More stuff added

Dissertation/ann_rep_brownian_motion_monte_carlo.tex

@@ -889,7 +889,7 @@ Let $t \in \lp 0,\infty\rp$ and $T \in \lp t,\infty\rp$. Let $\lp \Omega, \mathc
 		&=2\cdot \E\lb \lp \int^t_s d\cW_r^d\rp^2\rb \nonumber\\
 		&=2\cdot \E \lb \lp \cW^d_{t-s}\rp^2\rb 
 	\end{align} 
-\end{proof}
+
 Note now that:
 \begin{align}
 	\va \lb \cW^d_{t-s}\rb &= \E \lb \lp \cW_{t-s}^d\rp^2\rb - \E \lb \cW^d_{t-s}\rb^2 \nonumber \\
@@ -898,11 +898,27 @@ Note now that:
 \end{align}
 Now note that since $\cW^d_r$ are standard Brownian motions, and their expectation and variance are $\mymathbb{0}_d$ and $\mathbb{I}_d$ respectively. Whence it is the case that the probability density function for $\cW_{t-s}^d$ is:
 \begin{align}
-	\lp 2\pi\rp^{-\frac{d}{2}}\lp t-s\rp^{-\frac{1}{2}}\exp \lp \frac{-1}{2(t-s)}\mymathbb{e}_{1,d}\cdot \lb x \rb_*^2\rp
+	\ff_{\cW^d_{t-s}} \lp x\rp= \lp 2\pi\rp^{-\frac{d}{2}}\lp t-s\rp^{-\frac{1}{2}}\exp \lp \frac{-1}{2(t-s)}\mymathbb{e}_{1,d}\cdot \lb x \rb_*^2\rp
 \end{align}
-However $\cX^{d,t,x}_s$ is a shifted normal distribution
+However $\cX^{d,t,x}_s$ is a shifted normal distribution, specifically shifted by $x$. Its p.d.f. is thus:
+\begin{align}
+	\ff_{\cX^{d,t,x}_s}\lp \scrX \rp = \lp 2\pi\rp^{-\frac{d}{2}}\lp t-s\rp^{-\frac{1}{2}}\exp \lp \frac{-1}{2(t-s)}\mymathbb{e}_{1,d}\cdot \lb \scrX +x\rb_*^2\rp
+\end{align}
+The Law of the Unconscious Statistician then says that:
+\begin{align}
+	\E \lb \fu^T_d \lp \cX^{d,t,x}_s\rp\rb = \int_{\R^d}\fu^T_d\lp \scrX\rp\cdot \ff_{\cX^{d,t,x}_s}\lp \scrX\rp d\scrX
+\end{align}
+And further that:
+\begin{align}
+	\E \lb \alpha_d \lp \cX^{d,t,x}_s\rp\rb = \int_{\R^d} \alpha_d\lp \scrX\rp\cdot \ff_{\cX^{d,t,x}_s}\lp \scrX\rp d\scrX
+\end{align}
+
+\textcolor{red}{\textbf{Need to re-examine $\fu^T_d, \alpha_d$}}
+
+Now note this that
 
 
+\end{proof}