Deriving Binomial Option Value with a Riskless Hedge

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


# 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 $0<v <1 < u$. 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 $V^+ - \Delta u S$??

## Answer by user68819 (score 0)

https://quant.stackexchange.com/a/79430

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.

## Answer by Ricky Pang (score 0)

https://quant.stackexchange.com/a/79433

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....