Conditional Expectations of Brownian Motion in Derivative Pricing
Summary
The discussion corrects a misunderstanding about conditional expectations in a Brownian filtration. Since Brownian motion is a martingale, the conditional expectation of its future value given information at an earlier time equals its current value. At the current time, the process value is already measurable, so conditioning on that information leaves it unchanged rather than making it zero.
For a geometric Brownian motion, the pricing exercise concerns the expected terminal logarithm conditional on the current asset value. The relevant random component is the future increment of Brownian motion, whose conditional expectation is zero; the Brownian level itself is not zero in expectation conditional on the filtration. This distinction yields a conditional expected log value with drift accumulated over the remaining horizon. The explanation assumes the process is adapted to the stated filtration and clarifies that conditioning on the asset value differs from conditioning on all information in the filtration.
Key ideas
- Brownian motion is a martingale, so its future conditional expectation equals its present value.
- A process value measurable at the conditioning time remains unchanged under conditional expectation.
- The future Brownian increment has conditional mean zero, unlike the Brownian level.
- Derivative pricing must distinguish conditioning on the current asset value from conditioning on the full filtration.
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# Is a wiener proces measurable? (exercise from Bjork)
# Is a wiener proces measurable? (exercise from Bjork)
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
## Answer by Cettt (score 3, accepted)
https://quant.stackexchange.com/a/37726
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.
## Answer by Andrew (score 1)
https://quant.stackexchange.com/a/37721
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).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.