Why is there a negative in Joint probability density?

Question

From the book of O. Calin (An Informal Introduction to Stochastic Calculus with Applications)

Lemma 4.7.1. The joint density function of $(W_t,M_t)$ is given by $$f(a,b) = \dfrac{2(2b-a)}{\sqrt{2 \pi} t^{\frac{3}{2}}} e^{-\dfrac{(2b-a)^2}{2t}}, ~~~ a \leq b, b \geq 0.$$

Here, we define $W_t$ be a standard Brownian motion and $$M_t = \max_{0 \leq s \leq t} W_s.$$

Part of the proof of Lemma 4.7.1 The associated probability density $f(a,b)$ can be obtained by differentiation

My question is ``Why there is a negative sign before the limit?''

I found this on Wikipedia, to solve the joint probability density can be computed by getting the partial derivatives with the condition that all $X_n \leq x_n$. However, in my case is that, $M_t \geq b$.

Thank you for giving ideas.

Answer 1

Using conditional probability formula, \begin{eqnarray} P(W_t \leq a, M_t \geq b) &=& P(W_t \leq a) P(M_t \geq b | W_t \leq a) \\ &=& P(W_t \leq a) \Big( 1- P(M_t \leq b | W_t \leq a) \Big) \\ &=& P(W_t \leq a) - P(W_t \leq a) P(M_t \leq b | W_t \leq a) \\ &=& P(W_t \leq a) - P(M_t \leq b , W_t \leq a). \end{eqnarray} Differentiate $\Big(\dfrac{\partial^2}{\partial a \partial b}\Big)$, \begin{eqnarray} \dfrac{\partial^2}{\partial a \partial b} P(W_t \leq a, M_t \geq b) &=& \dfrac{\partial^2}{\partial a \partial b} P(W_t \leq a) -\dfrac{\partial^2}{\partial a \partial b}P(M_t \leq b , W_t \leq a) \\ &=& - f(a,b). \end{eqnarray}

Thanks to Ovidiu Calin for this answer.