I have been trying to find the expected shortfall for $X \sim Lognormal(\mu, \sigma^2)$. I have calculated the following;
$$ ES_X(p)=\frac{1}{p}\exp\left(\mu+\frac{\sigma^2}{2}\right)\left[1-\Phi\left(\Phi^{-1}(1-p)-\sigma\right)\right] $$
However, when comparing my result to those I have found online it appears as though the expected shortfall is:
$$ ES_X(\alpha)=\frac{1}{\alpha}\exp\left(\mu+\frac{\sigma^2}{2}\right)\left[\Phi\left(\Phi^{-1}(\alpha)-\sigma\right)\right], $$
where (I am fairly certain) $1-\alpha=p$. To me it appears as though these two answers are different, and if that is indeed the case does anyone happen to know where I went wrong in my calculations. I can't seem to figure out why these two answers do not correspond. For the record, the formula I am using is;
$$ ES_X(p)=\frac{E\left[X\mathbb{I}_{X>VaR_X(p)}\right]}{P(X>VaR_X(p))}=\frac{1}{p}\int\limits_{VaR_X(p)}^\infty xf(x;\mu,\sigma^2)dx $$
where $f(x;\mu,\sigma^2)$ is the probability density function of a Lognormal distribution.
I believe I came across the reason for my confusion. The solution I provided for the expected shortfall of a $Lognormal$ was under the assumption that $X$ was the loss for a portfolio. Whereas the results I came across online were for $X$, a payoff portfolio. I am fairly certain this is the cause for the slight discrepancies between the two solutions.