The difference of two Brownian Processes This is a homework problem. Please tell me if my solution is correct and if not please point out my error.
Problem
$W_{1}(t)$ and $W_{2}(t)$ are two independent Brownian processes with variance parameters $\alpha_{1}$ and $\alpha_{2}$. X(t) is a new random process s.t. $X(t) = W_{1}(t) - W_{2}(t)$ Find $R_{XX}(t_{1},t_{2})$ and the pdf $f_{X}(x,t)$
Solution
For $R_{XX}(t_{1},t_{2})$
This is the difference of 2 normal Random Variables so
$\mu_{X} = \mu_{W_1(t)} - \mu_{W_2(t)}$ and $\sigma_{X}^{2} = \sigma_{W_1(t)}^{2} + \sigma_{W_2(t)}^{2}$
$R_{XX}(t_{1},t_{2}) = E[(W_1(t_{1}) - W_2(t_{1}))(W_1(t_{2}) - W_2(t_{2}))] =$
$E[W_1(t_{1})W_1(t_{2}) - W_2(t_{1})W_2(t_{2}) - W_1(t_{1})W_2(t_{2})+W_2(t_{2})W_2(t_{1})]  =$
because $W_1$ and $W_2$ are independent I can separate the expectations of the products
$R_{W_1}(t_{1},t_{2}) - E[W_2(t_{1})]E[W_1(t_{2})] - E[W_1(t_{1})]E[W_1(t_{2})] + R_{W_2}(t_{1},t_{2}) =  $
$R_{W_1}(t_{1},t_{2}) - 0 \cdot 0 - 0\cdot 0+ R_{W_2}(t_{1},t_{2}) $
because the expectation of a Brownian Process is 0
$= \alpha_{1}min(t_{1}, t_{2}) + \alpha_{2} min(t_{1},t_{2})$
To find the pdf, X(t) is the linear combination of independent normal random variables
$\therefore X(t) \sim n(0, (\alpha_{1}+\alpha_{2})t) $
 A: $$R_{XX}(t,s)=E[X_tX_s]-\underbrace{E[X_t]E[X_s]}_{=0}=E[(W_{1,t}-W_{2,t})(W_{1,s}-W_{2,s})]= \\ =E[W_{1,t}W_{1,s}]-\underbrace{E[W_{1,t}W_{2,s}]}_{=0}-\underbrace{E[W_{2,t}W_{1,s}]}_{=0}+E[W_{2,t}W_{2,s}]$$
Assume wlog $s<t$. We have
$$E[W_{j,t}W_{j,s}]=E[((W_{j,t}-W_{j,s})+W_{j,s})W_{j,s}]=\alpha_js \ \ \ \ \ j=1,2$$
$$R_{XX}(t,s)=s(\alpha_1+\alpha_2)$$
therefore $R_{XX}(t_1,t_2)=\min(t_1,t_2)(\alpha_1+\alpha_2)$.
$$P(W_{1,t}-W_{2,t}\leq x)=\int_{\mathbb{R}}P(W_{1,t}\leq W_{2,t}+x|W_{2,t}=y_2)f_{W_{2,t}}(y_2)dy_2= \\ =\int_{\mathbb{R}}\int_{(-\infty,y_2+x]}f_{W_{1,t}}(y_1)f_{W_{2,t}}(y_2)dy_1dy_2= \\ =\int_{\mathbb{R}}\Phi\bigg(\frac{y_2+x}{\sqrt{t\alpha_1}}\bigg)f_{W_{2,t}}(y_2)dy_2$$
$$\partial_xP(W_{1,t}-W_{2,t}\leq x)=\int_{\mathbb{R}}f_{W_{1,t}}(y_2+x)f_{W_{2,t}}(y_2)dy_2$$
This is the autocorrelation of two Gaussian densities with mean zero. The Fourier tranform would be
$$\varphi_{X_t}(\omega)=\exp\{-0.5t(\alpha_1+\alpha_2)\omega^2\}$$
which transformed back results in
$$\partial_xP(W_{1,t}-W_{2,t}\leq x)=\frac{1}{\sqrt{2\pi t(\alpha_1+\alpha_2)}}\exp\bigg\{-\frac{x^2}{2t(\alpha_1+\alpha_2)}\bigg\}$$
