Under the Black-Scholes model that the stock prices moves as a [[Geometric Brownian Motion|geometric Brownian motion]], the Black-Scholes Formula provides the price of European options $V_{t}$ (whether it is a put or a call). ## Pricing with Feyman-Kac and the Risk-Neutral Measure Using the results developed in [[The Feyman-Kac Theorem and Risk-Neutral Measure|the Feyman-Kac note]] and/or properties of the [[Log-Normal Distribution]], we may model the stock and option prices as: $\begin{align*} dS_{t} &= r ~dt + \sigma ~d\tilde{W}_{t},\\[0.4em] V_{t} &= \mathbb{E}_{\mathbb{Q}}[V_{T} ~|~ \mathcal{F}_{t}] \cdot e^{-r\tau}, \end{align*}$where $\mathbb{Q}$ is the risk-neutral measure, and $\tilde{W}_{t}$ a Brownian motion under it. Now to utilize this probabilistic formula, rewrite the at-expiry value as: $V_{T}= \begin{cases} (S_{T}-K) \cdot\mathbf{1}_{S_{T} > K} &\text{for a call}, \\ (K-S_{T}) \cdot \mathbf{1}_{K > S_{T}} &\text{for a put}. \end{cases}$Now the pricing is reduced to computing $\mathbb{E}_{\mathbb{Q}}[S\mathbf{1}_{_{S>K}}],~ \mathrm{Prob}[S > K],~ \mathbb{E}_{\mathbb{Q}}[S\mathbf{1}_{_{S<K}}].$ > [!Exposition] Computing the expected values > Note that $S_{T} ~|~ S_{t}$ can be written as $S_{T} ~|~ S_{t} = S_{t}\exp\left[ \underbrace{\left( r-\frac{\sigma^{2}}{2} \right)\tau}_{\text{mean }=: ~a}+ \underbrace{\sigma \sqrt{ \tau }}_{\text{std }=:~b} \cdot Z \right]$for some $Z \sim N(0,1)$. Using this, $\begin{align*} \mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{S_{T}>K}]&= \int _{\mathbb{R}} S_{t}\exp[a+bz] \cdot\mathbf{1}_{a+bz > \log K / S_{t}} \cdot \phi(z, 0, 1) ~ dz\\ &= \int _{\mathbb{R}} S_{t}\exp\left[ a + b^{2} / 2\right]\cdot\mathbf{1}_{a+bz > \log K / S_{t}} \cdot \phi(z, b, 1) ~ dx\\ &= S_{t} \exp[a + b^{2} / 2] \cdot \left( 1-\Phi\left( \frac{\log K / S_{t} - a}{b} - b \right) \right)\\ &=S_{t} \exp[a + b^{2} / 2] \cdot \Phi\left( b + \frac{\log S_{t} / K + a}{b} \right) \\ &= S_{t} e^{r\tau} \cdot \Phi \underbrace{ \left(\frac{\log S_{t} / K +\left( r+\sigma^{2} /2 \right)\tau}{\sigma \sqrt{ \tau }} \right)}_{=: d_{+}}. \end{align*}$ > Similarly, $\begin{align*} \mathrm{Prob}_{\mathbb{Q}}[S_{T} > K] &= \mathrm{Prob}_{\mathbb{Q}}\Big[ \underbrace{\log \frac{S_{T}}{S_{t}}}_{\sim N(a,~ b^{2})} > \log \frac{K}{S_{t}} \Big ]\\ &= 1-\Phi\left(\frac{\log K / S_{t} - a}{b}\right)\\ &= \Phi\left( \frac{\log S_{t} / K + a}{b} \right)\\ &= \Phi \underbrace{\left( \frac{\log S_{t} / K + (r - \sigma^{2} / 2)\tau}{\sigma \sqrt{ \tau }} \right)}_{=: d_{-}} \end{align*}$ Now the option prices are simply $\begin{align*} C_{t}&= S_{t} \Phi(d_{+}) - Ke^{-r\tau}\Phi(d_{-}),\\[0.8em] P_{t} &= C_{t} - e^{-r\tau}F_{t} \\[0.4em] &= S_{t} \Phi(d_{+}) - Ke^{-r\tau}\Phi(d_{-}) - S_{t} + Ke^{-r\tau}\\[0.4em] &= Ke^{-r\tau}\Phi(-d_{-}) - S_{t}\Phi(-d_{+}). \end{align*}$ > [!info] Interpretation of $d_{-}$ > Roughly, they are "cut-offs" (in the Gaussian distribution scale) in order for a call to be in the money, as $\mathrm{Prob}-\mathbb{Q}[\mathrm{ITM}]=\Phi(d_{-})$ for a call. > - Therefore, large $d_{-}$ means stock goes up, i.e. more likely to be ITM. > - Schematically, $d_{-}=\frac{\text{log-returns}+ \text{expected log-drift}}{\text{volatility till expiry}}.