# Computing probabilities of standard Brownian motion

Let $$W_t$$ be a standard Brownian motion.

a. Find $$P(-3\leq2W_2-3W_5\leq5)$$

b. Find the variance of $$W_2-3W_3+2W_5$$

c. Find $$P(-2\leq W_2\leq 3\mid W_1=1)$$

d. Find $$P(-2 \leq W_3-W_4 \leq 1 \mid W_1=1)$$

e. Find $$\text{cov}(W_s, W_t^{2})$$

I have for part a)

$$W_2 \sim N(0,2) \implies 2W_2 \sim N(0,8)$$

$$W_5 \sim N(0,5) \implies 3W_5 \sim N(0,45)$$

Then $$2W_2-3W_5 \sim N(0,8+45-6(2))=N(0,41)$$, and so

$$P(-3\leq2W_2-3W_5\leq5)=P(-3/ \sqrt{41} \leq z \leq 5/ \sqrt{41})= .462854$$

For part b I have

\begin{align*} \text{var}(W_2-3W_3+2W_s)&=\text{var}(W_2)+9\text{var}(W_3)+4\text{var}(W_5)\\ &\quad -3\text{cov}(W_2,W_3)+2\text{cov}(W_2,W_5)-6\text{cov}(W_3,W_5) \\ &=\text{var}(W_2)+9\text{var}(W_3)+4\text{var}(W_5)\\ &\quad -3\text{var}(W_2)+2\text{var}(W_2)-6\text{var}(W_3) \\ &= 3\text{var}(W_3)+4\text{var}(W_5)=3(3)+4(5)\\ &=29 \end{align*}

For part e, I know that $$\text{cov}(X_s, X_t) = \sigma^{2} \min(s,t)$$ so then it should be $$\text{cov}(W_s, W_t^{2})= \sigma^{2} \min(s,t^{2})$$

I am not sure if this is the correct approach and how to do parts c and d

• Is $W_{5}$ independent of $W_{2}$? If not, how would that change your calculations? – Brian Borchers Mar 11 at 18:19
• $W_{t}^2 \neq W_{t^2}$. – user159517 Mar 11 at 20:05

For b) I recall that if $$X$$ and $$Y$$ are two random variables then $$Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)$$. Then \begin{align*} Var(W_2 - 3W_3 + 2W_5) = Var(W_2) + 9Var(W_3)+ 4Var(W_5) + 2\left(Cov(W_2,-3W_2) + Cov(W_2, 2W_5) + Cov(-3W_3,2W_5)\right) \end{align*} For c and d) perhaps we can use the Gaussian and independent increment properties of the Brownian motion, i.e. if $$s \leq t$$, then $$W_t - W_s \sim \mathcal{N}(0, t-s)$$ and is independent of $$W_s$$.
For e) suppose that $$s\leq t$$ \begin{align*} Cov(W_s, W_t^2) &= E[W_sW_t^2] - E[W_s]E[W_t^2] \\ &= E[W_t^2(W_t-W_s)] - E[W_t^3] \\ &= E[W_t^2]E[W_t-W_s] \\ &= 0 \end{align*} By symmetry, we have than $$s,t\in \mathbb{R}^+, Cov(W_s, W_t^2) = 0$$. I used the independence of increments.