Pricing a European Call from the Risk-Neutral Density
Summary
The note explains the integral notation used to price a European call. Writing the differential before the integrand is an equivalent way to write the usual integral; the density multiplies the call payoff over terminal prices above the strike. The payoff is zero below the strike, so the integral can be restricted to prices greater than the strike.
Under risk-neutral pricing, the discounted expected payoff gives the option value. With zero interest rates, discounting drops out, and integrating the payoff against the risk-neutral probability density is equivalent to taking its expectation. The explanation clarifies that the density is a probability density for the underlying's terminal price, expressed with respect to price increments. The argument applies beyond Black–Scholes when risk-neutral pricing assumptions hold, though the example assumes zero discount rates and a European call.
Key ideas
- A European call pays the positive difference between the terminal underlying price and its strike.
- Risk-neutral valuation expresses the call price as the discounted expected payoff.
- When the terminal price has a density, the expectation can be written as an integral against that density.
- The placement of the differential before or after the integrand is a notational convention.
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Full text
# What does $\int dS \phi (S - K)$ mean in Gatheral's book?
# What does $\int dS \phi (S - K)$ mean in Gatheral's book?
In Gatheral's book on stochastic volatility, he writes the price of an option as $$\int_K^\infty dS \phi (S - K)$$
where $\phi$ is a density.
Where does this come from?
I have multiple questions:
- Why does he write $dS$ before the integrand?
- Why does he use the word "option" and not "call option", since that's what it seems to be?
- Where does the formula come from? I get that in the usual black scholes setting, the price is $\int_K^\infty (S - K) dQ$ where $Q$ is the risk-neutral measure. Does this result also hold in general stochastic volatility setting, and if so, how do we get from $\int_K^\infty (S - K) dQ$ to $\int_K^\infty (S-K) \phi dQ$? Is he saying $\phi$ is the density of .. $S$ with respect to the measure $dS$??
## Answer by Quantuple (score 3)
https://quant.stackexchange.com/a/40944
- It's a mere a question of notation and you can consider that $$ \int f(x) dx \equiv \int dx f(x) $$
- It is indeed implied that the option under scrutiny is a (European) call option
- This comes from the definition of the risk-neutral measure $\Bbb{Q}$, as the measure under which the prices of self-financing portfolios are martingales when expressed under the money market numéraire $B_t$. Assuming zero discount rates, $B_t = 1, \forall t\geq 0$ and this becomes \begin{align} C_0 &= B_0 \Bbb{E}_0^\Bbb{Q}[B_T^{-1}C_T] \\ &= \Bbb{E}_0^\Bbb{Q}[ C_T] \\ &= \Bbb{E}_0^\Bbb{Q}[ (S_T - K)^+ ] \\ &=\int_{-\infty}^{+\infty}(S-K)^+ q(S) dS \\ &=\int_{K}^{\infty} (S-K) q(S) dS \\ &\equiv \int_{K}^{\infty} dS q(S) (S-K) \end{align} It's just that he denoted the risk-neutral pdf $q(S)$ by $\phi$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.