$ > > The required *log-returns* compared to the strike is $\log S_{t} / K$. > - If $S_{t}>K$, it is a positive start (quite literally), hence more likely to be ITM. > - Taking the log because the stock is log-normal. > > Adjustment by the *"expected" drift* $a$ under the risk-neutral world. > - The $a=(r-\sigma^{2} / 2)\tau$ thing also appears in the exponent of $\mathbb{E}_{\mathbb{Q}}[S_{T} ~|~ S_{t}]$, so high expected value = positive drift = more likely to ITM. > > Rescaled by *volatility-till-expiry*, $b=\sigma \sqrt{ \tau }$. > - Makes everything relative to the volatility-till-expiry. > - If the stock is moving like crazy (large $\sigma$), neither the starting point $\log S_{t} / K$ or the expected drift $a$ matters. > - If instead there is a long time till expiry (large $\tau$), then the starting point doesn't matter, but the drift still does (drift is $O(\tau)$, while volatility is $O(\sqrt{ \tau })$). ## Basic Properties and Greeks One property of European options (without dividends) is that *the call cannot fall below its intrinsic value $(S-K)_{+}$*, by looking at its pricing formula. However, *this is not true for puts*. - The difference between option price and intrinsic value is the **time value**. So the call always has positive time value, while the put can have negative time value (for $S < K$). - In particular, since $d_{+}-d_{-}=\sigma \sqrt{ \tau }$, an ITM put can have negative time value when this difference is tiny compared to the interest rate -- there is little volatility till expiry, making *paying the premium to get the exposure a worse investment than just saving the money*. The "obvious" formula for deltas are: $\begin{align*} \Delta_{t, \text{ call}}&= \Phi(d_{+}),\\ \Delta_{t, \text{ put}}&= -\Phi(-d_{+})=1-\Phi(d_{+}),\\ \end{align*}$*it is correct, but it is not obvious to prove since $d_{\pm}$ are dependent on $S_{t}$.* > [!proof]- Proof cuz why not > By put-call parity, it suffices to derive $\Delta_{t}:= \Delta_{t, \text{ call}}$. For simplicity, drop the subscript $t$ on $S$ since time is held constant: $\begin{align*} \Delta_{t}&= \frac{ \partial C_{t} }{ \partial S }\\ &= \Phi(d_{+}) + S \phi(d_{+}) \frac{ \partial d_{+} }{ \partial S } - Ke^{-r\tau} \phi(d_{-}) \frac{ \partial d_{-} }{ \partial S } \end{align*}$So we just need to prove that the last two terms cancel out. Now $\frac{ \partial d_{+} }{ \partial S }= \frac{ \partial }{ \partial S } \frac{\log S+ \text{some const.}}{b}=\frac{ \partial d_{-} }{ \partial S },$so it further reduces to proving $S\phi(d_{+})\overset{?}{=} Ke^{-r\tau}\phi(d_{-}).$Now that is just copium (still correct, but not fun to prove). > [!warning] Is Delta the probability of being in the money? > Not quite, first of all because $\mathrm{Prob}_{\mathbb{Q}}[S > K]=\Phi(d_{-})$, not $\Delta_{t}=\Phi(d_{+})$, but more importantly, this is the probability in the risk-neutral world, which does not follow the actual dynamics: *the probability of being in the money is very dependent on the drift $\mu$.