Conditional Brownian Motion Probabilities and Brownian Bridges
Summary
The document finds the conditional probability that Brownian motion at time 1 is below a threshold, given its value at time 2. One answer treats the two observations as a jointly normal vector and applies the conditional normal distribution, giving a mean of 1 and variance of one-half for the time-1 value under the stated condition. Another frames the same conditional path as a Brownian bridge pinned at the observed endpoint, whose intermediate values have a normal distribution.
A further answer constructs a linear combination of the two Brownian values that is uncorrelated with the endpoint, then uses joint normality to infer independence. The approaches illustrate Gaussian conditioning and bridge reasoning for stochastic processes. The discussion is narrowly focused on this probability calculation; it does not address trading applications, and the final response’s displayed probability expression appears to use a different threshold from the question, so readers should derive the target probability from the stated conditional distribution.
Key ideas
- Brownian motion values at different times form a jointly normal vector with covariance determined by the shared path.
- Conditioning on the value at time 2 gives a normal distribution for the value at time 1.
- A Brownian bridge describes the path between fixed starting and ending values.
- For jointly normal variables, zero covariance between a linear combination and the endpoint implies independence.
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Full text
# Browian motion: $P(B_1<4 | B_2 =1)$
# Browian motion: $P(B_1<4 | B_2 =1)$
I want to calculate $P(B_1<4 | B_2 =1)$ for the B.M.
What I tried: $P(B_1<4 | B_2 =1)=P(B_1 - B_2 < 3- B_2 | B_2 =1)$
but I cant use any independence to calculate further.
## Answer by Andrew (score 5)
https://quant.stackexchange.com/a/39395
Consider the multivariate normally distributed vector $(B_1,B_2)$, in particular $ \begin{pmatrix} B_1 \\ B_2 \\ \end{pmatrix} $ ~$N $$\left(\begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ \end{pmatrix} \right) $.
Then it holds that $B_1 |B_2=1$ ~$ N(1,1/2) $ (wikipedia).
## Answer by Richi Wa (score 4)
https://quant.stackexchange.com/a/39404
This is related to the concept of the Brownian Bridge.
If $B_t$ is Brownian motion then $W_t$ defined as $$ W_t = (B_t|B_T = 0) $$ is the Brownian bridge with end point $0$. There is version with $a$ and $b$ too.
You know where it starts and where it ends (that's why is the bridge) and there is uncertainty inbetween. One could read-up on the topic. If I just quote wikipedia then for $B_(t_1) = a$ and $B_(t_2) = b$ the Brownian bridge is normal with mean $$ a + \frac{t-t_1}{t_2-t_1}(b-a) $$ and covariance between $W(s)$ and $W(t)$ $$ \frac{(t_2-t)(s-t_1)}{t_2 - t_1}. $$
Thus in your case we tie the Brownian bridge at $B_0 = 0$ then $t_1 = 0$ and $a=0$, $t_2 = 2$ and $b = 1$ and you have to find the probability that the Brownian bridge at time $1$ is less than $4$.
## Answer by Focus (score 1)
https://quant.stackexchange.com/a/39408
Here is what I found:
$P(B_1<4 |B_2 =1) =P(2B_1 -B_2 < 7 |B_2 =1) =P(2B_1 -B_2<7 )= P(\sqrt{2} Z<7 ) = N(7/ \sqrt{2})$ where $Z$ is the standard normal r.v.. I have used the fact $2B_1 -B_2 $ is indep of $B_2$ since they are multivariate normal and their covariance is 0.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.