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Let $X(t), t \geq 0$ be a Brownian motion process with drift parameter $\mu = 2.5$ and variance $\sigma^{2} = 8$. If $X(0) = 20$, find

(a) $E(X(3))$

(b) $\mathrm{Var}(X(3))$

(c) $P(X(3) > 30)$

I am reading a book, and I am stuck on that exercise.


$X(3) - X(0)$ has a drift parameter of $7.5$ and variance of $24$ and it is normally distributed. Then, we can add $20$ to each side to get that $X(3)$ has an expectation of $27.5$? Is that right?

Then, $X(3) - X(0)$ has a variance of $24$. I don't know if I can just add $8$ to each side to separate the variances though, since it's not linear. If I can, it would just be $32$. Is that right?

I have no clue how to do $(c)$. I think it will have to do something with the CDF of the normal distribution though. Can someone please help me?

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  • $\begingroup$ From the formula for variance $E(X-EX)$ its clear that adjusting $X$ by a constant won't affect the variance (for b) $\endgroup$ – Calvin Khor Mar 4 at 11:52
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your first calculation is correct.

For the second one it could be helpful to write $X$ like this $$ X(t) = 20 + 2.5 \cdot t + \sqrt{8} \cdot B(t), $$ where $B$ is a standard Brownian motion. We know that $Var(B(t)) = t$ for all $t \geq 0$ hence $$ Var(X(3)) = Var(\sqrt{8} \cdot B(3)) = 8 \cdot 3 = 24. $$

For your third question just note that $X(3)$ has a normal distribution and according to the first two exercises we must have $X(3) \sim N(27.5, 24)$

Therefore,

$$ \Bbb P (X(3) > 30) = 1 - \Phi (\frac{30 - 27.5}{\sqrt{24}}). $$

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  • $\begingroup$ Hi Cettt, in the same problem, there is a part $(d)$ that asks how to find $P(X(0.5) > 10)$. How can I do this? I found that $\mathbb{E}[X(0.5)] = 21.25$, and using your method of rewriting $X$ as $X(t) = 20 + 2.5t + \sqrt{8}B(t)$, I got $\text{Var}(X(0.5)) = 8 \cdot 0.5 = 4$. Is this correct? $\endgroup$ – user614735 Mar 6 at 12:35

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