Distribution of “intermediate value” brownian motion.

brownian motionprobability theory

The question:

Suppose that $(B_t)_{t≥0}$ is a Brownian motion. Fix $0 ≤ s < t < ∞.$ Show that conditionally on
$ \{ B_s = x, B_t = z\}$ the intermediate value $B_{\frac{t+s}{
2}}$
has Gaussian distribution with mean $\frac{x+z}{
2}$
and
variance $\frac{t−s}
{4}$
.

My Attempt:

Let $Z_1 := B_t – B_{\frac{t+s}{2}}$ and $Z_2 := B_{\frac{t+s}{2}} – B_s$ (they are also independent)

Then by definition of brownian motion we have $Z_1 \sim N(0,\frac{t-s}{2})$ and $Z_2 \sim N(0,\frac{t-s}{2})$

Notice that $Z_3 := \frac{1}{2} (Z_2 – Z_1)$ is the linear combination of independent gaussian R.Vs and such is gaussian itself.
Infact we have $Z_3 \sim N (0,\frac{t-s}{4})$

However notice that by its definition $Z_3 = B_{\frac{t+s}{2}} – \frac{1}{2}(B_s+B_t)$

So we are left with $B_{\frac{t+s}{2}} = Z_3 + \frac{1}{2}(B_s+B_t)$ and then conditioned on the event $ \{ B_s = x, B_t = z\}$ we have :

$B_{\frac{t+s}{2}} = Z_3 + \frac{1}{2}(x+z)$

Notice $x \sim N(x,0)$ and $z \sim N(z,0)$

So finally $B_{\frac{t+s}{2}} \sim N(0,\frac{t-s}{4}) + N(\frac{x}{2} , 0 ) + N(\frac{z}{2} , 0 ) \sim N(\frac{x+z}{2},\frac{t-s}{4})$ (linear sum of gaussian's are guassian)

Is this all okay? Is $Z_3$ independent of our event on $B_t$ and $B_s$ ? Is there a quicker way of doing this ? 🙂

Thanks

Best Answer

This is OK, except that you still need to explain why $Z_3$ is independent of $B_s+B_t$ (Hint: Uncorrelated Gaussians are independent.) Also, there is no need to write constants as Gaussians with variance zero.

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