In continuation to this question

$p(h|D) = \frac{p(D|h)p(h)}{p(D)}$

$p(h) = $prior probability of hypothesis $h$

$p(D)$ = prior probability of training data $D$

$p(h|D)$ = probability of $h$ given $D$

$p(D|h)$ = probability of $D$ given $h$

I am inputting to a machine a training data $D$ in form of a table with $n$ training instances.

The machine read only $m$ training examples among $n$ training examples in order to get a hypothesis as output.

I have the following understanding: .

$p(h) = $prior probability of hypothesis $h$ = $\frac{\left\vert{Version- space}\right\vert}{\left\vert{Hypothesis-space}\right\vert}$

$p(D)$ = prior probability of training data $D$ = $\frac{\left\vert{Observed-training-data}\right\vert}{\left\vert{Training-data}\right\vert}$ = $\frac{{m}}{{n}}$

$p(D|h)$ = probability of $D$ given $h$ = $\frac{\left\vert{Training-examples-satisfied-by-hypothesis-h}\right\vert}{\left\vert{Training data}\right\vert}$ = $\frac{\left\vert{Training-examples-satisfied-by-hypothesis-h}\right\vert}{{n}}$

I do not know how to interpret

$p(h|D)$ = probability of $h$ given $D$

in the above forms.

Is my understanding about formulas for corresponding probabilities is correct and what is formula for $p(h|D)$?


1 Answer 1


The training data is already defined as the subset the machine is using for learning.

So I would redefine your parameters as follow :

Let $E$ be the dataset, and $D$ the subset of data used for training, with $card(E)=n$ and $card(D)=m$

then you have:

$p(h) = \frac{\vert{Instances-of-E-for-wich-h-is-satisfied}\vert}{card(E)}$

$p(D) = \frac{card(D)}{card(E)} = \frac{m}{n}$

$p(h/D) = \frac{\vert{Instances-of-D-for-wich-h-is-satisfied}\vert}{card(D)}$

$p(D/h) = \frac{\vert{Instances-of-D-for-wich-h-is-satisfied}\vert}{\vert{Instances-of-E-for-wich-h-is-satisfied}\vert}$

You can verify that you have $p(h/D) = \frac{p(D/h)p(h)}{p(D)}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.