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...


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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Alex C Dec 4 '16 at 0:35
  • $\begingroup$ first sentence of the quoted question complete red herring, bizarre to open with that, they're just trying to trick you :) $\endgroup$ – Mehness Dec 4 '16 at 1:09
  • $\begingroup$ Either follow Alex C's comment or use the damn formula for $V(s,t)$... $\endgroup$ – Lost1 Dec 4 '16 at 12:19

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.

  • $\begingroup$ 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. $\endgroup$ – Wolfy Dec 5 '16 at 2:17

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