Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to get the conditional distribution of $W(t/2)$ given $W(t)=x$ where $W(t)$ represents wiener process.

This was a problem in my exam and i couldn't think how to start :(

Any help!!

Thanks in advance.

share|improve this question
2  
What is stopping you in the usual approach? –  Did Sep 2 '12 at 12:48
1  
$W(t)=W(t/2)+(W(t)-W(t/2))$, where the two summands are independent and normally distributed with mean $0$ and variance $t/2$. Does that help? –  Harald Hanche-Olsen Sep 2 '12 at 12:59
    
@HaraldHanche-Olsen Not sure this is the most direct way. –  Did Sep 3 '12 at 9:41
    
@did: I could have added that when $X$ and $Y$ are independent normally distributed variables with mean 0 and the same variance, then $X+Y$ and $X-Y$ are normal and independent as well. If you assume this known, the desired result should be right around the corner. –  Harald Hanche-Olsen Sep 4 '12 at 11:01

2 Answers 2

Joint distribution of $(W(t/2), W(t))$ is a binormal distribution with zero means, and covariance: $$ \mathbb{Var}(W(t/2)) = \frac{t}{2}, \quad \mathbb{Var}(W(t)) =t, \quad \mathbb{Cov}(W(t/2),W(t)) = t/2 $$ translating into $\rho = \frac{\sqrt{2}}{2}$, $\sigma_1^2 = \frac{t}{2}$, $\sigma_2^2 = t$. Now, conditional distribution of binormal is well known: $$ W(t/2)|W(t)=x \sim \mathcal{N}\left(\frac{x}{2}, \frac{\sqrt{t}}{2} \right) $$ Alternatively, you could have used that the Wiener process conditioned upon $W(t)=x$ gives the Brownian bridge process.

share|improve this answer
    
Replace the variance t/2 by t/4 in the final formula. –  Did Sep 3 '12 at 9:33
    
As an aside, note that this is how Paul Lévy built Brownian motion on [0,1], using successive piecewise-linear approximations X_n: first choose W(1) centered normal with variance 1, this yields X_1(t)=tW(1) for every t in [0,1]. Second, choose W(1/2)-X_1(1/2) independently centered normal with variance 1/4, this yields X_2 linear on [0,1/2] and [1/2,1] and X_2(t)=W(t) for t=0, t=1/2, t=1. Repeat at 1/4 and 3/4 using variances 1/16, and so on. Then X_n(t) converges almost surely to some W(t) and the path t --> W(t) is indeed a Brownian motion. –  Did Sep 3 '12 at 9:38
    
@did Thank you for catching the typo. Using $\mathcal{N}\left(\mu, \sigma\right)$ notation, I should have written $\sqrt{t/4}$. Incidentally, the book of Yuval and Peres, "Brownian motion" (available from one of the author's homepage), constructs the Brownian motion exactly this way. –  Sasha Sep 3 '12 at 12:56
    
Interesting. I probably knew it then forgot it. (But Yuval and Peres are one same individual...) –  Did Sep 3 '12 at 14:40
    
@did An embarrassing slip. I should have said the book of Mörters and Peres. Sorry about that. –  Sasha Sep 3 '12 at 15:03

Here is the trick. Find $\alpha$ such that $W_{t/2}+\alpha W_t$ and $W_t$ are independent i.e. -since $W$ is a Gaussian process with zero mean- such that $$ \mathbb{E} \left[( W_{t/2}+ \alpha W_t) W_t\right] = 0 $$ You find $\alpha = -1/2$ using $\mathbb{E}\left[ W_a W_b \right] = a \wedge b$. Then

$$ \mathbb{E} \left[ W_{t/2} \,\big|\, W_t = x \right] = \mathbb{E} \left[ W_{t/2} - \frac{1}{2} W_t + \frac{1}{2} W_t \,\big|\, W_t = x \right] = \underbrace{\mathbb{E} \left[ W_{t/2} - \frac{1}{2} W_t \right] }_{=0}+ \frac{1}{2} x = \frac{1}{2} x$$

More generally we have (using the same trick) the famous interview question $$ t \le s, \quad \mathbb{E} \left[ W_t \,\big|\, W_s \right] = \frac{t}{s} W_s $$

share|improve this answer
    
Conditional expectation $\ne$ conditional distribution. You provide the former, which is not enough to deduce the latter. –  Did Sep 9 '12 at 17:27
    
Oh misread sorry. –  vanna Sep 10 '12 at 8:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.