Marginal and conditional distribution for continuous input and discrete output of a classifier

Question

Consider a one-dimensional classification problem with $X = \mathbb R$ and $Y = \{-1, +1\}$: $$p(y=-1)=\frac{3}{5} \qquad p(x \mid y=-1)=\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x+2)^2}{8}} \qquad N(-2, 2)$$ $$p(y=+1)=\frac{2}{5} \qquad p(x \mid y=+1)=\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2}} \qquad N(1, 1)$$

1. Find the marginal distribution $p(x)$ and the conditional distributions $p(y = -􀀀1 \mid x)$ and $p(y = +1 \mid x)$.
2. Guess from $p(y = 􀀀-1 \mid x)$ and $p(y = +1 \mid x)$ what the Bayes-optimal classifier is like.

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Could you please direct me on how to start with solving the first order or point me to a good easy explained document with examples about marginal distributions for continuous and discrete variables, and how could $N(-2,2), N(1,1)$ be used in solving this question?

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I started using the formula: $p(x) = p(x \mid y=-1) * p(y=-1) + p(x \mid y=+1) * p(y=+1)$ then replaced each term with its value from the question, then tried to simplify the formula but with no luck as it got more complex.

Answer 1

$$p(y=1|x) * p(x) = p(x|y=1) * p(y=1)$$ $$p(y=1|x) = \frac{p(x|y=1) *p(y=1)}{p(x|y=1)* p(y=1) + p(x|y=-1)* p(y=-1) }$$ $$p(y=1|X=x) = \frac{\frac{2}{5}*\frac{1}{\sqrt{2*\pi}}e^{-\frac{(x-1)^2}{2}}}{\frac{2}{5}*\frac{1}{\sqrt{2*\pi}}e^{-\frac{(x-1)^2}{2}}+\frac{3}{5}*\frac{1}{\sqrt{8*\pi}}e^{-\frac{(x+2)^2}{8}}}$$ I admit I used

http://www.wolframalpha.com/input/?i=2*exp%28-%28t-1%29%5E2%2F2%29%2F%282*exp%28-%28t-1%29%5E2%2F2%29%2B3%2F2*exp%28-%28t%2B2%29%5E2%2F8%29%29

to simplify :) $$p(y=1|x) = \frac{4}{4+3*e^{\frac{3}{8}*(x-4)*x}}$$

$$p(y=-1|x)=1-p(y=1|x)=\frac{3}{3+4*e^{-\frac{3}{8}*(x-4)*x}}$$

solving either equal to 1/2 gives you the (hint! there are 2!) transition points between most likely estimates of y