Let's consider a 1D Brownian motion Wt, and a real h > 0.Let's define the stopping time tau = inf(t/Wt >= h)and the following process:Wt' = Wt if t < tau, or 2h-Wt if t >= tauWe all know by the reflection principle that Wt' is also a Brownian motion.Can anyone tell me what is the joint law of Wt and Wt' ?

I'll call them W and Y to make notation simpler and drop the subscript t.let's assume h is positiveThe joint distribution isP(W< w, Y<y) = P(W<w, Y<y , t <tau ) + P(W<w, Y<y , t > tau )= P(W<w, W<y , t <tau) + P(W<w, W>2h-y, t > tau )= P(W<min(w,y), t <tau) + P(2h-y<W<w, t > tau )= P(W<min(w,y), MAX(W) <h) + P(2h-y<W<w, MAX(W) >h)where MAX(W) = the running max up to time t.the joint distribution of W and MAX(W) is in Karatzas and Shreve, page 95. For h<0, the derivation is similar but with the running min.Let me know if I screwed up.

Hi KO, thanks for the answer. I don't think you screwed up, but here is another one now.Let's assume that we have the two processes:dX = sigX(t).dWtdY = sigY(t).dWtand I define again:tau = inf (t/X>h)X' = X if t < tauX' = 2h-X if t > tauquestion: what is the joint probability of X and Y (and notice that the normal vols are different functions of time, i.e. different term structures).If sigY(t) = alpha.sigX(t) for all t, and alpha is constant, then it can be solved using the result from the previous question. The question is how to solve it in the more general case.Do you know how ?

If Xo and Yo are the same, then they are the same process.X(t) = X(o) exp[ - t sig^2/2 + sig W(t) ]Both are martingales. Seems you could use the optional sampling theorem with your stopping time. However, I doubt that X' is an "exponential martingale" (as is X).Perhaps just use the above solution and some algebra.What class are you enrolled in?

X is not an exponential martinale, but just a martingale.If you assume that sigX(t) is constant in time and equal to 1, i.e. Xt is just a Brownian motion, then X' is also a Brownian motion. And my initial question was to find the joint law between X and X', which was answered by yourself.Now in the general case I mention below, intuitively, I would say that X' is also a martingale, and I would even go all the way to say that there exists a Brownian motion Wt' such that dX' = sig(t).dWt'.Still intuitively, I would say that:t < tau => dWt' = dWtt >= tau => dWt' = - dWtI don't have rigorous proof of that though.Can you show me how you would use the optimal sampling theorem ?KO, I'm not enrolled in any class: are you ?

If X(o) <> zero, isn't this the solution to dX = sig X dWX(t) = X(o) exp[ - t sig^2/2 + sig W(t) ] (*)this falls within the class of (what I call) exponential martingales. In fact, I think any strictkly positive martingale has the exponential martingale form.Note, I didn't realize you were letting sig depend on X (the only way sig*X = constant). However, let's just consider the simple case sig = constant, in which case (*) is correct.X is not symmetric about h from the time X hits h onward (which is part of the proof of the reflection principle). Some paths go to h+K, but no paths go to h-K for large enough K.I doubt X' is a martingale as well. Perhaps a different adjustment after tau would make it one.If it is a martingale (wrt the filtration generated by W, which I bet is the same as that generated by W'), then by the martingale rep thm, there is an f such that dX' = f dW.the optional sampling theorem states that a stopped martingale is a martingale.....W(t min tau) is a martingale where tau is a stopping time. These aren't really stopped martingales, so it may not be helpful. I just know we used the result a lot in proving hitting time results. Not in any class. Asked since this seemed like HW questions rather than finance. What's the finance angle, if any?There is always brute force. Try simulation. This won't prove the result, but might disprove it with a particular example.

Sorry it is my fault, as I used confusing notations.You're right but this is not what I meant to write.The process is:dX=sig1(t).dWtdY=sig2(t).dWtX(0) = 0, Y(0) = 0and instead of using sig1 and sig2, I used sigX and sigY which you interpreted as sig.X and sig.Y, which is understandable.So again this is my mistake, and the processes I'm considering are indeed:dX=sig1(t).dWtdY=sig2(t).dWtwhere t->sig1(t), and t->sig2(t) are deterministic functions of time.So the question is:posit tau = inf(t/X(t)>h) for h > 0and X'(t) = X(t) if t < tauX'(t) = 2.h - X(t) if t >= tauthen the question is:what is the joint law of X' and Y.No wmy contention is to say that X' is also a martingale, and it exists a Brownian motion such that dX' = sig1.dW' where sig1 is the same local vol as used for X.More than that, I would even say that:for t < tau, dW' = dWfor t >= tau, dW'=-dWjust from intuition rather than hard proof.However this doesn't seem enough to find the joint law. What do you think ?The relationship to finance is finding a semi-analytic formula for barrier pricing in Black-Scholes with deterministic term structure interest rates and volatilities.

Ahhhhh, OK.Yes, they are definitely good ol' regular martingales. You can view sig1(t) and sig2(t) as speeding up or slowing down the brownian motion. i.e. a time change. Oksendal has some on this, also K&S page 174. The time change is related to the quadratic variation process, which in this case is just the integral of sig^2 up to time t.>>>>>>>for t < tau, dW' = dWfor t >= tau, dW'=-dWjust from intuition rather than hard proof.>>>>>>>>>since W' is also a BM, dW=dW' in a distributional sense in that the actual paths will be different, but the distributions of integrals against either of these will be the same since sig is just a function of time.I think the reflection principle should hold for X(t) = int[0,t] sig(s) dW(s)i.e. X and X' are the same distributionally (which is what is meant in the original reflection principle)Let's assume it holds. X(t), X'(t), and Y(t) are all normal random variables. I'd split the distribution up into t<tau and t>tau as before and then rewrite as a sum of conditional probsP(A) = P(AnB) + P(AnB' ) = P(A | B) P(B) + P(A | B' ) P(B' )where A = { X' <x, Y < y}and see how far you get. no more time right now. It seems like it can be done with the same "machinery".