Let $B(t)$ be a standard Brownian motion. For $t\geq 0$, define $$U(t) = e^{-t}B(e^{2t}).$$ The problem is to determine the covariance function of the process. Supposedly, the answer is $e^{-s-t}$.
Initially, I thought $U(t)$ followed a standard normal distribution because of scaling, but apparently I was mistaken. My next attempt was to let $0\leq s<t$ and \begin{align*} \text{Cov}[e^{-s}B(e^{2s}),e^{-t}B(e^{2t})] &= e^{-s-t}\text{Cov}[B(e^{2s}),B(e^{2t})]\\ &= e^{-s-t}E[\{B(e^{2s})\}^2] \end{align*} after a few steps, using the fact that independent blocks of time yield zero mean. If $e^{-s-t}$ is the correct answer, this suggests that $E[\{B(e^{2s})\}^2] = 1$.
My questions are
- Why isn't $U(t)$ standard normal?
- Why $E[\{B(e^{2s})\}^2] = 1$?
Thanks!
\begin{align*} \text{Cov}[U(s),U(t)] &= e^{-s-t}\text{Cov}[B(e^{2s}),B(e^{2t})]\\ &= e^{-s-t}(e^{2s}\wedge e^{2t})=e^{t\wedge s-t\vee s} \end{align*}
Also
$$U(t)\overset{d}{=}e^{-t}\sqrt{e^{2t}}B(1)=B(1)\sim N(0, 1)$$