Finance – Page 2 – Sisi Tang

December 17, 2019March 8, 2021

Feynman-Kac Formula

Feynman-Kac formula connects the solution to a SDE to the solution of a PDE. For example, the Black-Scholes formula is an application of Feynman-Kac.

Let $X(t)$ satisfies the following SDE driven by standard Brownian Motion:

$\textnormal{d}X_u = \beta(u,X_u)\textnormal{d}u + \gamma(u,X_u\textnormal{d}W_u).$

Let $h(y)$ be a Borel measurable function, and $t \in [0,T]$ . Define

$\begin{aligned} g(t,x) := \mathbf{E} [&\int_t^Te^{-\int_t^TV(\tau,X_{\tau})\textnormal{d}\tau}f(r,X_r)\textnormal{d}r \\ &+e^{-\int_t^TV(\tau,X_{\tau})}h(X_T)|X_t = x ] \end{aligned}.$

Then, $g(t,x)$ satisfies PDE

$g_t + \beta g_x + \frac{1}{2}\gamma^2 g_{xx} - Vg + f = 0.$

Proof.

To get an idea of how this formula is proved, we start with a base case. The general case is proved by the same manner with a little more complicated calculation.

Suppose $V=f=0$ .

Let $\mathcal{F}(t)$ be the natural filtration associated with the standard Brownian Motion in the SDE. By the Markov property of the solution to SDE, we have $\forall s \in [0,T]$ ,

$\mathbf{E}[h(X_T)|\mathcal{F}(s)] = \mathbf{E}[h(X_T)|\sigma(X_s)] = g(s,X_s).$

Then, for $0<s<t<T$

$\begin{aligned} \mathbf{E}[g(t,X_t)|\mathcal{F}(s)] & = \mathbf{E}[ \mathbf{E}[h(X_T)| \mathcal{F}(t) ] |\mathcal{F}(s)] \\ &= \mathbf{E}[h(X_T)|\mathcal{F}(s)] = g(s,X_s) \end{aligned}$

Hence, $g(t,X_t)$ is a martingale.

Then, we apply Ito’s formula to $g(t,X_t)$ . Its drift term should be zero because it is a martingale. This leads to the Feynman-Kac PDE.

$\begin{aligned} \textnormal{d}g(t,X_t) & = g_t\textnormal{d}t + g_x \textnormal{d} x + \frac{1}{2}g_{xx} \textnormal{d} X \textnormal{d} X \\ & = [g_t+\beta g_x + \frac{1}{2}\gamma^2 g_{xx}] \textnormal{d} t + \gamma g_x \textnormal{d} W\end{aligned}.$

By setting the $\textnormal{d} t$ term equal to $0$ , we get the PDE:

$g_t+\beta g_x + \frac{1}{2}\gamma^2 g_{xx} = 0.$

For the general case, we only need to factor out the $\int_0^t$ integral and make the integral inside the expectation only contains $\int_0^T$ term. We will get $G(t,X_t)$ is a martingale with $G(t,X_t)$ defined as:

$G(t,x):= e^{-\int_0^tV(\tau,X_{\tau})\textnormal{d}\tau}(g(t,x)+\int_0^t e^{-\int_t^rV (\tau,X_{\tau})\textnormal{d}\tau } f(r,X_r) \textnormal{d}r)$

Similarly, by applying Ito’s formula to above martingale and setting the drift term equal to 0, we will get the result.

April 18, 2019March 8, 2021

SIMM

International Swaps and Derivatives Association (ISDA) regulates the variation and initial margin (IM) and standardized it in a model called Standard Initial Margin Model(SIMM).

By far, there are 9 posts about SIMM by ISDA
https://www.isda.org/category/margin/isda-simm/

The most recent model is ISDA SIMM version 2.1 which is effective starting from Dec. 1st, 2018.
https://www.isda.org/2018/08/27/isda-publishes-isda-simm-2-1/
https://www.isda.org/a/zSpEE/ISDA-SIMM-v2.1-PUBLIC.pdf

March 8, 2019March 8, 2021

European and American Put-Call Parity

We assume no dividend and positive risk-free interest rate.

European put-call parity

European put and call option with same maturity $T$ and strike $K$ satisfy the put-call parity:

$C_E - P_E = S_0 - Ke^{-rT},$

where $C_E$ is the price of European call option, $P_E$ is the price of the European put option, $S_0$ is the price of the underlying asset at time $t=0$ .

$C_E - P_E$ can be seen as a forward contract with maturity $T$ and strike $K$ . A short proof of European put-call parity is as follows:

$(S_T-K)^+ + (K_T-S)^+ = S_T - K$

That is to say the terminal payoff of long call and short put is equal to that of forward (with the same maturity $T$ and strike $K$ ). Hence,

$\begin{aligned}& P(0,T)E[(S_T-K)^+] + P(0,T)E[(K-S_T)^+] \\ & = P(0,T)E[S_T - K],\end{aligned}$

where $P(0,T)$ is the discount factor from $T$ to $0$ , and $E$ is the expectation under the risk neutral measure. Above equation is equivalent to the European put-call parity formula.

Never prematurely exercise American call option

If we wait until maturity, the profit of the call option is $(S_T-K)^+$ . If we exercise the option at time $t$ , then we have $-K$ cash position and a stock the worth $S_t$ at this time. Then, at time $T$ , the total portfolio value would be $S_T-Ke^{r(T-t)}$ . That is

$C_E = C_A$

But if the underlying asset pays a dividend, then it might be optimal to exercise the call option early.

American put-call parity

American put and call option satisfies the following inequality:

$S_0 - K \leq C_A-P_A \leq S_0 - Ke^{-rT}$

For the first inequality,

Suppose at time 0, we have the following portfolio: long a call option, short a put option, short underlying asset, and have $K$ cash.

At $t=0$ , the portfolio value is $C_A - P_A - S_0 + K$ .

If the long position side of the put option decides to exercise the option, we then exercise our call option at the same time, otherwise, wait until maturity to decide exercise or not. With this strategy, call and put has the same exercise time and hence can be seen as a forward with maturity undetermined. Say, the maturity is $t \in [0,T]$ .
Then, at time $t$ , the value of our portfolio is $S_t-K$ for the option part and $-S_t + Ke^{rt}$ for the asset and cash part. The total value of the portfolio is $Ke^{rt} - K > 0$ .

By the non-arbitrage principal, we have

$C_A - P_A - S_0 + K \geq 0$

For the second inequality,

$\begin{aligned} &~ C_A - P_A \\ = &~ C_E - P_E + P_E - P_A \\ = &~ S_0 - K e^{-rT} + P_E - P_A \\ \leq &~ S_0 - Ke^{-rT}\end{aligned}$