VaR and Expected Shortfall Under Geometric Brownian Motion
Summary
The document examines loss risk for an asset modeled with geometric Brownian motion. It begins with exact expressions for Value at Risk and Expected Shortfall based on the asset’s lognormal price distribution, then asks how these relate to the common volatility-times-square-root-of-time VaR approximation. The response explains that this simpler expression comes from a local, delta-based approximation to portfolio value changes, using normally distributed log returns. It also notes that a gamma term can extend the approximation to account for curvature.
For Expected Shortfall, the response defines the measure as the negative conditional average of losses beyond the VaR threshold and derives an approximation using the mean of a truncated normal distribution. It distinguishes this from an exact result for a particular case. The material is useful for understanding the assumptions behind the formulas, but the response shifts from the original asset-loss setup to an option portfolio approximation. Notation and conventions differ, so the equations should be checked against the intended portfolio and confidence-level definition before use.
Key ideas
- Under geometric Brownian motion, asset prices are lognormally distributed and log returns are normally distributed.
- The familiar square-root-of-time VaR form is presented as a local delta approximation to value changes.
- Including option gamma adds a quadratic term to the local VaR approximation.
- Expected Shortfall can be derived by conditioning on losses beyond the VaR threshold.
- The option-based approximation and the exact asset formula apply under different setups and assumptions.
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# VaR and Expected Shortfall for Geometric Brownian Motion
# 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 $ES_{t,a}=S_0e^{rt}-E(S_t|S_t<x)$ 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:
- 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? ($r=0$)
- Are my Expected Shortfall definition and formula correct? Thanks in advance
## Answer by FunnyBuzer (score 2, accepted)
https://quant.stackexchange.com/a/58811
- 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.
- 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: $$ES^{1-\alpha}_t=S_{t-1}\left(1-\frac{\Phi(\Phi^{-1}(1-\alpha)-\sigma)e^{\mu+\frac{\sigma^2}{2}}}{1-\alpha}\right)$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.