# Finding todays price of a derivative

Today's market prices for European call options $$c(T;K)$$ and put options $$p(T;K)$$ with maturity T and any strike K. Let $$B_t = e^{rt}$$ be the price of the risk-free bond and St the price of the stock. a) Let $$f(x) = |5-x| + |10-x|$$. How would you calculate today's price of the derivative with payoff $$f(S_T)$$ at maturity T (in terms of call and put prices)? What if you can observe only market prices for European call options? I know that if I plot the function f(x), I get a payoff diagram that is a strangle. But I'm not quite sure on how I should proceed from here. I am also aware that the strike for one long call option is 5 and put option is 10.

• This thing is equivalent to a combination of 4 options: two puts and two calls.... Sep 25, 2019 at 2:59
• Hi Alex, it would be great if you could expand on that..
– Anon
Sep 25, 2019 at 3:56

Recall that $$|x|=\max\{x,-x\}=2\max\{x,0\}-x$$. Thus, \begin{align*} f(x)&=|5-x|+|10-x| \\ &= 2\max\{5-x,0\} +x-5 + 2\max\{10-x,0\} +x-10 \\ &= 2x-15+ 2\max\{5-x,0\} + 2\max\{10-x,0\} \\ \end{align*} Thus, by no-arbitrage, the time $$t$$ price of $$f(S_T)$$ is given by $$V(t,S_t)= 2S_te^{-q(T-t)}-15e^{-r(T-t)} + 2P(S_t,5,T) +2C(S_t,10,T).$$
If all you have available are call options, use the put-call parity to transforms puts into corresponding call options: $$P(S_t,K,T) = Ke^{-r(T-t)}-Se^{-q(T-t)}+C(S_t,K,T).$$
• All you need to do is to set $q=0$ Sep 28, 2019 at 15:05