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I have a question about formula in a script about mathematical finance in finite discrete time. We are looking at the Cox-Ross-Rubinstein model, i.e. We have a bank account with undiscounted value $S^0_k=(1+r)^k$, where $k=0,\dots,T$, and a risky asset (undiscounted) $S^1_k=\prod_{j=1}^kY_j$, where $Y_j$ describes the change of the risky asset in every step, i.e. $Y_j$ are iid under a measure $P$ and taking values $1+u$ or $1+d$ with probability $p$ and $1-p$. It's this usual binomial tree model. The discounted values are just given by $S^0_k/S^0_k$ and $S^1_k/S^0_k$. The sigma algebra is generated by $S^1$ or equivalently by the $Y$'s. In each step the risky asset $S^1$ can either go up $(1+u)$ or go down $(1+d)$ with corresponding probabilities.

We study the pricing of a new financial instrument, i.e. a r.v. $H\in \mathcal{F}_T$ which is positive. The price process of $H$ is then given by $$V_k=E_Q[H|\mathcal{F}_k]$$ where $Q$ is the unique equivalent martingale measure for $S^1$. The $Y_j$'s are again iid under $Q$ with $Q[Y_1=1+u]=q_1$

By definition, $V$ is a martingale under $Q$. Suppose we fix some node at time $k-1$ and let $v_{k-1}$ be the value of $V_{k-1}$ there. Then $V_k$ can take two values at the successor nodes, i.e. $v_k^u$ and $v_k^d$. Now here is my question: The say: The $Q$ martingale property says

$$v_{k-1}=q_1v^u_k+(1-q_1)*v_k^d$$

I do not see how we get this formula, which is just the expectation under $Q$. By definition $V_k$ is a r.v. but here we treat it as a number. This would be true, if we have independence of $\mathcal{F}_k$ which I do not see. It would be very helpful if someone could show me how to derive this equation. Thanks in advance.

hulik

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Which mathematical finance script are you seeing this in? It might help us if we can look at their exposition of the first part, also. –  Alexander Gruber Dec 18 '12 at 18:52
    
@AlexanderGruber The script is from a course at my university and not online available. –  user20869 Dec 18 '12 at 19:32
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