Brownian Bridges Do Not Change Vanilla Option Volatility

# Option pricing with Brownian Bridge


# Option pricing with Brownian Bridge












Say I have an asset following arithmetic Brownian motion $$ dX(t) = \sigma dW^\bot (t) $$ with $\sigma$ constant, and I have prices of vanilla options on $X$.

I introduce a Brownian bridge $$ dY(t) = \nu dW(t) - \frac{\nu}{T} W(T) dt $$ wih $\nu$ is constant, and I also have the orthogonality condition $$ dW(t)dW^\bot(t) = 0 $$

Since $Y(T) =0$ I will have at all times $t$ $$ E_t (X_T + Y_T - K)_+ = E_t(X_T - K)_+ $$

But how do I prove this explicitly?

EDIT

To be clearer, what I mean with prove explicitly is to start with the observation that since $$ \sigma dW^\bot (t) + \nu dW(t) = \sqrt{\sigma^2 + \nu^2} dZ(t) $$ The expectation can be written as $$ E_t(X_T + Y_T - K)_+ = E_t \left( \sqrt{\sigma^2 + \nu^2}Z(T) - \nu W(T) - K \right)_+ $$ Is there a way to evaluate the above expectation on the right hand side, keeping the volatility $\sqrt{\sigma^2 + \nu^2}$ and still satisfy the fact that it must equal the vanilla option price at time $t$?

If $W(T)$ were uncorrelated to $Z(T)$ then I could apply conditioning and evaluate it easily, but unless I am missing something I don't think it's possible to evaluate the above expectation without the $\nu$ dropping out again because $W(T)$ is not orthogonal to $Z(T)$, right?

Basically what I want to do is have an option, which is still delta-hedgeable with $X$ only, and which has as volatility $\sqrt{\sigma^2 + \nu^2}$, where $\nu$ is the historical volatility of $X$ and $\sigma$ is the future volatility of $X$. Vanilla options only contain the future volatility, so I thought this trick with Browian bridge might work, but not sure.

Any other ideas on how to achieve what I want with or without Brownian bridge would be welcome!

## Answer by Canardini (score 3)

https://quant.stackexchange.com/a/50036

To prove that $Y(T)=0$, you integrate the SDE :

$$\int_{0}^{T}dY(t) = \int_{0}^{T}\nu dW(t) - \frac{\nu}{T} \int_{0}^{T}W(T) dt$$

$$Y(T)-Y(0)=\nu\left( W(T)-W(0)\right)- \frac{\nu}{T}W(T)\left(T-0\right)$$

Therefore $$Y(T)=Y(0)$$

You forgot to write the initial conditions but I guess $Y(0)=0.$

EDIT :

The dependency in $\nu$ is irrelevant in pricing that option as $Y_T=0$. The second term is just a function of $\sigma$.

If you fix the price, then you cannot fix $\sigma$( Price/Bachelier is a bijection). Somehow you decided to introduce the Brownian bridge, and what you wrote is correct. However, $Z(T)$ and $W(T)$ are correlated, $$cov( \sqrt{\sigma^2 + \nu^2}Z(T) ,\nu W(T))=\nu^2 T$$, and if you derive the distribution of $\sqrt{\sigma^2 + \nu^2}Z(T) -\nu W(T)$. the $\nu$ just disappears. At then end of the day, your model is a Bachelier model. Your motivation is still unclear to me. The option price will give you the future volatility. The historical volatility gives you the past dynamics... which is irrelevant in pricing the option. Are you trying to decompose the implied volatility into a historical vol and another component?