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!