8
$\begingroup$

I am trying to understand EM and trying to infer parameters of this model using this technique but am having trouble understanding how to begin:

So, I have a weighted linear regression model as follows where I have observations $X = (x_i, x_2....x_n)$ and the corresponding observations $Y = (y_1, y_2....y_n)$. The model of the relationship between $X$ and $Y$ is a weighted linear regression model and the distributional assumptions are as follows:

$$ y_i \sim \mathcal{N}(\beta^Tx_i, \frac{\sigma^2}{w_i}) $$ $$ \beta \sim \mathcal{N}(0, \Sigma_\beta) $$ $$ w_i \sim \mathcal{G}(a, b) $$

Here $\beta$ are the regression parameters and the model allows for unequal variances by having the response variables to have individual weights on the variance. My goal is to find the most likely linear relationship given by the parameters $\beta$.

So, I can now write the log-posterior as following:

$$ \log P(Y, \beta, w|X) = \sum_{i=1}^n \big(\log P(y_i|x_i, \beta, w_i) + \log P(w_i)\big) + log P(\beta) $$

Now, I have been trying to understand EM and am not sure that my understanding is yet complete but as I understand it, in order to start estimating the parameters, I start by taking the expectation of the log-posterior distribution $\log P(Y, \beta, w|X)$ with respect to the latent/hidden parameters which in my case are $\beta$ and $w$. So this required expected value will be:

$$ \int\int P(\beta, w | X) * \log P(Y, \beta, w | X) dw \;d\beta $$

However, I have no idea how to proceed from here to compute this expectation. Would greatly appreciate any suggestions on what the next step should be. I am not looking for someone to derive me all the necessary things but just a nudge in the right direction on what should I be looking to solve in the next steps.

$\endgroup$
  • $\begingroup$ are you sure EM as in Expectation-Maximisation applies to your problem? $\endgroup$ – Xi'an Jan 20 '15 at 15:17
  • $\begingroup$ I think so. I am trying to understand a paper and they use EM for solving this weighted bayesian linear regression problem. $\endgroup$ – Luca Jan 20 '15 at 15:17
  • $\begingroup$ The latent variables cannot be $\beta$ and the $w_i$'s. If you are interested in $\beta$, the latent variables are presumably the $w_i$'s. In which case you have to find the expected complete log-likelihood $Q(\beta|\beta_0)$ function of the E-step and optimise it in $\beta$ in the M-step. $\endgroup$ – Xi'an Jan 20 '15 at 15:21
  • $\begingroup$ Thanks for your comment. If I may try and clarify, the paper does mention that we are interested in maximizing the incomplete log likelihood $\log p(Y|X)$ but we work with the complete data likelihood given by: $\log P(y, w, \beta|X)$, which to me looked like the posterior distribution in this setup. So, I assumed $\beta$ is being treated as a hidden bvariable in this setup. $\endgroup$ – Luca Jan 20 '15 at 15:45
  • 2
    $\begingroup$ How much do you already know about the EM algorithm? Which book or paper have you studied about it? Starting from scratch on a forum like this sounds like a bad idea. $\endgroup$ – Xi'an Jan 20 '15 at 15:50
3
$\begingroup$

Let me recall the basics of the EM algorithm first. When looking for the maximum likelihood estimate of a likelihood of the form$$\int f(x,z|\beta)\text{d}z,$$ the algorithm proceeds by iteratively maximising (M) expected (E) complete log-likelihoods, which results in maximising (in $\beta$)at iteration $t$ the function $$Q(\beta|\beta_i)=\int \log f(x,z|\beta) f(z|x,\beta_t)\text{d}z$$ The algorithm must therefore starts by identifying the latent variable $z$ and its conditional distribution.

In your case it seems that the latent variable is $\varpi$ made of the $w_i$'s while the parameter of interest is $\beta$. If you process both $\beta$ and $\varpi$ as latent variables there is no parameter left to optimise. However, this also means that the prior on $\beta$ is not used.

If we look more precisely at the case of $w_i$, its conditional distribution is given by$$f(w_i|x_i,y_i,\beta)\propto\sqrt{w_i}\exp\left\{-w_i(y_i-\beta^Tx_i)^2/2\sigma^2\right\}\times w_i^{a-1}\exp\{-bw_i\}$$ which qualiifies as a $$\mathcal{G}\left(a+1/2,b+(y_i-\beta^Tx_i)^2/2\sigma^2\right)$$distribution.

The completed log-likelihood being$$\sum_i \frac{1}{2}\left\{\log(w_i)- w_i(y_i-\beta^Tx_i)^2/\sigma^2\right\}$$ the part that depends on $\beta$ simplifies as$$-\sum_iw_i(y_i-\beta^Tx_i)^2/2\sigma^2$$and the function $-Q(\beta|\beta_t)$ is proportional to \begin{align*}\mathbb{E}\left[\sum_iw_i(y_i-\beta^Tx_i)^2\Big|X,Y,\beta_t\right]&=\sum_i\mathbb{E}[w_i|X,Y,\beta_t](y_i-\beta^Tx_i)^2\\&=\sum_i\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}(y_i-\beta^Tx_i)^2\end{align*} Maximising this function in $\beta$ amounts to a weighted linear regression, with weights $$\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}$$

$\endgroup$
  • $\begingroup$ Thanks for this and I will go through this rigorously. However, this work I am looking at does treat $\beta$ as a hidden variable too. They mention they take the expectation with the approximate form of posterior $Q(\beta, w)$ approximating it as $Q(w)Q(\beta)$. So this bit has me really confused... $\endgroup$ – Luca Jan 20 '15 at 17:07
  • 1
    $\begingroup$ If you treat both $\beta$ and $w$ as latent variables, there is no parameter left... $\endgroup$ – Xi'an Jan 20 '15 at 17:13
  • 1
    $\begingroup$ Perhaps what they have dome is the MAP estimate instead of ML estimate. If I try and reformulate this as the MAP estimate, I am guessing the prior distribution of $\beta$ would come into play? $\endgroup$ – Luca Jan 20 '15 at 17:22
  • 1
    $\begingroup$ One very quick thing...I am not sure if you see this but when you have the equation for the complete log-likelihood, is the first term not $log(\sqrt{w_i})$? Also, I am guessing the term you show is the log-likelihood proportional to a constant. I always get confused with this when stuff gets rolled up into constants. $\endgroup$ – Luca Jan 21 '15 at 17:46
  • 1
    $\begingroup$ correction made: I put $1/2$ in front of the whole expression. $\endgroup$ – Xi'an Jan 22 '15 at 5:09

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.