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Suppose we have some function y(x) (the interpretation of which will follow almost immediately, in Exercise 7.13, but for now simply think of it as positive) and we define a function u(x, t, T) by u(x, t, T) = E< exp y(X (s))ds

where we understand the expectation to be over the sample paths of X(s), starting from X(t) = x, satisfying the stochastic differential equation dX = b(X, t)dt + v/a(X, t)dW. When t = Tthe integral will be 0, we conclude that u(x, T, T) = 1 and that u(x, t, T) < 1 otherwise.

If we break the integral into a piece from t to t + dt and then a piece from t + dt to T, we have exp y(X (s))ds exp t+dt y(X (s))ds exp y(X (s))ds t+dt

and if we now Taylor expand the first term on the right hand side of Eq. (7.84) we have exp y(X (s))ds

The expectation over paths beginning at X(t) can be broken into two pieces: first the expectation of paths that go from X(t) = x to values of X(t + dt) = x + dX and then the expectation of paths starting at x + dX. In other words, we have shown that u(x, t, T) = EdXf[1 - y(x)dt + o(dt)]tt(x + dX, t + dt, T)g (7.86)

which we now Taylor expand and average in the usual manner to obtain a differential equation.

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