Question.

Let V be the value of a derivative written on an asset whose time t=o value is $\displaystyle S$ and whose value in the next period at time t=1 will be $\displaystyle Su$ or $\displaystyle Sd$, where $\displaystyle d<1=r<u$ and $\displaystyle r$ is the interest rate for the period. Suppose the derivative value in the next period is correspondingly $\displaystyle V_u$ and $\displaystyle V_d$. Show by constructing an appropriate portfolio that

$\displaystyle (1+r)V=\frac{(V_u-V_d)(1+r)+V_dd-V_ud}{u-d}$.

The background to this question is a chapter in Mathematics for Finance and Valuation entitled "One period, one risky asset and two states" where the risk-neutral probability was introduced and used to value the forward contract and call and put options.

I cannot get to that equation. My attempt at solution:

First, I find state probabilities (eg $\displaystyle q_u$ is the probability of "up" state and so on) from the current price of asset S:

$\displaystyle Suq_u+Sdq_d=S

q_u+q_d=1$

$\displaystyle Suq_u+Sd(1-q_u)=S$ eliminate S

$\displaystyle q_uu+d-q_ud=1$

$\displaystyle q_u=\frac{1-d}{u-d}$

$\displaystyle q_d=1-q_u=\frac{u-1}{u-d}$

Then I assume I can use the same probabilities to value the derivative claim on the asset S (can I?). The value of the derivative in one year's time V(1+r) should be

$\displaystyle V(1+r)=V_uq_u+V_q_d=V_u\frac{1-d}{u-d}+V_d\frac{u-1}{u-d}=\frac{V_u(1-d)+V_d(u-1)}{u-d}=\frac{(V_u-V_d)+V_du-V_ud}{u-d}$

So my answer is different from the instruction by one component: I have $\displaystyle V_du$ and they have $\displaystyle V_dd$. Any comments?

Another question, does interest r=1 make sense?... Is that 100%?