How does exercise price affect derivative formula?

Question

I was doing exercise on page $29$ of the book "The Mathematics of Finance: Modeling and Hedging Book by Joseph Gail Stampfli and Victor Goodman". The first exercise is quite easy to solve. Here's the text:

1. A unit of stock is valued at $\$110$. In one year the stock price will be either $\$130$ or $\$100$. Suppose that the corresponding derivative value will be $U = \$10$ or $D = \$0$. The current one-year riskless interest rate is $4\%$ ($e^{r\tau} = 1.04$). Find the derivative price at $t = 0$.

Here's my solution:

It is a direct application of derivative price formula:

$$a = \frac{U-D}{S_u-S_d}= \frac{10-0}{130-100}= \frac{1}{3}.$$

$$V_0 = aS_0 + (U-aS_u) e^{-r\tau }= \frac{1}{3} \times 110 + (10-\frac{1}{3} \times 130) \times \frac{1}{1.04} = 4.6153 .$$

However, I was not able to do the second questions. And I have no ideas on how to figure out $D$ and $U$.

2. Suppose that a stock has the same values as in Exercise | and the interest rate is $4\%$. Suppose that the derivative is a call option whose exercise price is $\$100$. Find the call option price at $t = 0$.

I am not savvy at the subject of Math Finance so any help will be appreciated! Wish you guys give any answer to me.

Answer 1

Hint: For a call option with strike $K$ ($\$100$ in your case), the payoff at expiry $T$ ($1$ year in your case) is:

$$ \max \{S - K, 0 \}, $$

where $S$ is the stock price at expiry. So, under your two scenarios:

$$ U = \max \{S_u - K, 0 \}, \; \; D = \max \{S_d - K, 0 \}. $$

Then you can use the same pricing formula.

Edit: Btw, if you plug in $a$ in your pricing formula and rearrange a bit, you get:

$$V_0 ={\rm e}^{-r\tau} \left[pU+(1-p)D \right],$$

where

$$p =\frac{S_0{\rm e}^{r\tau} -S_u}{S_u -S_d} \left( = \frac{{\rm e}^{r\tau} - S_uS_0^{-1}}{S_uS_0^{-1} - S_dS_0^{-1}}\right).$$

If you redo the calculation using this version of the pricing formula, you should get the same result.

(This is easy to remember as discounted expectation/average and equivalent to the risk neutral pricing mentioned in the other answer. Here, $p$ can be interpreted as risk neutral probability.)

Answer 2

Let me show you a different approach using Risk-Neutral Pricing.

Under no-arbitrage conditions, a risk neutral investor would be willing to buy the stock if the price $S_0$ is equal to the discounted expected payment at t=1:

$S_0 = \frac{1}{1+r}E_Q[S_1]$

The present and (possible) future prices of the stock define implicitly the probabilities that the risk neutral investor assigns to this events. This is what we call the risk-neutral probability measure $Q$.

Calculating:

$110 = \frac{1}{1.04}[130p + 100(1-p)] \implies p = 0.48$

Now for the derivative the same logic applies:

$V_0 = \frac{1}{1+r}E_Q[V_1] = \frac{1}{1.04}[10 \cdot 0.48 + 0 \cdot 0.52] = 4.61538$

The advantage of this approach is that once we have the risk-neutral measure, we can calculate easily the price of any derivative on the underlying asset. In the case of the call option you mention, when the stock goes up you can buy at \$$100$ sell at \$$130$ and get \$$30$, otherwise you get \$$0$. Then:

$C_0 = \frac{1}{1+r}E_Q[C_1] = \frac{1}{1.04}[30 \cdot 0.48 + 0 \cdot 0.52] = 13.84615$

Hope this helps. You can find more about risk neutral pricing at:

https://quant.stackexchange.com/questions/55239/what-is-the-risk-neutral-measure

Answer 3

I accepted ir7's answer, and I want to show the complete solution to the problem if there's any confusion.

$$a = \frac{U-D}{S_u-S_d}= \frac{ \max \{S_u - K, 0 \}- \max \{S_d - K, 0 \}.}{30}\\ = \frac{ \max \{130 - 100, 0 \}- \max \{100 - 100, 0 \}.}{30} = 1.$$

And we also have

$$V_0 = aS_0 + (U-aS_u) e^{-r\tau }= 1 \times 110 + (30-1 \times 130) \times \frac{1}{1.04} = 13.846 .$$

And this coincides with papi's answer.