Conditional Distribution Brownian process $X(t) = W^{2}(t)$ This is a homework problem. Please tell me if my solution is correct, if I am not correct please point out my error. If there is a better way to solve the problem, please let me know.
Problem
W(t) is a standard Brownian process $X(t) = W^{2}(t)$
a) find $f_{X}(x;t)$
b) find $f_{X}(x_{2}|x_{1};t_{2}. t_{1})$
Solution:
a)
$$ f_{W(t)} = \frac{1}{\sqrt{2 \pi t}}e^{-\frac{W(t)^{2}}{2t}} \\
g(w) = w^{2} \\
w_{1} = \sqrt{x} \hspace{5mm} w_{2} = -\sqrt{x} \\
f_{X(t)}(x) = \frac{f_{W(t)}(\sqrt{x})}{|2 \sqrt{x}|} + \frac{f_{W(t)}(-\sqrt{x})}{|-2 \sqrt{x}|} 
= \frac{1}{\sqrt{2 \pi t X(t)}}e^{- \frac{X(t)}{2t}} \hspace{3mm}, X(t) \in (0, \infty)
$$
b)
Assume $ t_{1} < t_{2}$
let $W_{1} = X(t_{1}) \hspace{3mm}W_{2} = X(t_{2}) $
W(t) is Brownian so $W_{1}$ and $W_{2}$ are jointly normal $\therefore$
$$ 
E[W_{2}|W_{1} = w] = \mu_{W_{2}} + \rho \sigma_{W_{2}}\frac{w - \mu_{W_{1}}}{\sigma_{W_{1}}} \\
Var(W_{2} | W_{1} = w) = (1- \rho^{2})\sigma^{2}_{W_{2}} \\
\rho = \frac{Cov(W_{1}, W_{2})}{\sigma_{W_{1}} \sigma_{W_{2}}} \\
W_{1} \sim N(0, t_{1}) \hspace{3mm} W_{2} \sim N(0, t_{2}) \\
\therefore 
\rho = \sqrt{\frac{t_{1}}{t_{2}}} \\
$$
$Cov(W_{1}, W_{2}) = t_{1}$ because $t_{1} < t_{2}$
$ X_{1}(t) = W_{1}^{2}(t) $ , $ X_{2}(t) = W_{2}^{2}(t)$
$\pm \sqrt{X_{1}(t)} = W_{1}(t) $  ,   $\pm \sqrt{X_{2}(t)} = W_{2}(t) $
for a given $(w_{1}, w_{2}) $ we have four points
$( \sqrt{x_{1}}, \sqrt{x_{2}}), ( -\sqrt{x_{1}}, \sqrt{x_{2}}), ( \sqrt{x_{1}}, -\sqrt{x_{2}}), ( -\sqrt{x_{1}}, -\sqrt{x_{2}})$
$$
\frac{\partial x_{1}}{\partial w_{1}} = 2w_{1} \hspace{3mm}\frac{\partial x_{1}}{\partial w_{2}} = 0 \\
\frac{\partial x_{2}}{\partial w_{1}} = 0 \hspace{3mm}\frac{\partial x_{2}}{\partial w_{2}} = 2w_{2} \\
J(w_{1}, w_{2}) = det \begin{bmatrix}2w_{1} & 0 \\0 & 2w_{2} \end{bmatrix} = 4w_{1}w_{2}
$$
$$ 
f_{X(t_{1}),X(t_{2})}(x_{1}, x_{2}) = \frac{1}{\pi \sqrt{1- \rho^{2}}}\frac{1}{|4 \sqrt{x_{1}} \sqrt{x_{2}}|}[exp[- \frac{1}{2(1- \rho^{2})}(x_{1} - 2 \rho \sqrt{x_{1}}\sqrt{x_{2}}+x_{2}) + exp[- \frac{1}{2(1- \rho^{2})}(x_{1} + 2 \rho \sqrt{x_{1}}\sqrt{x_{2}}+x_{2})]]
