I will claim $$E[W(T) \vert F_t] = 0$$ for $t<T$. Anyway, in an exercise in Bjork the results requires that $$E[W(t) \vert F_t] = 0$$ But why? Isn't $W(t)$ measurable at time $t$ and hence not necessarily $0$? $W$ is of course a Wiener Process.

More precisly: $$F(t,x)=E[2*\ln(x) \vert F_t],$$ $$X(T)=\exp \left \{\ln(X(t)) + c(T-t)+\sigma[W(T)-W(t)] \right \}$$ where $c$ is come constant and $X(t)=x_t$. The result is then $$F(t,x)=2\ln x_t + 2c(T-t)$$ This can only be true if $E[W(t) \vert F_t] = 0$. Why is that?

I am talking about Exercise 5.9 in Bjork, Arbitrage Theory Continous Time Finace and the result is sketched here on page 8 http://www.maths.lth.se/matstat/kurser/fmsn25masm24/ht11/Bjork_sol.pdf


I think you mixed several things up. I will try to help you out.

Everything started with your claim that $\Bbb E \bigl[W(T) \mid \mathcal F_t \bigr] = 0$ which is wrong!

if $W$ is a Brownian notion, then

$$ \Bbb E \bigl[W(T) \mid \mathcal F_t \bigr] = W(t), \quad t\leq T. $$ This follows from the fact that Brownian motions are martingales. Here and in everything that follows, I assume that $\mathcal F$ is the filtration such that $W$ is adapted to $\mathcal F$.

This brings us to the second issue. $$ \Bbb E \bigl[W(t) \mid \mathcal F_t \bigr] = W_t, $$ which is just a special case of my first equation. Alternatively, you could also argue that $W(t)$ is $\mathcal F_t$ measurable.

The next smaller issue is the definition of your function $F$. Your definition of $F$ and the definition from the pdf file differ. In the pdf file we have that

$$ F(t,x) = \Bbb E ^{t,x} \Bigl[2 \ln \bigl(X(T)\bigr) \Bigr] = \Bbb E\Bigl[2 \ln \bigl(X(T)\bigr) \mid X(t) = x\Bigr]. $$

Your definition of $F(t,x)$ would simplify to $F(t,x) = 2 \ln(x)$.

Last but not least I will show that $F(t,x) = 2 \ln (x) + 2(\mu - \frac 12\sigma^2) (T-t)$.

Therefore, note that $$ \ln (X(T)) = x + (\mu - \frac 12\sigma^2) (T-t) + \sigma(W(T)-W(t)), $$ so it remains to show that $$ \Bbb E^{t,x} \Bigl[W(T) - W(t) \Bigr] = 0. $$ But this follows (almost) from my first and second equation.

  • $\begingroup$ Thanks! Yeah I messed up by not being aware of martingale principle. I get it now $\endgroup$
    – Sanjay
    Jan 13 '18 at 21:18

From where do you know that $E[W(T)|F_t]=0$? When $W(t)$ is a Wiener Process with respect to $F_t$ it holds that $E[W(T)|F_t]=W(t)$ (because $W(t)$ is a martingale with respect to that filtration).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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