Expectation and autocorrelation function of a Wiener process (Brownian motion)

brownian motionexpected valueprobabilitystochastic-processes

Let $\{W(t): t \geqslant 0\}$ be a Wiener process (Brownian motion) with variance parameter $σ^2 = 1$.
Define the process $\{Z(t): t \geqslant 0\}$ by $Z(t) = e^{-ct} W(e^{2ct})$, with $c > 0$.

  1. We must compute the expectation function of $Z(t)$.
    I was wondering if $Z(t)$ is also a Wiener process, which would imply that its expectation function is always equal to $0$.

  2. We must compute the autocorrelation function of $Z(t)$.
    I was wondering if I can use the independent increments property of a Wiener process to split the formula of autocorrelation as follows : $R(t, t+T) = E(Z(t) Z(t+T)) = E(Z(t)) E(Z(t+T))$.

Best Answer

For the first question, we have \begin{align*} \mathbb{E}[Z(t)] &= \mathbb{E}[e^{-ct}W(e^{2ct})] = e^{-ct} \mathbb{E}[W(e^{2ct})] = 0. \end{align*}

For the second question, we have \begin{align*} \mathbb{E}[Z(t)Z(t+T)] &= \mathbb{E}[(e^{-ct}W(e^{2ct}))\cdot (e^{-c(t+T)}W(e^{2c(t+T)})) ] \\ &= e^{-2ct-cT} \mathbb{E}[W(e^{2ct})W(e^{2c(t+T)})]. \end{align*} Now, to compute $\mathbb{E}[W(e^{2ct})W(e^{2c(t+T)})]$, we have \begin{align*}\mathbb{E}[W(e^{2ct})W(e^{2c(t+T)})]&= \mathbb{E}[W(e^{2ct})(W(e^{2c(t+T)})-W(e^{2ct}) + W(e^{2ct}))] \\ &= \mathbb{E}[W(e^{2ct})(W(e^{2c(t+T)})-W(e^{2ct}))] + \mathbb{E}[W(e^{2ct})^2]. \end{align*} Since Brownian motion has independent increments, we know that $W(s)$ is independent of $W(S)-W(s)$ for all $0 \le s \le S$. Applying this to the first term with $s = e^{2ct}$ and $S = e^{2c(t+T)}$, we see

\begin{align*} \mathbb{E}[W(e^{2ct})(W(e^{2c(t+T)})-W(e^{2ct}))] &= \mathbb{E}[W(e^{2ct})] \mathbb{E}[W(e^{2c(t+T)})-W(e^{2ct})] = 0. \end{align*}

Since $W(e^{2ct}) \sim N(0,e^{2ct})$, we know that $\mathbb{E}[W(e^{2ct})^2] = e^{2ct}$, and so \begin{align*} \mathbb{E}[W(e^{2ct})W(e^{2c(t+T)})]&= e^{2ct}. \end{align*} Thus, we have

\begin{align*} \mathbb{E}[Z(t)Z(t+T)] &= e^{-2ct-cT} \mathbb{E}[W(e^{2ct})W(e^{2c(t+T)})] \\ &= e^{-2ct - cT} (e^{2ct}) \\ &= e^{-cT}. \end{align*}

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