# Proof of the value of an option using hedging and no-arbitrage [ Paul Wilmott Chapter 3.12.2]

I encounter a difficulty in understanding the proof of finding the value of an option. Before going into the proof, let's talk above the assumptions and parameters of the model. Assume that we know the value of an option at $$t + \delta t$$ and we now construct a portfolio to at time $$t$$ consisting one option and short $$\Delta$$ underlying. At time t, the portfolio has value

$$\Pi = V - \Delta S \tag{1}$$ where $$V$$ is the option value worths at time $$t$$ which is also an unknown. At time $$t+\delta t$$, there are two possible outcomes, either the asset rises $$V^+$$ or falls $$V^-$$. Therefore, the portfolio becomes

$$V^+ - \Delta u S ~~\text{or} ~~ V^- - \Delta v S \tag{2}$$ where $$0. Using hedging, we know that the $$\Delta$$ is $$\Delta = \frac{\text{Price Range of Option}}{\text{Price range of Underlying}} = \frac{V^+ - V^-}{(u-v)S} \tag{3}$$ Hence, we can rewrite $$V^+ - \Delta u S$$ and $$V^- - \Delta v S$$ as

\begin{align} V^+ - \Delta u S = V^+ - u\frac{V^+ - V^-}{(u-v)} \tag{4} \\ V^- - \Delta v S = V^- - v\frac{V^+ - V^-}{(u-v)} \tag{5} \\ \end{align}

The last step is to use the no-arbitrage argument to show that in change in portfolio equals to the interest earned at risk-free rate. Therefore, the change in portfolio is

$$\delta \Pi = r \Pi \delta t \tag{6}$$

Now, the following proof is that part that I don't understand. Paul Wilmott tried to represent the portfolio at $$t + \delta t$$ in terms of $$V^+, V^-, V, S, u, v, r$$. First, he rewrites the portfolio value into $$V^+ - \Delta u S$$ by

\begin{align} (1+ r \delta t) \Pi = V^+ - u\frac{V^+ - V^-}{(u-v)} \tag{7}\\ \Pi = V - \Delta S = V - \frac{V^+ - V^-}{(u-v)} \tag{8} \\ \end{align}

Originally, I thought he would plugin eq(8) into eq(7) and that's it. However, he used $$V^- - \Delta v S$$ instead.

$$(1 + r \delta t) \Big [ V - \frac{V^+ - V^-}{(u-v)} \Big ] = V^- - v\frac{V^+ - V^-}{(u-v)} \tag{9}$$

At this moment, I cannot follow his logic and don't understand why he does not plugin the definition of $$\Pi$$ in eq(8) into eq(7). Could anyone explain to me that why he switch to use $$V^- - \Delta v S$$ instead of $$V^+ - \Delta u S$$??

• Shouldn't there be a solutions textbook for these kind of problems? Commented May 19 at 15:08
• Make an argument as to why you think (8) should go into (7), try to justify and we can show you why you could be wrong. For the time being, this source by Rouah, gives 4 derivations of the black-scholes and maybe it can help you with the intuition. Commented May 19 at 20:59
• @THATSMYQUANTMYQUANTITATIVE I now seem understand why did Paul Wilmott doing in this way. I guess what he was trying to do is that by assuming risk-neutrality and no-arbitrage, no matter at $t+ \delta t$ the underlying is up or down, there should no arbitrage and risk-free and therefore we have two equation for up and down. By solving this system of linear equation, we can find out what V worth at $t$ Commented May 20 at 6:19

Going to go out on a limb here - without having done the maths, isn't the point of his argument that the constructed portfolio is riskless (locally). Therefore, in the next time step it can only earn the risk free rate of return "r". And further, whether the market goes up or down the return of the portfolio is the same.

Ignoring the gamma of the option, if the market ticks up, your option makes money but your delta hedge loses money. If the market ticks down, the option loses money but your delta hedge makes money. Net risk less (if there are only two states of the world). Therefore, if the portfolio is riskless it can only earn the risk free rate of return.

it seems that I did not really understand the meaning of no-arbitrage and risk-free portfolio mean. If I assume no-arbitrage and risk-less, similar to Binomial asset model, the portfolio at time $$t$$ should as same as $$t+ \delta t$$ in whatever situation. If the underlying will up at $$t + \delta t$$, we can get eq(7) and eq(8)

\begin{align} \Pi ( 1 + r \delta t) = V^+ - u \frac{V^+ - V^-}{u - v} \tag{7} \\ (1 + r \delta t) \Big [ V - \frac{V^+ - V^-}{u - v} \Big ] = V^+ - u \frac{V^+ - V^-}{u - v} \tag{8} \end{align}

Similarly, if the underlying falls at $$t + \delta t$$, then \begin{align} \Pi ( 1 + r \delta t) = V^- - v \frac{V^+ - V^-}{u - v} \tag{9} \\ (1 + r \delta t) \Big [ V - \frac{V^+ - V^-}{u - v} \Big ] = V^- - v \frac{V^+ - V^-}{u - v} \tag{10} \ \ \end{align}

Now, we have two equations which is enough to solve $$V$$ by adding these equation.

$$2 ( 1 + r\delta t) \Big [ V - \frac{V^+ - V^-}{u - v} \Big ] = V^+ + V^- - (u + v) \frac{V^+ - V^-}{u - v} \tag{11}$$

For the R.H.S: \begin{align} V^+ + V^- - (u + v) \frac{V^+ - V^-}{u - v} = \frac{(u-v)(V^+ + V^-) - (u+v)(V^+ - V^-)}{u - v } \end{align}

After simplifying the numerator, we get \begin{align} \frac{(u-v)(V^+ + V^-) - (u+v)(V^+ - V^-)}{u-v} = \frac{2uV^- - 2vV^+}{u-v} \end{align}

Therefore, we turn eq(11) to the following \begin{align} V &= \frac{1}{(1 + r\delta t)} \frac{(uV^- - vV^+)}{u-v} + \frac{V^+ - V^-}{u-v} \\ &= \frac{1}{(1 + r\delta t)} \frac{(u - 1 - r\delta t)V^- + (-v + 1 + r\delta t)V^+}{u-v} \\ \end{align}

Recall that $$u = 1 + \sigma \sqrt{\delta t}$$ and $$v = 1 - \sigma \sqrt{\delta t}$$, then \begin{align} \frac{u-1-r\delta t}{u -v } = \frac{(\sigma \sqrt{\delta t} - r \delta t )}{u-v} = \frac{1}{2} - \frac{r \sqrt{\delta t}}{2 \sigma} = 1 - p' \\ \frac{-v+1+r\delta t}{u -v } = \frac{(\sigma \sqrt{\delta t} + r \delta t )}{u-v} = \frac{1}{2} + \frac{r \sqrt{\delta t}}{2 \sigma} = p' \end{align}

Lastly, we can rewrite V everything in terms of $$V^+, V^-, p', r$$ as \begin{align} V = \frac{1}{1+ r \delta t} \Big [ p' V^+ + (1-p') V^- \Big] \end{align}

Once I arrive into the last step of this equation, I realise that the price is just a risk-neutral expected return with a discount factor by the interest rate....