How to find the Expectation and Covariance of a Process?

Question

How would one find the expectation and thus the covariance of a process, say,

$$X(t)=\alpha +B(t)-tB(1)$$

is the process and $t\in[0,1]$ and $\alpha\in\mathbb{R}$?

When considering this process would you take a discrete time method at both end points or a continuous time over the entirety of $[0,1]$?

I have assumed that $(B(t))_{t\in [0,1]}$ is a Brownian motion in my mathematics.

Any help explaining the lines are arguments would be greatly appreciated thanks!

Answer 1

Basically, we calculate the expected value, the variance and the autocovariance of a process by calculating the expected value and the variance of the random variable $X(t)$ and the covariance between the variables $X(s)$ and $X(t)$.

Hence, the expectation of the process is given by $$ \operatorname EX(t)=\alpha+\operatorname EB(t)-t\operatorname EB(1)=\alpha $$ since $\operatorname EB(t)=0$ for each $t\in[0,1]$. The variance of the process is given by \begin{align*} \operatorname{Var}X(t) &=\operatorname{Var}[\alpha+B(t)-tB(1)]\\ &=\operatorname{Var}[B(t)-tB(1)]\\ &=\operatorname E[B(t)-tB(1)]^2\\ &=\operatorname E[B^2(t)-2tB(t)B(1)+t^2B^2(1)]\\ &=t(1-t) \end{align*} using the fact that $\operatorname EB^2(t)=t$ and $\operatorname E[B(s)B(t)]=\min\{s,t\}$. Finally, \begin{align*} \operatorname{Cov}[X(s),X(t)] &=\operatorname E[(B(s)-sB(1))(B(t)-tB(1))]\\ &=\operatorname E[B(s)B(t)]-t\operatorname E[B(s)B(1)]-s\operatorname E[B(1)B(t)]+st\operatorname EB^2(1)\\ &=\min\{s,t\}-st \end{align*} for $s,t\in[0,1]$.

Answer 2

If $\{X(t), t\geq 0\}$ be an independent increment process, then it has the following properties:

1. $Cov(X(s),X(t)) = min\{s,t\}$.
2. $Var(X(t)-X(s)) = Var(X(t))-Var(X(s))$.

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By your supposition, $\{B(t), t\geq 0\}$ is a standard Brownian motion process, for $t \in [0,t]$ and a process is defined by $X(t)=\alpha + B(t) - tB(1)$. By the definition of standard Brownian motion process, we know that, it has stationary and independent increments and

$B(t)\sim N(0,t)$, meaning that, $E(B(t))=0$ and $Var(B(t))=t$.

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Expectation:

\begin{equation} E(X(t))=E(\alpha + B(t) - tB(1)) = \alpha. \end{equation} since $E(B(t))=0$ and $E(tB(1))=tE(B(1))=0$.

Variance:

\begin{eqnarray*} Var(X(t)) &=& Var(\alpha + B(t) - t\;B(1)) = Var( B(t) - t\;B(1))\\ &=& Var(B(t)) - t^2 Var(B(1)) = t - t^2 = t(1-t).\quad \mbox{by prop-2 above} \end{eqnarray*}

Covariance: \begin{eqnarray*} Cov\{X(s),X(t)\} &=& Cov\{B(s)-s\;B(1), B(t)-t\;B(1)\} \\ &=& Cov\{B(s),B(t)\} - Cov\{B(s),t B(1)\} - Cov\{sB(1),B(t)\} + Cov\{sB(s),tB(t)\}\\ &=& Cov\{B(s),B(t)\} - t\; Cov\{B(s), B(1)\} - s\; Cov\{B(1),B(t)\} + st\; Cov\{B(s),B(t)\}\\ &=& min\{s,t\} - t\; min\{s,1\} - s\; min\{1,t\} + st\; min\{s,t\}\\ &=& min\{s,t\} - ts - st + st\\ &=& min\{s,t\} - st. \end{eqnarray*}