# What is the value this “special” forward contract at maturity?

Background Information:

I am not sure this is relevant:

Terminal value pricing:

If the derivative $$X$$ equals $$f(S_T)$$, for some $$f$$ then in the value of the derivative at time $$t$$ is equal to $$V_t(S_t,t)$$, where $$V(s,t)$$ is given by the formula

$$V(s,t) = \exp{(-r(T-t)E_{\mathbb{Q}}(f(S_T)|S_t = s)}$$

And then the trading strategy is given by $$\phi_t = \frac{\partial V}{\partial s}(S_t,t)$$.

or perhaps I need t apply this formula to the question below:

$$V_t(X) = B_tE_t = B_t E_{\mathbb{Q}}[B_T^{-1} X| \mathcal{F}_t]$$

I am not sure...

Question:

Consider a Black-Scholes model $$S_t = \exp{(\sigma W_t + \mu t)}$$, $$B_t = \exp{(rt)}$$, where $$W_t$$ is Brownian motion with respect to a given measure $$\mathbb{P}$$.

Suppose you hold a forward contract obligating you to purchase $$1$$ share of stock for $$2$$ dollars at time $$t = 5$$.

What is the value $$X$$ of this contract at maturity $$t = 5$$? Express your answer in terms of $$S_5$$.

I am not sure how to solve this. Any suggestions is greatly appreciated.

• At time t=5 you must PAY 2 dollars and you will RECEIVE something worth $S_5$. Clearly the value of this $-2+S_5$: you can immediately sell the stock and reimburse yourself for the 2 you had to pay, leaving you holding $-2+S_5$ in cash. – Alex C Dec 4 '16 at 0:35
• first sentence of the quoted question complete red herring, bizarre to open with that, they're just trying to trick you :) – Mehness Dec 4 '16 at 1:09
• Either follow Alex C's comment or use the damn formula for $V(s,t)$... – Lost1 Dec 4 '16 at 12:19

## 1 Answer

It seems part of the instruction is there to trouble you.

If you have a contract forcing you to buy a stock $S$ at $t=5$ for 2\$, then the value of your contract at maturity is by definition$S_5 -2$. My guess is the question has a follow-up where they as you what the value is at time$t=0$. In this case you can simply create a replicating portfolio, buy buying the stock and borrowing 2\$, which has a value of $S_0 - 2 \exp(-rt)$.

If the contract became optional then the value at $t=5$ would not change but the value at $t=0$ becomes the value of a call on $S$ with maturity $T=5$ and strike price $K=2$, which you can find using Black-Scholes.

• Thank you, if asked to make the price $S_t$ in terms of Brownian motion $\tilde{W_t}$ with respect to the risk-neutral measure $\mathbb{Q}$ making the discounted process a martingale would we just say since $\tilde{W}_t = W_t + \gamma t$ then $W_t = \tilde{W}_t - \gamma t$ thus $S_t = \exp{\sigma(\tilde{W}_t - \gamma t) + \mu t}$. Then do we let $Z_t = B_t^{-1}S_t$ find the SDE $d Z_t$, see if it is driftless then we have the discounted stock a martigale? Perhaps I should turn this into a question. – Wolfy Dec 5 '16 at 2:17