Distribution (FDD) of random process Xt

Hello everyone!

The question I have is:

$\displaystyle \xi$ and $\displaystyle \eta$ are indpt standard normal random variables, we have a random process $\displaystyle X_t$ ,where:

$\displaystyle X_t=(\xi-\eta)\sqrt{t}, t\geq0$

I would like to find out the distribution (one dimension FDD) of $\displaystyle X_t$ ,for a fixed $\displaystyle t>0$

Here is what I did:

$\displaystyle F_t(x)=P((\xi-\eta)\sqrt{t}\leq x)$

$\displaystyle =\int P(\xi\leq\frac{x}{\sqrt{t}}+s|\eta=s)dF_\eta(s)$ by Total Probability Formula

$\displaystyle =\int P(\xi\leq\frac{x}{\sqrt{t}}+s)dF_\eta(s)$ by independence

$\displaystyle =\int F_\xi(\frac{x}{\sqrt{t}}+s)\cdot\frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2} s^2} ds$

here is where I stuck...how should I deal with $\displaystyle F_\xi(\frac{x}{\sqrt{t}}+s)$, or is there other ways to this kind of things?

What about n dimensional FDD?

Could anyone help me? :)

Re: Distribution (FDD) of random process Xt

I found a similar Q with answer, just post it on if it helps...

$\displaystyle \xi$ and $\displaystyle \eta$ are indpt standard normal random variables

$\displaystyle X_t=(\xi+\eta)t, t\geq0$

Find the n dimensional FDD.

and the answer looks like:

if$\displaystyle 0=t_1<t_2<...<t_k$,

$\displaystyle F_{t_1,...,t_k}(x_1,...,x_k) = \left\{\begin{matrix}0,&\mbox{ if }x_1<0,\\\Phi(\frac{1}{\sqrt{2}}min\{\frac{x_2}{t_ 2},...,\frac{x_k}{t_k}\}),&\mbox{ if }x_1\geq 0,\end{matrix}\right$

if$\displaystyle 0<t_1<t_2<...<t_k$,

$\displaystyle F_{t_1,...,t_k}(x_1,...,x_k) = \Phi(\frac{1}{\sqrt{2}}min\{\frac{x_2}{t_2},..., \frac{x_k}{t_k}\})$

I understand where the $\displaystyle min$ comes from, take 2D case, like above:

$\displaystyle P(\xi\leq\frac{x_1}{t_1}-\eta\wedge\frac{x_2}{t_2}-\eta)$

again using TPF:

$\displaystyle \int P(\xi\leq\frac{x_1}{t_1}-s \wedge\frac{x_2}{t_2}-s|\eta = s)dF_\eta(s)$

but how does this becomes

$\displaystyle F_{t_1,t_2}(x_1,x_2) = \Phi(\frac{1}{\sqrt{2}}min\{\frac{x_1}{t_1},\frac{ x_2}{t_2}\})$???

I believe the previous should have the similar FDD, so could anyone explain this to me??

Thank you for your help :)

Re: Distribution (FDD) of random process Xt

Looks like I am answering my own question again...

After a bit study, I found it is a very simple problem...maybe people won't type so much for a easy Q like this :P

$\displaystyle \xi$ and $\displaystyle \eta$ are indpt standard normal random variables,

therefore $\displaystyle (\xi - \eta) \sim N(1, (\sqrt{2})^2)$, so that $\displaystyle X_t \sim N(1, (\sqrt{2})^2)$

the 1D FDD is

$\displaystyle F_t(x)=P((\xi-\eta)\leq \frac{x}{\sqrt{t}})$

$\displaystyle =P(\frac{(\xi-\eta)}{\sqrt{2}}\leq \frac{1}{\sqrt{2}}\frac{x}{\sqrt{t}})$

$\displaystyle =P(z\leq \frac{1}{\sqrt{2}}\frac{x}{\sqrt{t}})$

$\displaystyle = \Phi(\frac{1}{\sqrt{2}}\frac{x}{\sqrt{t}}})$

this could be easily extend to nD FDD