$$
$$
f_{X(t_{2})|X(t_{2})}(x_{2}| x_{1}) = \frac{\sqrt{2 \pi t_{1}x_{1}}}{\pi \sqrt{1- \rho^{2}}}\frac{e^{\frac{x_{1}}{2t_{1}}}}{|4 \sqrt{x_{1}} \sqrt{x_{2}}|}[exp[- \frac{1}{2(1- \rho^{2})}(x_{1} - 2 \rho \sqrt{x_{1}}\sqrt{x_{2}}+x_{2}) + exp[- \frac{1}{2(1- \rho^{2})}(x_{1} + 2 \rho \sqrt{x_{1}}\sqrt{x_{2}}+x_{2})]]
$$
 A: (a)
$$P(W_t^2\leq x)=P(-\sqrt{x}\leq W_t\leq \sqrt{x})=\int_{[-\sqrt{x},\sqrt{x}]}f_{W_t}(y,t)dy=\Phi\bigg(\sqrt{\frac{x}{t}}\bigg)-\Phi \bigg(-\sqrt{\frac{x}{t}}\bigg) $$
where $\Phi$ is the standard normal cdf. Thus
$$f_{X_t}(x,t)=\frac{1}{2\sqrt{xt}}\phi \bigg(\sqrt{\frac{x}{t}}\bigg)+ \frac{1}{2\sqrt{xt}}\phi \bigg(-\sqrt{\frac{x}{t}}\bigg)= \frac{1}{\sqrt{xt}}\phi \bigg(\sqrt{\frac{x}{t}}\bigg) $$
where $\phi$ is the standard normal pdf.
(b)
Consider for $s < t$
$$P(X_t\leq x_1,X_s\leq x_2)=P(W_t\in [-\sqrt{x_1},\sqrt{x_1}],W_s\in [-\sqrt{x_2},\sqrt{x_2}])$$
we have
$$f_{W_t|W_s}(y_1,y_2)=\frac{1}{\sqrt{2\pi (t-s)}}\exp\bigg\{-\frac{(y_1-y_2)^2}{2(t-s)}\bigg\}$$
thus
$$P(X_t\leq x_1,X_s\leq x_2)=\int_{[-\sqrt{x_2},\sqrt{x_2}]}\bigg(\Phi\bigg(\frac{\sqrt{x_1}-y_2}{\sqrt{t-s}}\bigg)-\Phi\bigg(\frac{-\sqrt{x_1}-y_2}{\sqrt{t-s}}\bigg)\bigg)f_{W_s}(y_2)dy_2$$
so
$$\partial_{x_1}P(X_t\leq x_1,X_s\leq x_2)=\int_{[-\sqrt{x_2},\sqrt{x_2}]}\frac{1}{2\sqrt{x_1(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}-y_2}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{-\sqrt{x_1}-y_2}{\sqrt{t-s}}\bigg)\bigg)f_{W_s}(y_2)dy_2$$
and
$$\partial_{x_1,x_2}P(X_t\leq x_1,X_s\leq x_2)=\frac{1}{4\sqrt{x_1x_2(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{-\sqrt{x_1}-\sqrt{x_2}}{\sqrt{t-s}}\bigg)\bigg)f_{W_s}(\sqrt{x_2})+ \\ 
+\frac{1}{4\sqrt{x_1x_2(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{-\sqrt{x_1}+\sqrt{x_2}}{\sqrt{t-s}}\bigg)\bigg)f_{W_s}(-\sqrt{x_2})= \\
=\frac{1}{2\sqrt{x_1x_2(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{t-s}}\bigg)\bigg)\frac{1}{\sqrt{s}}\phi\bigg(\frac{\sqrt{x_2}}{\sqrt{s}}\bigg)$$
So ultimately
$$f_{X_t|X_s}(x_1,x_2)=\frac{1}{2\sqrt{x_1(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{t-s}}\bigg)\bigg)$$
