In Tomas Bjork's Arbitrage Theory in Continuous Time (or here), $\exists$ these propositions

Prop 2.24 Prop 2.25

How does the first formula follow from from the algorithm? I get that $\Pi(0;X) = V_0(0)$, but I don't really get what $E^{Q}[X]$ means...is that equal to $q_uu+q_dd$? Anyway using the algorithm I got $V_0(k) = \frac{1}{(1+R)^T} \sum_{l=0}^{T} V_T(k+l)q_u^{T-l}q_d^{l}$...is $\sum_{l=0}^{T} V_T(l)q_u^{T-l}q_d^{l}$ supposed to be $=q_uu+q_dd$?


1 Answer 1


I believe this is the way that Björk proposes, however I believe "my" way below is more elegant. The trick in Björk's case is to realize that in each "iteration" we get an expeted value:

$$V_0(0) = \frac{1}{1+R}\left(q_u V_{1}(1) + q_d V_1(0) \right) \\ = \frac{1}{(1+R)^{2}} \left( q_u^2 V_2(2) + 2q_uq_d V_2(1) + q_d V_2(0) \right) \\ = \frac{1}{(1+R)^{2}}E^Q[V_2].$$

Continuing in the same fashion you will arrive at

$$V_0(0) = \frac{1}{(1+R)^{T}}E^Q[V_T],$$

however to make this formal you should make some kind of induction argument.

My Method:

My methods does not use Proposition 2.24 but instead the fact that we already know the single period Binomial Model and the Law of total expectation. We already know that Proposition 2.25 holds true if $T=1$ since this reduces to the single-period Binomial model. So assume that $T \geq 2$ and assume that Proposition 2.25 holds true for $T-1$ periods. We when know from the induction assumption that.

$$ \Pi(1; X) = \frac{1}{(1+R)^{T-1}}E^Q[\Phi(S_T)|Z_1] $$


$$ \Pi(0; X) = \frac{1}{(1+R)} E^Q[\Pi(1; X)],$$

and hence

$$ \Pi(0; X) = \frac{1}{(1+R)} E^Q[\Pi(1; X)] = \frac{1}{(1+R)^{T}} E^Q[E^Q[\Phi(S_T)|Z_1]] \\ = \frac{1}{(1+R)^{T}} E^Q[\Phi(S_T)].$$

This concludes the proof of Proposition 2.25

What does $E^Q[X]$ mean:

If $Z_1,...,Z_T$ are independent random variables

$$E[f(Z_1,...,Z_T)] = \sum_{z_1,...,z_T=\text{u or d}} f(z_1,...,z_T)P(Z_1=z_1)\cdots P(Z_T=z_T) \\ = \sum_{z_1,...,z_T=\text{u or d}} f(z_1,...,z_T)p_{z_1} \cdots p_{z_T}. $$

This sum means that we sum over all possible outcomes/paths. Since we often use a martingale/risk neutral probability measure it is convenient to introduce the notation to denote the expectation under the probability measure $Q$.

$$E^Q[f(Z_1,...,Z_T)] = \sum_{z_1,...,z_T=\text{u or d}} f(z_1,...,z_T)Q(Z_1=z_1)\cdots Q(Z_T=z_T) \\ = \sum_{z_1,...,z_T=\text{u or d}} f(z_1,...,z_T)q_{z_1} \cdots q_{z_T}.$$

In your case $X=\Phi(S_T)$ which is a function of $Z_1,...Z_T$. Note also that $S_T = su^Yd^{T-Y}$ where $Y$ is the number of up-moves. The sum above over all outcomes can also be written as in Björk, using binomial coefficients:

$$E^Q[f(Z_1,...,Z_T)] = \sum_{z_1,...,z_T=\text{u or d}} f(z_1,...,z_T)q_{z_1} \cdots q_{z_T} \\ = \sum_{j=0}^T {T \choose j}q_u^j q_d^{T-j}\Phi(su^kd^{T-k})$$.

I hope it all checks out, I'm used to a different notation when working with the Binomial Model!

  • 1
    $\begingroup$ Is $V_T$ the value of the replicating portfolio at maturity? I guess it's equal to the payoff at T since it replicates the claim? Thanks. I think I'll stick with the induction argument since total expectation was not discussed in this class. $\endgroup$
    – BCLC
    Commented Jul 11, 2014 at 21:33
  • 1
    $\begingroup$ Yes, exactly! A naive version of total expectation is usually introduced in a first class in probability theory. See Wikipedia for more info, if you're interested: en.wikipedia.org/wiki/Law_of_total_expectation $\endgroup$ Commented Jul 12, 2014 at 15:18
  • $\begingroup$ Sorry for the confusion. It was discussed in one of our stat classes but not in this class so I guess we must prove it some other way. Thanks! $\endgroup$
    – BCLC
    Commented Jul 12, 2014 at 15:19
  • $\begingroup$ Is $E[V_i]$ the expected value of the replicating portfolio at time i under martingale measure? $\endgroup$
    – BCLC
    Commented Jul 12, 2014 at 15:38
  • 1
    $\begingroup$ Usually you write $E^Q[V_i]$ to denote the expectation under the martingale measure, if it is not clear from the context. $\endgroup$ Commented Jul 12, 2014 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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