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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:

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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)

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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

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  • $\begingroup$ Again thanks for the response. It must be that reading various resources on the topic may have confused me more than helped. I'm carrying on about variance. In Ross' book section 10.3.1 begins about brownian motion with a drift as follows: "We say that {X(t), t >= 0} is a Brownian motion process with drift coefficient μ and variance parameter σ^2 if:" And then they carry on about the three requirements. Here variance is not t but σ^2. In the original question B(1/t) is immediately mentioned as having variance 1/t. This is the point of confusion. Likewise, wouln't X(t) have variance t? $\endgroup$ – fullofquestions Nov 27 '16 at 19:20
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If $var(X(t))=t $, then $var(aX(t))=a^2t$, $var(a+X(t))=t$, $var(X (at))=at$, $var(X (t+a))=t+a$.

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  • $\begingroup$ Thank you. OK, I see that the input in the '()' tells you what the variance is for a standard brownian motion. $\endgroup$ – fullofquestions Nov 27 '16 at 6:57
  • $\begingroup$ I see you added the question (b) about $E[B(u)B(v)]=min(u,v)$. I haven't worked it out because it is difficult, but I think here they are implying u and v are dependent random variables. Otherwise $E[B(u)B(v)] = E[B(u)]E[B(v)]=0* 0 = 0$. $\endgroup$ – Alex Nov 27 '16 at 7:37

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