I'm studying for an exam and I came across a problem where I'm not sure about a few pieces of the solution. The problem goes as follows:
---start
Let $\{B(t), t \ge 0\}$ be a standard Brownian motion.
Let$ Y(t)= tB(1/t),t \gt 0, Y(0)=0$.
(a) What is the distribution of Y(t)?
(b) Compute Cov(Y(s),Y(t))
Solution:
a) "Since B(1/t) has a normal distribution with mean 0 and variance 1/t, we have"
$P(Y(t) <= y) = .... P(B(1/t) \le y/t)$
Then they normalize dividing by $\sqrt{1/t}$ to use the Phi standard normal lookup table
b) Since E[Y (t)] = 0 and E[B(u)B(v)] = min(u, v),
$Cov(Y(s),Y(t)) = E[Y(s)Y(t)] - E[Y(s)]E[Y(t)]$
$= ...$
$= stE[B(1/s)B(1/t)]$
= $st*min(1/s,1/t)$
= min(t,s)
---end
Question 1. Regarding part a). Since B(t) is a standard Brownian motion I believe that it is Normally distributed as follows ~ N(0,t), i.e. with mean 0 and variance t. In this book, Ross' Introduction to probability models, a normal distribution is defined as ~$N(0,tc^2)$ and a standard brownian motion has c = 1. Why is the variance 1/t? Shouldn't it be t? I Perhaps it is a coincidence but it seems that they are assigning variance to the input parameter, i.e. 1/t, to the standard Browninan motion.
Question 2:
In part b the solution makes the statement 'E[B(u)B(v)] = min(u, v)'. Where does that come from? It is useful though...
Question 3:
This is regarding part c. Don't want to bore you with the details but part of it goes:
$Cov(Y(s), Y(t)-Y(s)) = Cov(Y(s), Y(t))- Cov(Y(s),Y(s))$
$ = min(s, t)-s = 0, s \lt t$
I thought that Cov(X,X) = 1 but here it is s. Can you clarify why? Also I believe that the min(s,t) expression here is related to my question 2 above.
Thanks

If $var(X(t))=t $, then $var(aX(t))=a^2t$, $var(a+X(t))=t$, $var(X (at))=at$, $var(X (t+a))=t+a$.