2020年6月26日星期五

Strong Markov property for Brownian Bridge

The strong Markov property for Brownian motion is well known and it is also naturally true for the Brownian Bridge. In fact, since Brownian Bridge ended at $(T, 0)$ is thought as a Brownian Motion conditionned the end point, or thought as a perturbation for a linear interpolation, it is natural when we restart from another mid-point.

To prove it rigorously, for example, the Markov property for the Brownian Bridge, we have to do some calculus. Let $(W_t)_{t \geq 0}$ be the standard Brownian motion issued from $0$, and we have $(B_t)_{s \leq t \leq T}$ defined as 
$$ B_t = x + W_t - W_s - \frac{t-s}{T-s}(x + W_T -W_s) ,$$
a Brownian Bridge between $s, T$ and at $s$ it is $x$ and at $T$ its value is $0$. One way to see this formula is that the term $x + W_t - W_s$ is the Brownian Motion while we have to reduce the term at the endpoint $T$. Some simple calculus shows that it is equal to 
$$B_t = \frac{T-t}{T-s}(x + W_t - W_s) - \frac{t-s}{T-s}(W_T - W_t).$$

A Markov property is very simple but requires calculus: we would like to show that for $s < r < t < T$ we have
$$B_t = B_r + W_t - W_r - \frac{t-r}{T-r}(B_r + W_T -W_r). \quad (\star)$$ 
Now we prove it. An intermediate step tells us 
$$ - \frac{t-r}{T-r}(B_r + W_T -W_r) = - \frac{t-r}{T-s}(x + W_T - W_s)$$ and we put it into the formula that 
$$RHS (\star) = x + W_r - W_s - \frac{r-s}{T-s}(x + W_T -W_s) + W_t - W_r - \frac{t-r}{T-s}(x + W_T - W_s)\\ = x + W_t - W_s - \frac{r-s}{T-s}(x + W_T -W_s) - \frac{t-r}{T-s}(x + W_T - W_s) \\ = x + W_t - W_s - \frac{t-s}{T-s}(x + W_T -W_s).$$ 
This proves the Markov property. Then the strong Markov property is just an approximation and the regularity of the trajectory. 

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