How to find the Expectation and Covariance of a Process? 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!
 A: 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 $s,t\in[0,1]$.
A: If $\{X(t), t\geq 0\}$ be an independent increment process, then it has the following properties:


*

*$Cov(X(s),X(t)) = min\{s,t\}$.

*$Var(X(t)-X(s)) = Var(X(t))-Var(X(s))$.



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 $Var(B(t))=t$.

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*}
