I have to show that:

\begin{equation} P_{t,T}(K)=e^{-r(T-t)} \int_0^{\infty}\left(K-S\right)^+ q_T^S(S)dS \end{equation}

is equivalent to: \begin{equation} P_{t,T}(K)=e^{-r(T-t)}\int_{-\infty}^{K}\left(\int_{-\infty}^y q_T^S(z)dz\right)dy \end{equation}

Breeden and Litzenberger have shown that using Leibniz integration rule and differentiating the first equation twice leads to: \begin{equation} q_T^S(K)=e^{rf(T-t)}\frac{\partial^2P_{t,T}(K)}{\partial K^2}\vert_{K=S_T} \end{equation}

However, I have difficulties to directly go from the first to the second equation in an elegant way. Does anyone have an idea how this can be achieved?

Many thanks for the help!


1 Answer 1


The first equation expresses the option price as a discounted expected value of the payoff contingent on an asset price $S \geqslant 0$. Without loss of generality, we assume that the probability density function has support in $[0,\infty)$, and rewrite as

$$\begin{align} P_{t,T}(K) &=e^{-r(T-t)} \int_{-\infty}^{\infty}\left(K-S\right)^+ q_T^S(S)\,dS \\ &= e^{-r(T-t)} \int_{-\infty}^{K}\left(K-S\right) q_T^S(S)\,dS \end{align} $$

Integrating by parts with $u = K-S$ and $dv = q_T^S(S)\,dS $, we have $du = -dS $ and

$$v = \int_{-\infty}^S q_T^S(z) \, dz,$$

which with vanishing boundaries terms yields the result

$$P_{t,T}(K) = e^{-r(T-t)} \int_{-\infty}^K \left(\int_{-\infty}^Sq_T^S(z) \, dz \right) \, dS$$


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