Breenden-Litzenberg#
General#
Consider call price \(C(K) = (S-K)^+\), where \(S\) is the stock price at expiry
We clearly have
\[
E_Q[ (S - K)^+] = (S-K) \cdot P(S > K)
\]
And
\[\begin{align*}
\frac{\partial}{\partial K} C(K)
=& \frac{\partial}{\partial K} \mathbb{E} \left( (S(T) - K)^+ \right) \\
=& \frac{\partial}{\partial K} (S-K) \cdot P(S > K)\\
=& - P(S > K)
\end{align*}\]
In other words, the probability that the call option expires ITM, is simply
\[\begin{align*}
P(S > K) = - \frac{\partial C}{\partial K}
\end{align*}\]
Denote the CDF of S as \(F_S(x) = P(S \leq x)\), then
\[\begin{align*}
P(S \leq K) = F_S(K) = 1+ \frac{\partial C}{\partial K}
\end{align*}\]
Differentiating again we have
\[\begin{align*}
f_S(K)
=& \frac{\partial^2}{\partial K^2} C(K)
\end{align*}\]
Which is the market implied density of the stock. Similarly for puts:
\[\begin{align*}
\frac{\partial Put(K) }{\partial K}
=& \frac{\partial}{\partial K} \mathbb{E} \left( (K - S(T))^+ \right) \\
=& \frac{\partial}{\partial K} (K - S) \cdot P(S \leq K)\\
=& P(S \leq K)
\end{align*}\]
And
\[\begin{align*}
\frac{\partial^2}{\partial K^2} p(K) = f_S(K)
\end{align*}\]
So in conclusion:
\[\begin{align*}
P(S \leq K) =& \frac{\partial P}{\partial K} = 1+ \frac{\partial C}{\partial K} \\
f_S(K) =& \frac{\partial^2}{\partial K^2} C(K) = \frac{\partial^2}{\partial K^2} P(K)
\end{align*}\]
Black Scholes Framework#
Under BS, we have
\[
C = S_0 \Phi(d_1) - K e^{-rT} \Phi(d_2)
\]
\[
d_1 = \frac{\ln\left(\frac{S_0}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)T}{\sigma \sqrt{T}}, \quad d_2 = d_1 - \sigma \sqrt{T}
\]
Considering only near-dated options, we can assume \(r = 0\), we have
\[
\frac{\partial C}{\partial K} = -\Phi(d_2)
\]
so that
\[\begin{align*}
P(S > K) = - \frac{\partial C}{\partial K} = \Phi(d_2)
\end{align*}\]
Note that delta is given by \(\Phi(d_1)\) which is very close to \(\Phi(d_2)\), as \(T\) is very small for near-dated options.
Hence for short dated ATM call options tend to have delta = 0.5, as that is its N(d1) = N(d2) = probability expiring ITM.
i.e. For ATM (\(S = K\)), as \( T \to 0\):
\[
d_{1,2} = \frac{\left(r \pm \frac{\sigma^2}{2}\right)T}{\sigma \sqrt{T}}
\to \frac{\sigma \sqrt{T}}{2} \to 0
\]
\[
\phi(d_1), \phi(d_2) \to 0.5
\]