Sisi – Page 2 – Sisi Tang

July 4, 2020March 8, 2021

Interest Rates Basic

Suppose $b>a>0$ . Let $P(a,b)$ be the discount factor discounting 1 dollar paid at $b$ to time $a$ . Let $T_1, T_2, \ldots, T_{N+1}$ be the floating rate reset date of the floating leg in a swap. For the case of LIBOR 3M, $T_i = \frac{i}{4}$ if we assume the day count convention is that 3 months is $\frac{1}{4}$ of a year. Define $\alpha_n = T_{n+1} - T_{n}$ .

Forward LIBOR Rate

Let L_n(t) be the forward LIBOR rate for period $[T_n, T_{n+1}]$ seen at time $t$ for $t < T$ . Then,

$L_n(t,T) := \frac{1}{\alpha_n}\left(\frac{P(t,T_n)}{P(t,T_{n+1})}-1\right).$

This definition makes sense if we seen it as:

$1+\alpha_nL_n(t) = \frac{1}{P(T_n, T_{n+1})} = \frac{P(t,T_n)}{P(t,T_{n+1})}.$

Instantaneous Forward Rate

Instantaneous forward rate is the forward rate with time interval goes to infinity.

$\begin{aligned}f(t,T)&=\lim_{\delta\rightarrow 0}\frac{1}{\delta}\left(\frac{P(t,T)}{P(t,T+\delta)}-1\right) \\ & = -\frac{\partial \log P(t,T)}{\partial T}\end{aligned}.$

Short Rate

Short rate $r(t)$ is an instantaneous interest rate such that if $r$ is deterministic

$P(t,T) = e^{-\int_t^Tr(s)\textnormal{d}s}.$

Or if $r$ is stochastic

$P(t,T) = \mathbf{E}\left[e^{-\int_t^Tr(s)\textnormal{d}s}|\mathcal{F}_t\right].$

Forward Par Swap Rate

In finance, ‘par’ usually means ‘equal’. Par swap rate is the fixed rate such that the floating leg and the fixed leg of the swap make the contract has value 0. We denote $S_n(t)$ by the forward par swap rate on tenor $[T_n, T_{N+1}]$ . Suppose the pay frequency of the fixed leg is the same as the floating leg (otherwise redefine the below annuity as one that matches the fixed leg pay frequency). Let $A_n(t)$ represents the present value of annuity, i.e.,

$A_n(t) = \sum_{k=n+1}^{N+1} \alpha_{k-1}P(t,T_k).$

Then the present value of the fixed leg is

$PV_{\mbox{fix}} = S_n(t)A_n(t)$

The present value of the floating leg is

$\begin{aligned}& PV_{\mbox{float}} \\ = & \sum_{k=n}^N L_k(t)\alpha_k P(t,T_{k+1}) \\ =& \sum_{k=n}^N \frac{1}{\alpha_k}\left(\frac{P(t,T_k)}{P(t,T_{k+1})}-1\right) \alpha_k P(t,T_{k+1}) \\ =& \sum_{k=n}^N \left(P(t,T_k) - P(t,T_{k+1})\right) \\ =& P(t,T_n) - P(t,T_{N+1}) \end{aligned}$

By letting $PV_{\mbox{fix}} = PV_{\mbox{float}}$ , we have that

$S_n(t) = \frac{P(t,T_n) - P(t,T_{N+1})}{A_n(t)}.$

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.

August 3, 2019March 8, 2021

Skewness and Tailedness

Skewness describes how symmetric distribution is. It is defined as the third moment of the distribution after normalization:

$\mbox{Skew}[X]:= \mathbb{E}[(\frac{X-\mu}{\sigma})^3]$

Kurtosis is a measure of tailedness. The heavier the tail is, the larger its Kurtosis is. Mathematically, it is defined as

$\mbox{Kurt}[X]:= \mathbb{E}[(\frac{X-\mu}{\sigma})^4]=\frac{\mu_4}{\sigma^4}=\frac{\mu_4}{\mu_2^2},$

where $\mu$ is the mean, and $\sigma$ is the standard deviation of the distribution of $X$ ; and $\mu_k$ is $k$ -th central moment.

It can be shown that the Kurtosis of Gaussian distribution is $\mbox{Kurt}[N(0,1)] = 3$ . People usually use excess Kurtosis as the extra Kurtosis of a distribution compared with standard Gaussian distribution. Namely,

$\mbox{Excess Kurt}[X] := \mbox{Kurt}[X] - 3.$

We can create a plot with the square of Skewness as its x-axis and Kurtosis as its y-axis. This plot is called Cullen and Frey graph.

This graph helps us to determine which distribution our data is closest to.