kalman filter update equation

Assume that futures price $F(t,T)$ follows the Ito process as described by the following stochastic process

$$ln F(t,T)=lnF(0,T)+(Z_1(t)e^{-k(T-t)}+Z_2(t))-(1/4k)[(1-e^{-2kT})(h_1^2+h_2^2))+4h_1h_0(1-e^{-kT})+2h_0^2tk]$$

where $Z_1(t)$ and $Z_2(t)$ are state variables following:

$$dZ_1(t)=-kZ_1(t)dt+h_1dW_1(t)+h_2dW_2(t)$$

$$dZ_2(t)=h_0dW_1(t)$$

Assume $W_1$ and $W_2$ are independent. How to derive kalman filter update/measurement equations to estimate the parameters $h_0,h_1,h_2,k$? Observations are $lnF(t,T)$. Many Thanks

• Maybe Euler discretisation? – Egodym Jun 19 '16 at 20:25
• You may want to have a look of the book "Time Series Analysis" by Hamilton. – Gordon Jun 20 '16 at 14:53

1 Answer

Consider historical observation dates $t_0 < t_1 < \cdots < t_n$. From the state variable equations \begin{align*} dZ_t^1&=-kZ_t^1dt+h_1dW_t^1+h_2dW_t^2,\\ dZ_t^2&=h_0dW_t^1. \end{align*} We obtain that, for $i=1, \ldots, n$, \begin{align*} Z_{t_i}^1 &= e^{-k \Delta t_i} Z_{t_{i-1}}^1 + h_1 \int_{t_{i-1}}^{t_i}e^{-k (t_i-s)}dW_s^1 + h_2 \int_{t_{i-1}}^{t_i}e^{-k (t_i-s)}dW_s^2,\tag{1}\\ Z_{t_i}^2 &= Z_{t_{i-1}}^2 + h_0 \int_{t_{i-1}}^{t_i}dW_s^1.\tag{2} \end{align*} Let \begin{align*} \pmb{x}_{t_i} &= [Z_{t_i}^1, \ Z_{t_i}^2]^T, \end{align*} and \begin{align*} F &= \left(\! \begin{array}{cc} e^{-k\Delta t_i} & 0\\ 0 & 1 \end{array} \!\right). \end{align*} Moreover, let \begin{align*} \pmb{v}_{t_i} = \bigg[h_1 \int_{t_{i-1}}^{t_i}e^{-k (t_i-s)}dW_s^1 + h_2 \int_{t_{i-1}}^{t_i}e^{-k (t_i-s)}dW_s^2, \ h_0 \int_{t_{i-1}}^{t_i}dW_s^1\bigg]^T \end{align*} be a two dimensional normal random vector with zero mean and covariance matrix \begin{align*} Q = \left(\! \begin{array}{cc} \frac{h_1^2+h_2^2}{2k}\big(1-e^{-2k \Delta t_i} \big) & \frac{h_0h_1}{k} \big(1-e^{-k\Delta t_i} \big) \\ \frac{h_0h_1}{k} \big(1-e^{-k\Delta t_i} \big) & h_0^2 \Delta t_i \end{array} \!\right). \end{align*} Then, based on $(1)$ and $(2)$, \begin{align*} \pmb{x}_{t_i} = F \pmb{x}_{t_{i-1}} + \pmb{v}_{t_i}, \end{align*} which is the transition equation or state equation.

Let $T_1 < \cdots < T_m$ be the futures maturities. From the equation \begin{align*} \ln F(t,T)&=\ln F(0,T)+\left(Z_t^1 e^{-k(T-t)}+Z_t^2\right) \\ &\qquad -\frac{1}{4k}\left[(1-e^{-2kT})(h_1^2+h_2^2))+4h_1h_0(1-e^{-kT})+2h_0^2tk\right], \end{align*} we obtain that \begin{align*} \ln F(t_i,T_j)&=\ln F(0,T_j)+\left(Z_{t_i}^1 e^{-k(T_j-t_i)}+Z_{t_i}^2\right) \\ &\qquad -\frac{1}{4k}\left[(1-e^{-2kT_j})(h_1^2+h_2^2))+4h_1h_0(1-e^{-kT_j})+2h_0^2t_ik\right].\tag{3} \end{align*} For observation time $t_i$, let $\pmb{y}_{t_i}$ be an $m$-dimensional observation vector defined by \begin{align*} \pmb{y}_{t_i} = \left(\! \begin{array}{c} \ln F(t_i, T_1)\\ \vdots\\ \ln F(t_i, T_m) \end{array} \!\right), \end{align*} and $\pmb{d}_{t_i}$ be an $m$-dimensional deterministic vector defined by \begin{align*} \pmb{d}_{t_i} = \left(\! \begin{array}{c} \ln F(0,T_1) - \frac{1}{4k}\left[(1-e^{-2kT_1})(h_1^2+h_2^2))+4h_1h_0(1-e^{-kT_1})+2h_0^2t_ik\right]\\ \phantom{\frac{\frac{1}{1}}{\frac{1}{1}}} \vdots \phantom{\frac{\frac{1}{1}}{\frac{1}{1}}}\\ \ln F(0,T_m) - \frac{1}{4k}\left[(1-e^{-2kT_m})(h_1^2+h_2^2))+4h_1h_0(1-e^{-kT_m})+2h_0^2t_ik\right] \end{array} \!\right). \end{align*} Moreover, let $H$ be an $(m \times 2)$ matrix defined by \begin{align*} H = \left(\! \begin{array}{cc} e^{-k(T_1-t_i)} & 1\\ \phantom{\frac{\frac{1}{1}}{\frac{1}{1}}}\vdots\phantom{\frac{\frac{1}{1}}{\frac{1}{1}}} & \vdots\\ e^{-k(T_m-t_i)} & 1 \end{array} \!\right), \end{align*} and $\pmb{w}_{t_i}$ is an $m$-dimensional normal random vector with zero mean and a constant covariance matrix $V$, to be defined below. Then, based on Equation $(3)$, $$\label{spot_forward_measurement_eqn} \pmb{y}_{t_i} = \pmb{d}_{t_i} + H \pmb{x}_{t_i} +\pmb{w}_{t_i},$$ which is the measurement equation or observation equation. Here $\pmb{w}_{t_i}$ is a $m$-dimensional vector of normal random variables. The $(m \times m)$ covariance matrix $V$ of $\pmb{w}_{t_i}$ is determined, together with the model parameters, as part of the maximum likelihood estimation.

• Gordon, appreciated. I 'm learning the book you specified and will get back to your answer later. – snowave Jun 20 '16 at 21:24
• @snowave: The format of my answer is based on that book. – Gordon Jun 20 '16 at 21:25
• Your vector c_ti is zero, right? – snowave Jun 21 '16 at 21:36
• @snowave: yes. Revised. – Gordon Jun 21 '16 at 22:56