Prove that $X(t)=e^{-t}B(e^{2t})$ is a Markov process and identify the transition kernel.

Question

Let $\{B(t):t\geq 0\}$ be a Brownian motion. Define $X(t)=e^{-t}B(e^{2t})$ for all $t\geq 0$. Show that $X(t)$ is a Markov process and identify the Markov transition kernel.

I started with the Markov property: $P\left[X_{t+s} \in A | \mathcal{F}_{s}\right]=p(t, X(s), A)$.

$X(t+s)=e^{-(s+t)}(B(e^{2(s+t)})-B(e^{2s}))+e^{-t}X(s)$.

So, if $X(t)$ is a Markov process, the transition kernel should satisfy:

$p(t, x, y)=P(e^{-(s+t)}(B(e^{2(s+t)})-B(e^{2s}))+e^{-t}x=y|\mathcal{F}_s)=P(B(e^{2(s+t)})-B(e^{2s})=e^{(s+t)}(y-e^{-t}x)|\mathcal{F}_s)=P(B(e^{2(s+t)})-B(e^{2s})=e^{(s+t)}(y-e^{-t}x))$

Thus, $p(t,x,y)=p_{e^{2(s+t)}-e^{2s}}(e^{(s+t)}(y-e^{-t}x))$, where $p_{e^{2(s+t)}-e^{2s}}(y)$ is the density function of $\mathcal{N}(0,e^{2(s+t)}-e^{2s})$.

But the answer is not right. And it seems that the transition density should be the density function of $\mathcal{N}(e^{-t}x, 1-e^{-2t})$.

Any help will be appreciated! Thank you!

Answer 1

I see.

It should be:

$P(e^{-(s+t)}(B(e^{2(s+t)})-B(e^{2s}))+e^{-t}x\in A|\mathcal{F}_s)=P(e^{-(s+t)}(B(e^{2(s+t)})-B(e^{2s}))+e^{-t}x\in A)$.

Obviously, $e^{-(s+t)}(B(e^{2(s+t)})-B(e^{2s}))+e^{-t}x$ follows $\mathcal{N}(e^{-t}x, 1-e^{-2t})$.

And then I need to verify that the density function of $\mathcal{N}(e^{-t}x, 1-e^{-2t})$ is indeed a transition density.