# What is the effect of mean-reversion on an upper barrier knock-out call option?

Consider a mean-reverting normal model for an underlying

$dX^{(1)}_t=-\kappa X^{(1)}_tdt+\sigma^{(1)} dW^{(1)}_t$,

for fixed time-independent constants, $\kappa$ (mean-reversion) and $\sigma^{(1)}$ (volatility) and Brownian motion, $W^{(1)}_t$. Suppose that using this model, I calculate options prices for all $t$, then calibrate the time-dependent local vol, $\sigma_t^{(2)}$, of a second normal model (without mean-reversion)

$dX^{(2)}_t=\sigma_t^{(2)} dW^{(2)}_t$,

so that the two models give the same prices for vanilla options at all times.

Will a continuous upper barrier knock-out call option be cheaper in the first or second model?

For simplicity, take $X_0=Y_0=0$, and assume that the upper barrier, $B$, is larger than the strike, $K$.

• Is $X_t$ the time-$t$ value of a tradable asset $X$? – Quantuple Apr 24 '16 at 21:09
• Yes. The asset has different dynamics due to the different models, so I denote this by $X_t^{(1)}$ and $X_t^{(2)}$ for model 1 and model 2 respectively. – Jase Uknow Apr 24 '16 at 21:24
• Yes, I got your notations :) I am only asking this because, in that case, the two processes should exhibit the same drift under the risk-neutral measure $\mathbb{Q}$ (discount asset prices should emerge as $\mathbb{Q}$ -martingales). And with the same drift, the time-dependent volatility $\sigma^{(2)}_t$ will actually be equal to the constant volatility $\sigma^{(1)}$ (it is similar to deriving a local volatility surface corresponding to a flat Black-Scholes volatility surface). – Quantuple Apr 25 '16 at 13:22
• I see your point, actually I don't want to assume a simple relationship between forward and spot. For example, if we assume we are modelling an underlying commodity, then the no-arbitrage forward-spot relationship doesn't hold, and in that case we can have mean-reversion in the model. – Jase Uknow Apr 28 '16 at 21:54
• You're thinking about non storable commodity I presume? I am not very familiar with the commodities world so I am just asking out of curiousity how could this relationship not hold. Regarding your issue: have you tried a numerical approach (when intuition fails, seems like a good idea to have a first feel of the result) – Quantuple Apr 28 '16 at 22:22

When you are solving for the local vol in the non mean reverting model, you will find that it also depends on strike. Thus, you can only match vanilla options prices between the two models for a single strike.

Let's say that you pick a strike K>0 for which you match the vanilla option price. The you will find that for strike B, where B>K, the mean reverting model produces a lower option price, since it has less probability of an extreme move. Now, the barrier option viewed simplistically consists of short optionality at the B strike - hence the barrier option is worth more under the mean reverting model, assuming that the K strikes have been matched.

## First this is not a full answer, but it might help you.

You probably hit $B$ quickly with $(1)$ than with $(2)$.

## Hint of previous assertion

I might reformulate your question.

I suppose your pricing condition is $$\left\langle X^{(2)}\right\rangle_t=\left\langle X^{(1)}\right\rangle_t$$ so you get :

$$X^{(1)}_t = \int_0^t\sigma^{(1)}e^{-\kappa(t-u)}dW^{(1)}_u$$

and

$$X^{(2)}_t = \int_0^t\sigma^{(1)}e^{-\kappa u} dW^{(2)}_u$$

You want then to study if there exists order between $\tau^1_B$ and $\tau^2_B$

$$\tau^i_B=\inf\left\{t\geq 0:X^{(i)}_t\geq B\right\}$$

Setting $V(t)=\left\langle X^{(2)}\right\rangle_t=\left\langle X^{(1)}\right\rangle_t$ and $$Y^i_t(\lambda) = e^{\lambda X^i_t -\frac{\lambda^2}{2}V(t)}$$

you get (prove it (I think I have the proof but I am not sure) ) for $\lambda\geq 0$

$$\mathbb{E}[Y^2_t(\lambda)|Y^2_s(\lambda)]=Y^2_s(\lambda)$$ where as $$\mathbb{E}[Y^1_t(\lambda)|Y^1_s(\lambda)]\geq Y^1_s(\lambda)$$

so you get :

$$\mathbb{E}[e^{-\frac{\lambda^2}{2}V(\tau^2_B)}]\leq \mathbb{E}[e^{-\frac{\lambda^2}{2}V(\tau^1_B)}]$$

which invalidates $\tau^2_B>\tau^1_B$ a.s since $V$ is increasing.

I know this is not a proof of first assertion but I hope it can help you.

• Hi this looks interesting, but I'm not sure how it solves the original question. What is the meaning of the angular brackets? – Jase Uknow Apr 28 '16 at 21:59
• @MJ73550, I fail to see (1) how identical quadratic variations $\left\langle X^{(2)}\right\rangle_t=\left\langle X^{(1)}\right\rangle_t$ $\iff$ equivalent terminal distributions (ie same vanilla option prices) between the 2 models (2) $Y_t^i$ as you define it seems like a Doléans-Dade exponential to me, hence a martingale regardless of $i$, thus I don't get the different treatment for $i=2$ (martingale) and $i=1$ (supemartingale). I'm presumably missing a point. – Quantuple Apr 28 '16 at 22:17