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What is the covariance function for $U(t)$ if $U(t) = e^{-t}B(e^{2t})$ for $t \geq 0$ where $B(t)$ is standard Brownian motion.

Thanks for the help!

1 Answers 1

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Note that for $s \leq t$

$\Bbb{E}[U_t U_s] = e^{-t} e^{-s}\Bbb{E}[B(e^{2t}) B(e^{2s})] = e^{-t} e^{-s} e^{2s} = e^{-(t-s)}$

For general case

$\Bbb{E}[U_t U_s] = e^{-|t-s|}$

So you can see that your process is a gaussian process with mean zero, variance 1 and covariance given by $C(t,s) = \Bbb{E}[U_t U_s] = e^{-|t-s|} $

Compare this with the definition of Walsh

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