1
$\begingroup$

Tsay, R. S. (2010), Analysis of Financial Time Series, 2nd Edition, discusses on page 504 the estimation of local trend models (state space). The measurement and the transition equations are as follows

(1) $Y_t = \mu_t +e_t$
(2) $\mu_t = \mu_{t-1} + u_t$

with $e_t$ ~ $N(0,\sigma^2_e)$ and $u_t$ ~ $N(0,\sigma^2_u)$!

Based on properties of forecast errors, the Kalman filter provides an efficient way to evaluate the likelihood function of the data for estimation. Specifically, the likelihood function under normality is

(3) $p(y_1.....y_T|\sigma_e^2,\sigma_u^2) = P(y_1\sigma_e^2,\sigma_u^2){\displaystyle \prod_{t=2}^{T}(y_t|F_{t-1},\sigma_e^2,\sigma_u^2)}$ $= P(y_1\sigma_e^2,\sigma_u^2){\displaystyle \prod_{t=2}^{T}(v_t|F_{t-1},\sigma_e^2,\sigma_u^2)}$

where $y_1$ ∼ $N(\mu_{1|0},V_1)$ and $v_t = (y_t −\mu_{t|t−1})$∼$N(0,V_t)$.

Consequently,assuming $\mu_{1|0}$ and $\Sigma_{1|0}$ are known, and taking the logarithms, we have

(4) $\ln[L(\sigma_e^2, \sigma_u^2)] = -\frac{T}{2}\ln(2\pi)-\frac{1}{2} \sum_{t=1}^{T}(\ln(V_t)+\frac{v_t^2}{V_t})$

where ${v_t^2}$ is $y_1 −\mu_{1|0}$.

My question is, how did Tsay arrive from (3) to (4)? To be more precise, where is $(\ln(V_t)+\frac{v_t^2}{V_t})$ coming from?

$\endgroup$
  • $\begingroup$ Could you give a full reference for Tsay (2010)? $\endgroup$ – Richard Hardy May 20 '16 at 9:29
  • $\begingroup$ You have strange notation here $\endgroup$ – Taylor May 20 '16 at 20:50
2
$\begingroup$

The density of a normal distribution is $\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}$. With $\mu=0, \sigma^2=V_t, x=v_t$ and the sum of it (log of product gets to sum of logs) you get

$$ \log\left\{\prod_{t=1}^T\frac{1}{\sqrt{2\pi V_t}}\exp\left\{-\frac{v_t^2}{2V_t}\right\}\right\} = \sum_{t=1}^T \left(-\frac{1}{2}\log\{\sqrt{2\pi}\})-\frac{1}{2}\log\{V_t\}-\frac{1}{2}\frac{v_t^2}{V_t}\right) $$

And that equals $-\frac{T}{2}\log\{\sqrt{2\pi}\}-\frac{1}{2}\sum_{t=1}^T\left(\log\{V_t\}+\frac{v_t^2}{V_t}\right)$. Where $\log$ equals $\ln$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.