Question: We have a spread option with payoff: $$\max (P_{T} - HR\times G_T, 0)$$, where $$P$$, $$G$$ are underlying prices and $$HR$$ is a constant.

At time zero only contract $$G$$ is available for trading. The contract $$P$$ will only open trading at $$0 < t_1 < T$$.

What's the (optimal i.e. risk neutral expectation based) price of the contract assuming joint lognormality.

I am a bit confused on where to start with this question. Unless I am incorrect are there different cases to consider before we price the contract? Any suggestions would be appreciated.

At $$t_1$$, this payoff can be priced using the Margrabe formula as used for pricing an exchange option.

See Margrabe Formula here

Using the notations in the question and those used the hyperlinked document above -

$$Price_{t_1} = P_{t_1}e^{(\mu_P-r)\tau}\Phi(d_+) -HR \times G_{t_1}e^{(\mu_G-r)\tau}\Phi(d_-) \tag{1}$$

$$Price_0$$ is the discounted value of $$Price_{t_1}$$ using the discount factor $$e^{-rt_1}$$

$$P_{t_1}$$ is assumed to be known at time $$0$$ - e.g. $$P$$ begins trading at $$t_1$$ at par, such that we know that $$P_{t_1} = 100$$. Otherwise $$P_{t_1}$$ needs to be estimated by other means.

So, the only unknown we are left with in $$(1)$$ is $$G_{t_1}$$.

$$G_{t_1}$$ is random at $$t_1$$ but it's distribution is known.

Hence, the brute force method would be to find the expectation of $$Price_{t_1}$$ by numerical integration of $$(1)$$ over the known probability distribution of $$G_{t_1}$$ (lognormal distribution).

$$Price_0 = e^{-rt_1}\mathbf{E}(Price_{t_1})$$

$$Price_0 = e^{-rt_1}\int_{-\infty}^{+\infty}Price_{t_1}pdf(G_{t_1}(x))dx$$

• Gordon in the house woot woot!! Guessing that means this solution is sound :) – Wolfy May 15 '19 at 19:12
• What does the pde() function stand for? Probability density estimation? – Wolfy May 15 '19 at 19:18
• Yes, i meant it to be probability density function. "pdf" would have been a better acronym to use. – bhutes May 16 '19 at 2:51
• Any chance you know this one: quant.stackexchange.com/questions/45640/…? – Wolfy May 16 '19 at 22:16