* ## Continuous Dividends If the stock pays a continuous dividend of $DS_{t}dt$ over time, it affects both the stock's and hedged portfolio's dynamics: $\begin{align*} \frac{dS_{t}}{S_{t}}&= (\mu-D)dt + \sigma ~dW_{t},\\ dP_{t}&= dV_{t}-\Delta_{t}dS_{t}-D\Delta_{t}S_{t}dt. \end{align*}$Redoing [[Black-Scholes Equation|Black-Scholes]] by applying [[Ito's Lemma]] to $dV_{t}$ and equating $dP_{t}=rP_{t}dt$ gives $rV=\frac{ \partial V }{ \partial t } +(r-D)S\frac{ \partial V }{ \partial S } +\frac{\sigma^{2}S^{2}}{2}\frac{ \partial^{2} V }{ \partial S^{2} }, $so simply replacing $r$ with $r-D$ in the no-div Black-Scholes equation. Now solving the equation is the usual business: > [!exposition] Feyman-Kac with continuous dividends > The stock dynamics under the risk-neutral measure also changes with dividends: $\frac{dS_{t}}{S_{t}}=(r-D)dt + \sigma~d\tilde{W}_{t},$so the distribution of $S_{T} ~|~ S_{t}$ becomes $S_{T} ~|~ S_{t} =S_{t} \exp\left[ \left( r-D-\frac{\sigma^{2}}{2} \right)\tau + \sigma \sqrt{ \tau }Z \right]$where $Z \sim N(0,1)$. Therefore all the expectations/probabilities just involve replacing $r$ with $r-D$, but the discounting is still with $e^{-r\tau}$ (without the dividend). That is, $\begin{align*} \mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{S_{T} > K}]&= S_{t}e^{(r-D)\tau}\Phi \underbrace{\left( \frac{\log S_{t} / K + (r-D+\sigma^{2} / 2)\tau}{\sigma \sqrt{ \tau }} \right)}_{=: d_{+}(D)},\\ \mathrm{Prob}_{\mathbb{Q}}[\mathbf{1}_{S_{T}> K}]&= \Phi \underbrace{\left( \frac{\log S_{t} / K + (r-D-\sigma^{2} / 2)}{\sigma \sqrt{ \tau }} \right)}_{=: d_{-}(D)}, \end{align*}$and the no-div $d_{\pm}$ are just special cases $D=0$. Plugging those into the option payoffs, $\begin{align*} C_{t}&= Se^{-Dt}\cdot\Phi(d_{+}(D)) - Ke^{-r\tau}\cdot\Phi(d_{-}(D)),\\ P_{t}&= Ke^{-r\tau}\Phi(-d_{-}(D)) - Se^{-D\tau} \cdot \Phi(-d_{+}(D)). \end{align*}$ ## Discrete Dividends Suppose at time $t_{d}$, the underlying stock goes ex-div by a certain proportion: say $DS_{t_{d}}$. Let $V_{d}(S, t)$ denote the new option prices. Then *by no-arb, the option price should not jump across $t_{d\pm}$* (i.e. the moments before and after the dividend), else buying/selling across the discontinuity is an arbitrage. Mathematically, any stock price $S$, $V_{d}(S, t_{d-})=V_{d}((1-D)S, t_{d+}).$That is, there is discontinuity (in the analysis sense) in both the $S$ and $t$ directions, but when both evolve across $t_{d}$, the process is continuous. Since Black-Scholes is solved backwards (explicitly with the PDE, and implicitly when deriving $\mathbb{E}[V_{T} ~|~ S_{t}]$ for Feyman-Kac), the solution is schematically: ![[DiscreteDivSolution.png#invert|w60|center]] - For the pre-dividend period ($t < t_{d}$), since if $V(S,t)$ is a solution to the Black-Scholes PDE then so is $V(\lambda S, t)$, we see that $V_{d}(S, t)=V_{v}((1-D)S, t),$where $V_{v}$ is the vanilla option written on a stock without dividends, both solves the PDE and satisfies the boundary conditions as $t_{d}$. - After the dividend, everything is business as usual. - In conclusion, $V_{d}(S, t)=\begin{cases} V_{v}((1-D)S, t) &t<t_{d}, \\ V_{v}(S, t) & t_{d}<t<T. \end{cases}$