1
$\begingroup$

Suppose you have a D-dimensional data vector $x$ = ($x_1$, ..., $x_n$) and associated class variable $y \in \{0, 1\}$, which is Bernoulli with parameter $\alpha$. Assume the dimensions of $x$ are conditionally independent given $y$, and that the conditional likelihood of each $x_i$ is Gaussian with $\mu_{i0}$ and $\mu_{i1}$ as the means of the two classes and $\sigma_i$ as their shared standard deviation.

Use Bayes' rule to show that $p(y=1|x)$ takes the form of the logistic function.

i.e.

$ p(y=1|x) = \frac{1}{1 + exp\{-\sum_{i=1}^{D} w_ix_i - b\}} $

Where $w = (w_1, ..., w_n)$ is the weight vector and $b$ is the bias.

I started with

$ p(y=1|x) = \frac{p(x | y = 1)p(y=1)}{p(x)}$ by Bayes rule

$=\frac{\alpha p(x | y = 1)} {\alpha p(x | y = 1) + (1 - \alpha) p(x | y = 0)}$ by the Product rule

And here is where I got stuck. The dimensions of $x$ are conditionally independent given $y$, so can I somehow split $p(x | y = k)$ by dimensions?

$\endgroup$
  • $\begingroup$ $D$ is actually $n$? $\endgroup$ – user91011 Oct 4 '13 at 13:09
2
$\begingroup$

Well, since you know that given $\left\{y=i\right\}$ (for $i=0,1$), the sequence of random variables $(x_i)_i$ are independent, you know that $$ P(x\mid y=i) = \prod_{j=1}^nP(x_j\mid y=i) $$ Now, simply use the fact that $P(x_j\mid y=i)$ represents a Gaussian density with the respective mean and variance (according to the class), namely, $$ P(x_j\mid y=i) = \frac{1}{\sqrt{2\pi\sigma_j^2}}\exp\left(-\frac{1}{2\sigma_j^2}(x_j-\mu_{j,i})^2\right) $$ BTW, you may need to use the following simple "observation" \begin{align} p(y=1|x) &=\frac{\alpha p(x | y = 1)} {\alpha p(x | y = 1) + (1 - \alpha) p(x | y = 0)}\\ &=\frac{1}{1+\frac{1-\alpha}{\alpha}\frac{p(x | y = 0)}{p(x | y = 1)}}. \end{align}

$\endgroup$
  • $\begingroup$ Does this mean that we assume that our sample is from gaussian to obtain the density above? $\endgroup$ – Erik Hambardzumyan Nov 23 '18 at 8:06

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.