Oksendal Chapter 3 Solutions

Exercise 3.1

We are given:

$\sum \Delta (s_j B_j)=\sum s_j \Delta B_j + \sum B_{j+1}\Delta s_j$

Simply take the limit of both sides, and note that the right/left endpoint does not matter in the sum on the far right, because this is a standard Riemann sum of a continuous functionn (path of Brownian motion). Thus we have:

$tB_t = \int_0^t s\ dB_s + \int_0^t B_s\ ds$

Rearranging we conclude:

$\int_0^t s\ dB_s = tB_t - \int_0^t B_s\ ds$

Exercise 3.2

We want to show:

$\int_0^t B_s^2\ dB_s = \frac{1}{3} B_t^3 - \int_0^t B_s\ ds$

By the binomial theorem we have:

$B_{j+1}^3 = (B_j + \Delta B_j)^3 = B_j^3 + 3 B_j^2 \Delta B_j + 3B_j (\Delta B_j)^2 + (\Delta B_j)^3$

Subtracting $B_j^3$ from both sides we have:

$B_{j+1}^3 -B_j^3 = 3 B_j^2 \Delta B_j + 3B_j (\Delta B_j)^2 + (\Delta B_j)^3$

$\sum (B_{j+1}^3 -B_j^3) = \sum 3 B_j^2 \Delta B_j + \sum 3B_j (\Delta B_j)^2 + \sum (\Delta B_j)^3$

Taking the limit of the sum on the left side gives $B_t^3$ . The first sum on the right side becomes $3\int_0^t B_s^2 \ dB_s$ . In the middle sum, we have the quadratic variation of Brownian motion, which becomes $ds$ when passed to the limit, so the second sum on right side becomes $3\int_0^t B_s\ ds$ . Finally, the last sum on the right side has the term $(\Delta B_j)^3= \Delta B_j (\Delta B_j)^2$ . Recall that when we pass this term to a limit, it goes to 0. We can use the “multiplication rule” that $(dB_s) (dB_s)= ds$ but $(ds)(dB_s)=0$ . So we have:

$B_t^3 = 3 \int_0^t B_s^2\ dB_s + 3 \int_0^t B_s\ ds$

We have our result be rearranging the above.

Exercise 3.3-a

Clearly, by properties of condtional expectation we have:

$X_s = E[X_s | \mathcal H_s^{(X)}]$

But since $X$ is a martingale with respect to $N_t$ we have that $X_s = E[X_t | \mathcal N_s]$ . So:

$X_s= E\bigg{[} E[X_t| \mathcal N_s] \bigg{|} \mathcal H_s^{(X)}\bigg{]}$

On the other hand, recall that $\mathcal H_t^{(X)}$ is the smallest $\sigma$ -algebra for which $X$ is measurable, and when considering conditional expectation, “the smallest $\sigma$ -algebra wins”. Thus,

$E\bigg{[} E[X_t| \mathcal N_s] \bigg{|} \mathcal H_s^{(X)}\bigg{]} = E[X_t| \mathcal H_s^{(X)}]$

Thus:

$X_s= E [X_t| \mathcal H_s^{(X)}]$

The other conditions for $X_t$ to be a martingale w.r.t $\mathcal H_t^{(X)}$ follow easily from fact that it is a martingale with respect to $\mathcal N_t$ .

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