Mean Reversion and Upper Barrier Knock-Out Call Prices

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


# 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, $K$.

## Answer by dm63 (score 1)

https://quant.stackexchange.com/a/26292

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.

## Answer by M. Jeunesse (score 0)

https://quant.stackexchange.com/a/25633

### 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.