Skip to main content
Tweeted twitter.com/#!/StackStats/status/255186051046465536
fixed link; added tags; light editing & formatting
Source Link
gung - Reinstate Monica
  • 147.5k
  • 89
  • 406
  • 717

This is a problem from a machine learning pset that I'm self-learning from (http://www.seas.harvard.edu/courses/cs281/assignment-1.pdf).

Suppose we are provided with a hierarchy of three distributions: $p(\alpha)$, $p(\theta | \alpha)$ and $p(y | \theta, \alpha)$. Write expressions to compute the following related distributions:

$p(y|\theta)$, $p(y|\alpha)$, $p(y)$, $p(\theta|y,\alpha)$, $p(\alpha|y,\theta)$, $p(\theta)$, $p(\theta|y)$, $p(\alpha|y)$

I see that I integrate over the joint distribution to get the marginal probability

$p(y|\theta) = \int p(y|\theta,\alpha)p(\alpha)d\alpha$$$p(y|\theta) = \int p(y|\theta,\alpha)p(\alpha)d\alpha$$

This gives me the first expression in terms of the givens.

However, I don't know how to get the others. I know I can set up various equations using Baye'sBayes' Theorem

$p(\theta|\alpha) = p(\alpha|\theta) * p(\theta)/p(\alpha)$$$p(\theta|\alpha) = p(\alpha|\theta) * p(\theta)/p(\alpha)$$

But the above equation has two unknowns ($p(\theta)$ and $p(\alpha|\theta)$). Is there another property that I can exploit besides marginalizing the joint distribution and utilizing Baye'sBayes' Theorem?

Thanks!

This is a problem from a machine learning pset that I'm self-learning from (http://www.seas.harvard.edu/courses/cs281/assignment-1.pdf).

Suppose we are provided with a hierarchy of three distributions: $p(\alpha)$, $p(\theta | \alpha)$ and $p(y | \theta, \alpha)$. Write expressions to compute the following related distributions:

$p(y|\theta)$, $p(y|\alpha)$, $p(y)$, $p(\theta|y,\alpha)$, $p(\alpha|y,\theta)$, $p(\theta)$, $p(\theta|y)$, $p(\alpha|y)$

I see that I integrate over the joint distribution to get the marginal probability

$p(y|\theta) = \int p(y|\theta,\alpha)p(\alpha)d\alpha$

This gives me the first expression in terms of the givens.

However, I don't know how to get the others. I know I can set up various equations using Baye's Theorem

$p(\theta|\alpha) = p(\alpha|\theta) * p(\theta)/p(\alpha)$

But the above equation has two unknowns ($p(\theta)$ and $p(\alpha|\theta)$). Is there another property that I can exploit besides marginalizing the joint distribution and utilizing Baye's Theorem?

Thanks!

This is a problem from a machine learning pset that I'm self-learning from http://www.seas.harvard.edu/courses/cs281/assignment-1.pdf.

Suppose we are provided with a hierarchy of three distributions: $p(\alpha)$, $p(\theta | \alpha)$ and $p(y | \theta, \alpha)$. Write expressions to compute the following related distributions:

$p(y|\theta)$, $p(y|\alpha)$, $p(y)$, $p(\theta|y,\alpha)$, $p(\alpha|y,\theta)$, $p(\theta)$, $p(\theta|y)$, $p(\alpha|y)$

I see that I integrate over the joint distribution to get the marginal probability

$$p(y|\theta) = \int p(y|\theta,\alpha)p(\alpha)d\alpha$$

This gives me the first expression in terms of the givens.

However, I don't know how to get the others. I know I can set up various equations using Bayes' Theorem

$$p(\theta|\alpha) = p(\alpha|\theta) * p(\theta)/p(\alpha)$$

But the above equation has two unknowns ($p(\theta)$ and $p(\alpha|\theta)$). Is there another property that I can exploit besides marginalizing the joint distribution and utilizing Bayes' Theorem?

Source Link
gjx
  • 33
  • 5

Properties of conditional probability distributions

This is a problem from a machine learning pset that I'm self-learning from (http://www.seas.harvard.edu/courses/cs281/assignment-1.pdf).

Suppose we are provided with a hierarchy of three distributions: $p(\alpha)$, $p(\theta | \alpha)$ and $p(y | \theta, \alpha)$. Write expressions to compute the following related distributions:

$p(y|\theta)$, $p(y|\alpha)$, $p(y)$, $p(\theta|y,\alpha)$, $p(\alpha|y,\theta)$, $p(\theta)$, $p(\theta|y)$, $p(\alpha|y)$

I see that I integrate over the joint distribution to get the marginal probability

$p(y|\theta) = \int p(y|\theta,\alpha)p(\alpha)d\alpha$

This gives me the first expression in terms of the givens.

However, I don't know how to get the others. I know I can set up various equations using Baye's Theorem

$p(\theta|\alpha) = p(\alpha|\theta) * p(\theta)/p(\alpha)$

But the above equation has two unknowns ($p(\theta)$ and $p(\alpha|\theta)$). Is there another property that I can exploit besides marginalizing the joint distribution and utilizing Baye's Theorem?

Thanks!