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Let $B(t)$ denote the standard Brownian motion and let $X(t)$ denote a Brownian motion with $X(0)=0$, drift $0$ and variance $9$.

  • Find the distribution of $aB(s)+bB(t)$, where $a,b,s,t$ are real numbers and $0<s<t$

  • Find $\Bbb P(X(2) -2X(3) ≤ 4)$

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hint: Suppose that $t>s$ (reverse the logic otherwise). Try $B(t)=B(t)-B(s)+B(s)$. What can you say about $B(t)-B(s)$ and $B(s)$? –  Alex R. Nov 3 '12 at 18:56
    
Please do not deface your question cancelling parts of it after some answers mention them. –  Did Nov 13 '12 at 19:57

1 Answer 1

up vote 1 down vote accepted

We write $$aB(s)+bB(t)=b(B(t)-B(s))+bB(s)+aB(s)=b(B(t)-B(s))+(a+b)B(s).$$ As $B(t)-B(s)$ and $B(s)$ are independent, and normally distributed of respective laws $N(0,t-s)$, and $N(0,s)$ (definition of Brownian motion), $aB(s)+bB(t)$ is normally distributed, of mean $0$ and variance $$b^2(t-s)+(a+b)^2s=b^2t-sb^2+a^2s+2abs+b^2s=b^2t+a^2s+2abs.$$

The second question almost follows from the first one.

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Thanks! I got it. The second question I did myself and got it. –  Alvin Nov 3 '12 at 19:25
1  
The variance is incorrect. –  Did Nov 13 '12 at 19:58
    
@did Thanks for pointing out the miscomputation. –  Davide Giraudo Nov 13 '12 at 20:01

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