Recovering the Risk-Neutral Density From Call Prices
You observe the prices $c(K)$ of European calls on a stock for every strike $K > 0$, all with the same maturity $T$. The risk-free rate $r$ is constant, but you do not know the stochastic process followed by the stock (it need not be lognormal).
Show how to recover the risk-neutral probability density $f(s)$ of the terminal stock price $S_T$ from the call price curve $c(K)$. Also state what the first derivative $\partial c/\partial K$ tells you.
Hints
- Under the risk-neutral measure, $c(K) = e^{-rT}\int_K^\infty (s - K)\,f(s)\,ds$, whatever the process, as long as $f$ is the density of $S_T$.
- Differentiate under the integral sign with respect to $K$ (Leibniz rule): the boundary term vanishes because the integrand $(s - K)$ is zero at $s = K$.
- The first derivative is $-e^{-rT}[1 - F(K)]$; differentiate once more to isolate the density.
Worked Solution
How to Think About It: Write the call as an integral against the unknown density and notice that $K$ appears both as a limit of integration and inside the integrand. Differentiate twice using the Leibniz rule. Every model-specific detail lives inside $f$, so the derivation is model-free; only discounting at $r$ is needed.
Quick Estimate: In discrete form, $f(K) \approx e^{rT}\,\dfrac{c(K + h) - 2c(K) + c(K - h)}{h^2}$: a butterfly spread with strikes $K - h$, $K$, $K + h$ costs $c(K-h) - 2c(K) + c(K+h)$ and pays a triangle of height $h$ around $K$ with area $h^2$, so its price divided by $h^2$ is the discounted probability density near $K$. Sanity check with Black-Scholes ($S = 100$, $r = 5\%$, $\sigma = 20\%$, $T = 1$): $e^{rT}c''(100) = 0.0197$, which matches the lognormal density of $S_T$ at $100$.
Formal Solution:
*Step 1 -- Model-free pricing identity.* Under the risk-neutral measure, with $f$ the density and $F$ the CDF of $S_T$, $$c(K) = e^{-rT}E^Q[(S_T - K)^+] = e^{-rT}\int_K^\infty (s - K)\,f(s)\,ds.$$
*Step 2 -- First derivative (Leibniz rule).* $K$ enters as the lower limit and in the integrand: $$\frac{\partial c}{\partial K} = e^{-rT}\left[-(K - K)f(K) + \int_K^\infty \frac{\partial}{\partial K}(s - K)\,f(s)\,ds\right] = -e^{-rT}\int_K^\infty f(s)\,ds = -e^{-rT}[1 - F(K)].$$ So $-e^{rT}\,\partial c/\partial K = P^Q(S_T > K)$, the risk-neutral probability of finishing in the money (equivalently, $-\partial c/\partial K$ is the price of a digital call paying $1$ if $S_T > K$).
*Step 3 -- Second derivative.* Differentiating once more, $$\frac{\partial^2 c}{\partial K^2} = e^{-rT}f(K) \quad\Longrightarrow\quad f(K) = e^{rT}\,\frac{\partial^2 c}{\partial K^2}.$$
*Step 4 -- Consistency checks.* Because $f \ge 0$, the call price must be convex in $K$; because $\int f = 1$, $\partial c/\partial K$ must run from $-e^{-rT}$ at $K = 0$ to $0$ as $K \to \infty$. Violations of either are static arbitrages (butterfly or call-spread arbitrage). With Black-Scholes prices plugged in, the formula returns the lognormal density exactly, so it is a genuine generalization rather than a new model.
*Step 5 -- The same for puts.* $p(K) = e^{-rT}\int_0^K (K - s)f(s)\,ds$ gives $\partial p/\partial K = e^{-rT}F(K)$ and $\partial^2 p/\partial K^2 = e^{-rT}f(K)$, consistent with put-call parity ($c - p$ is linear in $K$).
Answer: $f(K) = e^{rT}\,\dfrac{\partial^2 c}{\partial K^2}$, obtained by differentiating $c(K) = e^{-rT}\int_K^\infty (s - K)f(s)\,ds$ twice with respect to $K$ (Breeden-Litzenberger). The first derivative gives the risk-neutral tail probability: $\partial c/\partial K = -e^{-rT}P^Q(S_T > K)$.
Intuition
A call is the discounted expectation of $(S_T - K)^+$, a hinge function of $S_T$. Differentiating a hinge once gives a step, and twice gives a delta function, so the second strike derivative of the call price integrates the density against a spike at $K$: it is the density itself, up to discounting. This is the Breeden-Litzenberger result, and it is used daily to extract market-implied distributions from option chains, to price butterflies (which are discrete second differences of calls), and to build local-volatility surfaces. The first derivative gives the risk-neutral probability of finishing in the money, which is also the price of a digital call.