3
$\begingroup$

I am basically trying to do a weighted linear regression in a bayesian way. This is to ensure that the I can take care of the heretoscedastic noise.

So, my model is like:

$$ y_i \sim \mathcal{N}(\beta^Tx_i, \sigma^2/w_i) $$

Here, $y_i$ are each of the output for the input $x_i$. Now, I have the following distributional assumptions on $\beta$ and $w_i$.

$$ \beta \sim \mathcal{N}(\beta_0, \Sigma_0) $$

$$ w_i \sim \textrm{Gamma}(a, b) $$

Now, I want to approximate the posterior $p(\beta, w|Y, X)$ using VB. So the full log joint model can be written as:

$$ \ln p(Y, \beta, w|X) = \sum_{i=1}^N \ln p(y_i|w_i, \beta, x_i) + \sum_{i=1}^N \ln p(w_i) + \ln p(\beta) $$

Now, looking at variational inference, I need to minimise the KL-divergence between some appropriately chosen distribution $q(\beta, w)$ and the joint model i.e.

$$ E_q\bigg[\ln \frac{p(Y, \beta, w|X)}{q(\beta, w)}\bigg] $$

We can expand this as:

$$ E_q\bigg[\sum_{i=1}^N \ln P(y_i|w_i, \beta, x_i)\bigg] + E_q\bigg[\sum_{i=1}^N \ln P(w_i)\bigg] + E_q\bigg[\ln p(\beta)\bigg] - E_q\bigg[\ln q(\beta, w)\bigg] $$

I can now apply the mean field approximation i.e. $q(\beta, w) \approx q(\beta) q(w)$.

$$ E_{q_{\beta,w}}\bigg[\sum_{i=1}^N \ln P(y_i|w_i, \beta, x_i)\bigg] + E_{q_w}\bigg[\sum_{i=1}^N \ln P(w_i)\bigg] + E_{q_{\beta}}\bigg[\ln p(\beta)\bigg] - E_{q_{\beta}}\bigg[\ln q(\beta)\bigg] - E_{q_{w}}\bigg[\ln q(w)\bigg] $$

I think the reasoning so far is correct but now I am quite lost as to how to proceed. I am not looking for a full solution here but an idea into how to proceed and what tools I would need to generate the update equations for estimating the parameters of the $\beta$ and $w$ variables.

$\endgroup$
2
$\begingroup$

Check this paper Variational Bayesian inference for linear and logistic regression from Jan Drugowitsch.

$\endgroup$
1
$\begingroup$

You derived the variational lower bound in your equation. Have a look at the wiki page for variational approximation. https://en.wikipedia.org/wiki/Variational_Bayesian_methods

$\endgroup$
  • $\begingroup$ Thanks for your reply. Which wiki page are you referring to? $\endgroup$ – Luca Oct 28 '15 at 10:46
  • $\begingroup$ Please, provide more details in your answer and correct spelling, so one can understand your answer. $\endgroup$ – Alexey Zaytsev Oct 28 '15 at 11:16
0
$\begingroup$

tried to calculate it without any warranty.

$P(y|x\beta,\frac{\sigma}{w_i})P(\beta|\beta_0,\Sigma_0)$ at one point I have $-\beta(zy\sum{x_i }+ \Sigma_0\beta_0)+0.5(z\sum{x_i}+\Sigma_0)\beta^2$ where I made $z=\sum{\frac{\sigma}{w_i}}$ can you figure out the update equations ? I guess its like(hope R is fine): $var = diag(N)*w X'X + \Sigma_0 \\ \beta = solve(var)*(diag(N)*w XY + \Sigma_0 *\beta_0 )$

for the gamma normal update: $P(Y|x\beta)P(w_i|a,b)$ separated for $ln w_i$ and $w_i$ i got stuff like $((a-1)-0.5\frac{\sigma}{w^2_i}) ln w_i + (\frac{\sigma}{w^2_i}( y^2-yx\beta + (x\beta)^2)) \frac{w_i}{2} $ ok this might very well be wrong but here the update equations I got. $a_i =a_0 +0.5*\frac{\sigma}{w^2_i}$ $b_i =b_0 +0.5*(\frac{\sigma}{w^2_i} ((y_i- x_i\beta)^2 + var(\beta) ))$ let me know where I made mistakes. and if you understand this. this $ln\frac{\sigma}{w_i}$ term confuses me. on second thought there should not be a $w_i$ value in the update equation. on second thought $\frac{\sigma}{w^2_i}$ could be very well be wrong and only be $\sigma$

$\endgroup$
  • $\begingroup$ Thanks for the reply. I will have a look tonight and report back! $\endgroup$ – Luca Oct 29 '15 at 10:43
  • 1
    $\begingroup$ @Luca: is your problem solved? I can help regarding the derivations... $\endgroup$ – Sandipan Karmakar Sep 15 '17 at 13:38
  • $\begingroup$ @SandipanKarmakar Thank you for your kind offer. I never got very far with it. I will update when I get back to it someday. $\endgroup$ – Luca Sep 15 '17 at 13:42

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.