# VaR and Expected Shortfall for Geometric Brownian Motion

Given that $$dS_t=\mu S_tdt+\sigma S_tdW_t$$ ,a risk free rate r and defining Value at Risk and Expected Shortfall as $$VaR_{t,a}=S_0e^{rt}-x$$ where $$x$$ is the amount such that $$P(S_t\leq x)=1-a$$ ($$a:$$confidence level) and $$ES_{t,a}=S_0e^{rt}-E(S_t|S_t I found

$$VaR_{t,a}=S_0e^{rt} - S_0e^{\sigma\sqrt{t}N^{-1}(1-a)+(\mu-\frac{\sigma^2}{2})t}$$ and $$ES_{t,a}=S_0e^{rt}-\frac{S_0e^{\mu t}N[N^{-1}(1-a)-\sigma \sqrt{t}]}{1-a}$$

I have two questions:

1. A popular VaR formula is $$S_0\sigma \sqrt{t}N^{-1}(1-a)$$. Is this obtained by taking the Taylor expansion and ignoring any power of $$t\geq 1$$ as well as ignoring the time value of money? ($$r=0$$)
2. Are my Expected Shortfall definition and formula correct? Thanks in advance

1. We know the formula to price a call option in the Black-Scholes-Merton model: $$C=S_0\Phi(d_1)-e^{rt}K\Phi(d_2)$$ with $$d_1=\frac{\log\frac{S_0}{K}-T(r+\frac{\sigma^2}{2})}{\sigma\sqrt T}$$ and $$d_2=d_1-\sigma\sqrt T$$, assuming the underlying stock pays no dividends. The option delta is given by: $$\Delta:=\frac{\partial C}{\partial S}=\Phi(d_1)$$ Note that when the maturity gets shorter the delta diverges, i.e. $$\Delta_{\text{ATM}}\rightarrow 1/2$$, $$\Delta_{\text{ITM}}\rightarrow 1$$ and $$\Delta_{\text{OTM}}\rightarrow 0$$. Let us consider a portfolio consisting of a long call option and express the dollar value change of the portfolio: $$V^{\\\}_{t}=C_{t}-C_0$$ For small changes in the underlying, we can approximate the delta: $$\Delta\approx\frac{C_{t}-C_0}{S_{t}-S_0}$$ Hence, $$V^{\\\}_{t}\approx \Delta(S_{t}-S_0)\approx \Delta S_0\log\frac{S_{t}}{S_0}.$$ Moreover, we know that log-returns are conditionally normally distributed: $$\log\frac{S_{t}}{S_0}\sim\mathcal{N}\left(\left(\mu-\frac{\sigma^2}{2}\right)t,\sigma^2t\right)$$ The variance of this portfolio is: $$Var(V^{\\\}_{t})\approx\Delta^2S_0^2\sigma^2t,$$ So, the VaR of this portfolio will be: $$VaR^{1-\alpha}_{t}=-\sqrt{Var(V^{\\\}_{t})}\Phi^{-1}(1-\alpha)\approx -\Delta S_0\sigma\sqrt t\Phi^{-1}(1-\alpha).$$ This is the delta approximation that you provided. Considering also the option gamma, one could extend this VaR approximation by considering the quadratic term in the Taylor expansion.
2. Let's start with the definition of the expected shortfall: $$ES^{1-\alpha}_t=-\mathbb E_0\left[V^{\\\}_t\Big|V^{\\\}_t<-VaR^{1-\alpha}_t\right]$$ Note that, we can write $$\log\frac{S_t}{S_0}$$ in terms of a standard normal variable as above: $$\log\frac{S_t}{S_0}=\left(\mu-\frac{\sigma^2}{2}t\right)+\sigma\sqrt tZ_t,\text{ where }Z_t\sim\mathcal N(0,1)$$ So, just like before, we obtain the approximation: $$V_t^{\\\}\approx\Delta S_0\log\frac{S_t}{S_0}=\Delta S_0\left(\left(\mu-\frac{\sigma^2}{2}t\right)+\sigma\sqrt tZ_t\right)$$ The standard normal distribution truncated at a threshold $$K$$ is defined as $$\phi_K(z|z\leq K)=\frac{\phi(z)}{\Phi(K)}\text{ and }\mathbb E[z|z\leq K]=-\frac{\phi(K)}{\Phi(K)}.$$ This means that \begin{align*} ES^{1-\alpha}_t &= -\mathbb E_0\left[V^{\\\}_t\Big|V^{\\\}_t<-VaR^{1-\alpha}_t\right] \\ &= -\mathbb E_0\left[\Delta S_0\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma\sqrt t Z_t\right)\Bigg|\Delta S_0\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma\sqrt t Z_t\right)<-VaR^{1-\alpha}_t\right] \\ &= -\Delta S_0\left(\mu-\frac{\sigma^2}{2}\right)t-\Delta S_0\sigma\sqrt t\mathbb E_0\left[Z_t\Bigg|Z_t<-\frac{VaR^{1-\alpha}_t}{\Delta S_0\sigma\sqrt t}\right] \\ &=-\Delta S_0\left(\mu-\frac{\sigma^2}{2}\right)t+\Delta S_0\sigma\sqrt t\frac{\phi\left(-\frac{VaR^{1-\alpha}_t}{\Delta S_0\sigma\sqrt t}\right)}{\Phi\left(-\frac{VaR^{1-\alpha}_t}{\Delta S_0\sigma\sqrt t}\right)} \end{align*} From 1. we know that in the case of normal distribution $$VaR^{1-\alpha}_{t}=-\sqrt{Var(V^{\\\}_{t})}\Phi^{-1}(1-\alpha)$$ Therefore, $$ES^{1-\alpha}_t = -\Delta S_0\left(\mu-\frac{\sigma^2}{2}\right)t+\Delta S_0\sigma\sqrt t\frac{\phi(\Phi^{-1}(1-\alpha))}{1-\alpha}$$
To conclude, these results use the delta approximation. But there is also an exact result available for the case of GBP: $$ES^{1-\alpha}_t=S_{t-1}\left(1-\frac{\Phi(\Phi^{-1}(1-\alpha)-\sigma)e^{\mu+\frac{\sigma^2}{2}}}{1-\alpha}\right)$$
• A very insightful answer to the first question, thanks a lot. About the second question, the exact ES formula is actually the lognormal ES formula. Now we know that $S_t \sim lnN(lnS_0 + (\mu - \sigma^2 /2)t, \sigma^2 t)$. By substituting $\mu$ for the mean of $lnS_t$ and $\sigma^2$ for the variance of $lnS_t$ and setting $r=0$ in the formula you provided I obtain the formula I posted, so I guess I got the ES correct. Oct 19 '20 at 23:03