I've been asked to consider a system using Bayes' Theorem. It has a Boolean input vector x and binary output $y$ = {$0$,$1$}. i.e., $x = {x_1, x_2, ..., x_n}$.
Given equations:
$p(y=1|x) = \frac{1}{1+exp[-v(x)]}$
$v(x)$ = $log\frac{p(x|y=1)}{p(x|y=0)}$ + $log \frac{p(y=1)}{p(y=0)}$
I'm asked to show that the above equations hold... Is the question asking to find the probability the $y = 1$ given any $x_i$ input? If so, I am not certain how the first equation was calculated.
I was thinking along the lines of using something like $P(y|x) = P(y)\frac{P(x|y)}{P(x)}$ to attempt to prove the first equation correct. That said, I am unsure what $-v(x)$ is in the denominator.
I was able to find in my text the first equation as being the logistic function. So if I am to plot this from some arbitrary values, like -5 to 5, the $v(x)$ value determines the steepness of the curve and that it would cross the y axis at 0.5. But how does $v(x)$ affect steepness? Would something like Excel be useful in this scenario to get an accurate graph or could I eyeball it?
I consulted a text I found on Google books and was able to understand $v(x)$ a bit more. It is $log\frac{p(y=1|x)}{p(y=0|x)}$. So $v(x)$ is the log posterior ratio which, as defined above, is the log likelihood ratio + the log prior ratio.
Ok, so I believe I now understand what is being asked as well as the equations given. Are my thoughts above a valid approach or do I need to look in a different direction? If a different direction is needed, any guidance into what that may be is greatly appreciated.
I think you're reading more into these equations than there is. This is just a disguised form of
$$ p(y=1\mid x)=\frac{p(y=1\mid x)}{p(y=0\mid x)+p(y=1\mid x)}\;, $$
which is what you get if you plug the result you already obtained, $v(x)=\log\frac{p(y=1\mid x)}{p(y=0\mid x)}$, into the upper equation and multiply the fraction through by $p(y=1\mid x)$.