Question is based on the paper Maximum likelihood for blind separation and deconvolution of noisy signals using mixture models pdf download link

Assume the model is a FIR system of order 2 expressed as

$$y(n) = \sum_{m=0}^M \theta_mx(n-m) + v(n) \tag{1}$$.

The model can be rewritten as $$y(n) = \mathbf{\theta^T}x(n) + v(n) \tag{2}$$

The state space representation would be: $$y(n) = \theta_0 x(n) + \theta_1 x(n-1) + \theta_2 x(n-2) \tag{3}$$, and

$$ y(n) = \theta^T x(n) + v(n) \tag{4} $$

Assuming the signal $x(n)$ is an $i.i.d.$ sequence of $(n \times 1)$ random vectors with independent components. $v(n)$ is an $i.i.d$ additive zero-mean Gaussian noise of unknown variance, $\sigma^2_v$. Here it still counts as a state space model, because $x(n)$ is still (albeit trivially) markov.

Question : Problem formulation : $\{x(n)\}$ (n by 1) is the unobserved input and $\{v(n)\}$ is the unobserved additive noise. The MLE solution will based on Expectation Maximization (EM). In the paper, I cannot understand how the likelihood expression would look like. Why do the Authors say mixture models?

What will be the complete likelihood and log-likelihood expression ?

This is how I proceeded.

I have used that fact that sum of 2 independent Gaussians is, again, Gaussian: $\theta^T x(n-M:n) + v(n) \sim \mathcal{N}(\theta^T \mu_x, \theta^T \Sigma_x \theta + 1)$.

Now, $\theta^T x(n-M:n)$ is Gaussian, too: a linear transformation of a Gaussian vector is still Gaussian. $\theta^T x(n-M:n) \sim \mathcal{N}(\theta^T \mu_x, \theta^T \Sigma_x \theta)$.

The variance of $y(n)$ is affected by $\theta$.

PDF of the observations $\mathbf{y}$ conditioned on the data sequence $x$ is

$$f(\mathbf{y|x,\theta}) = {\left(\frac{1}{\sqrt{\left(2\pi \sigma^2_v\right)}}\right)}^N \exp\left[-\frac{{\left(y-\theta^T x\right)}^2}{2\sigma^2_v}\right] \tag{1}$$ PDF of the complete data $\xi = [x, y]^T$:

$$f(\xi|\theta) = f(y|x,\theta)f(x) \tag{3}$$

  • $\begingroup$ What is WGN? It looks as if your example has observable $\mathbf{H}$, but do you observe $u_n$? In a regular moving average model such as MA(1), $y_n=\varepsilon_n+\theta_1 \varepsilon_{n-1}$, the regressors are unobserved, therefore OLS does not work (at least directly). Maximum likelihood still does, though, and the maximum likelihood estimation of MA(1) and MA(q) processes is covered in a number of time series textbooks; Hamilton's "Time series analysis" should have that. $\endgroup$ Nov 30, 2015 at 20:12
  • $\begingroup$ WGN = white Gaussian noise. I want to apply OLS technique to estimate $\theta$. Will the expression reduced to the same = $p\sigma^2_u/\sigma^2_v$ (Given in Steven Kay Book: Chapter 3: Estimation Theory of Signals Vol1) where $p$ is the MA order number and irrespective of the distribution of information signal $u$ is Bernoulli or not?In a supervised/non-blind setting how can I obtain the ML estimates off $\theta$? $\endgroup$
    – SKM
    Dec 2, 2015 at 17:27
  • $\begingroup$ Taking into account what I wrote in my previous comment, what is still unclear? I do not have the book you are citing, and I do not have the time to read it (so it would not help much even if I had the book). Of course, that does not say anything about other users, maybe someone will check it. $\endgroup$ Dec 2, 2015 at 19:10

1 Answer 1


"In the paper, I cannot understand how the likelihood expression would look like."

If they are using EM to estimate the model, then they are not using the (incomplete) likelihood function $f(y|\theta)$. They are using (evaluating) the complete likelihood function, which is the joint density of all $y$s and all $x$s. To get this, you would have to multiply your density $f(y|x, \theta)$ by $f(x|\theta)$. Edit: you have it on your last line in a general form, but you didn't write it out specifically for this model.

That's the beauty of EM. You can maximize something that depends on unobserved data using a clever iterative procedure. This is not what the contribution of the paper is, however.

"Why do the Authors say mixture models?"

It's because the observed data, which you denote $y(n)$ (different from the paper), is a mixture of the unobserved inputs, which you denote $x(n)$ (this is also different notation than the paper uses). If the dimension of $x(n)$ is one, however, this name might be slightly misleading. In either case, the observed data is a linear transformation of the unobserved data, with some extra noise added on top.

Edit: here is the complete data likelihood, as per request, with your notation (sort of).

\begin{align*} f(y_{1:T},x_{1:T};\theta) &= f(y_{1:T}|x_{1:T};\theta)f(x_{1:T};\theta) \\ &= \prod_{t=2}^T f(y_t|x_t) f(x_t|x_{t-1}) f(x_1) \\ &= \prod_{t=2}^T f(y_t|x_t) f(x_t) f(x_1) \\ &= \prod_{t=1}^T f(y_t|x_t) f(x_t) \end{align*}

with $f(y_t|x_t) = \text{Normal}(\theta ^T x_t, \Sigma_v)$

  • $\begingroup$ Thank you for taking the time out to go through the paper. I am unable to write out the expression for the complete likelihood and it would really be helpful if you can mention it in your answer. $\endgroup$
    – SKM
    Feb 3, 2016 at 3:53
  • $\begingroup$ Okey dokey. I'll get to work on it in a bit. $\endgroup$
    – Taylor
    Feb 3, 2016 at 15:02
  • $\begingroup$ Just a friendly reminder that you had said you would be helping me in getting the expression for the log-likelihood. $\endgroup$
    – SKM
    Feb 8, 2016 at 21:25
  • $\begingroup$ @SKM sorry for the delay. Is this what you're looking for? I couldn't find the distribution of each $x_t$, but you can fill that in yourself $\endgroup$
    – Taylor
    Feb 9, 2016 at 19:52
  • $\begingroup$ The log of what I just added seems to be the third formula for section 2 $\endgroup$
    – Taylor
    Feb 9, 2016 at 19:55

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