Does there exist time varying Gaussian model?

To be specific, a 2D Gaussian model whose mean and covariance matrix is varying with another parameter called time. The $\mu$(mean) and $\sigma$(covariance matrix) is changing with according to time.

Then, we say the distribution is $p(x_1, x_2, t)$ with $\mu(t)$ and $\sigma(t)$. Given $(x_1, x_2, t)$, we can get its probability at $(x_1, x_2)$ with specific $t$.

For original Gaussian distribution, if I have i items: $(x_1^1, x_2^1), (x_1^2, x_2^2) \dots (x_1^i, x_2^i)$, and I want to calculate the probability of a specific item $P(x_1^i, x_2^i)$, I can first calculate the mean and covariance matrix of all the items, and then calculate the probability by applying Gaussian distribution.

Now, the problem is that I have the data: $(x_1^1, x_2^1, t^1), (x_1^2, x_2^2, t^2) \dots (x_1^i, x_2^i, t^i)$, and want the mean and covariance matrix can be different values according to $t$, say $\mu(t)$ and $\sigma(t)$. Which means at different time $t$, the Gaussian distribution has different $\mu$ and $\sigma$. And at certain $t$, the distribution is still a normal Guassian distribution.

$(x_1, x_2)$ are two Double value, and $t$ is radian. I hope to cope with $t$ in directional statistics way.

If there exists such kind of model, how to inference its parameters, the relationship between its mean and covariance matrix with time?

  • 1
    $\begingroup$ Seems like you are describing Gaussian process. What kind of data do you have? $\endgroup$ – mpiktas Dec 23 '14 at 7:32
  • $\begingroup$ @mpiktas I've re-edited the question, is it suitable to use Gaussian process? $\endgroup$ – 宇宙人 Dec 23 '14 at 7:42

Take a look at Wiener process, Brownian motion and Ito calculus. For instance, a particular form of the process with time-varying mean and variance is Heston model used a lot in finance: $dS_t=\mu S_tdt+\sqrt{\nu_t}S_tdW_t^S$

$d\nu_t=\kappa (\theta-\nu_t)dt+\xi\sqrt\nu_tdW_t^\nu$.

Here, in the first equation you see how both mean and the variance of $dS_t$ changes with time.

  • $\begingroup$ Thanks, last year I thought this approach was too difficult for me to solve, so I walked around the difficulty with some other type of model which was much simpler. But recently I met another similar problem, and I recalled that I asked the question here. This time I will take effords to learn these models any way. $\endgroup$ – 宇宙人 Jun 24 '16 at 2:22
  • $\begingroup$ That's the right approach. Nothing's too difficult if you are ready for it. $\endgroup$ – Aksakal Jun 24 '16 at 11:43

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.