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).


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