma^2$.
Here is my attemp. Lower values of $X$ are "squashed", and conversely higher values of $X$ are "dispersed" in $\text{exp}(X)$.
But there must be more than this (e.g. why is $\sigma^2$ halved?).

You can understand the $\sigma^2/2$ term from Taylor series:
\begin{align}
e^X
&\sim e^\mu(1+(X-\mu)+\frac12(X-\mu)^2)\\
&= e^\mu(1+(X-\mu)+\frac12(X^2-2\mu X+\mu^2))\\
E[e^X]
&\sim e^\mu(1+(E[X]-\mu)+\frac12(Var(X)+E[X]^2-2\mu E[X]+\mu^2))\\
&= e^\mu(1+\frac12\sigma^2)\\
&\sim e^\mu e^{\sigma^2/2}\\
&= e^{\mu+\sigma^2/2}
\end{align}
The final expression is simple because $E[X]$ cancels with $\mu$.
