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This is a Bayes formula incorporating 2 random variables. The final expression seems a bit tricky to simplify the exponents and I'm still not so confident with my algebra (pardon me ;)). Can you have a go ?

From here :

$ \frac{\frac{1}{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y-1)^2}}{\frac{1}{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y+1)^2} + \frac{1}{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y-1)^2}} $

To there :

$ \frac{1}{1 + e^{-2y}} $

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3 Answers 3

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Hint

Multiply by

\begin{equation} \frac{2\sqrt{2\pi}}{2\sqrt{2\pi}} \frac{\mathrm{e}^{\frac{1}{2}(y-1)^2}}{\mathrm{e}^{\frac{1}{2}(y-1)^2}}. \end{equation}

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    $\begingroup$ Thx, gonna toy with it a bit :) $\endgroup$
    – keeepkool
    Commented Jul 1, 2015 at 9:07
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Note that $\exp\{x\} = e^x$.

\begin{align} &\frac{\frac{1}{2}\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}(y-1)^2\right\}}{\frac{1}{2}\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}(y+1)^2\right\} + \frac{1}{2}\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}(y-1)^2\right\}}\\\\ &\qquad= \frac{\frac{1}{2}\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}(y-1)^2\right\}}{\frac{1}{2}\frac{1}{\sqrt{2\pi}}\left(\exp\left\{-\frac{1}{2}(y+1)^2\right\} + \exp\left\{-\frac{1}{2}(y-1)^2\right\}\right)}\\\\ &\qquad = \frac{\exp\left\{-\frac{1}{2}(y-1)^2\right\}}{\exp\left\{-\frac{1}{2}(y+1)^2\right\} + \exp\left\{-\frac{1}{2}(y-1)^2\right\}}\\\\ &\qquad = \frac{1}{\exp\left\{-\frac{1}{2}(y+1)^2+\frac{1}{2}(y-1)^2\right\}+1}\\\\ &\qquad = \frac{1}{\exp\left\{-\frac{1}{2}(y^2+2y+1)+\frac{1}{2}(y^2-2y+1)\right\}+1}\\\\ &\qquad = \frac{1}{\exp\left\{\frac{1}{2}(-y^2-2y-1+y^2-2y+1)\right\}+1}\\\\ &\qquad = \frac{1}{\exp\left\{\frac{1}{2}(-4y)\right\}+1}\\\\ &\qquad = \frac{1}{e^{-2y}+1} \end{align}

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$$ \frac{\frac{1}{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y-1)^2}}{\frac{1}{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y+1)^2} + \frac{1}{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y-1)^2}} $$ $$= \frac{\frac{1}{2}\frac{1}{\sqrt{2\pi}}\ \cdot\ e^{-\frac{1}{2}(y-1)^2}}{\frac{1}{2}\frac{1}{\sqrt{2\pi}}\ \cdot\ \left(e^{-\frac{1}{2}(y+1)^2} + e^{-\frac{1}{2}(y-1)^2}\right)}$$ $$= \frac{e^{-\frac{1}{2}(y-1)^2}}{e^{-\frac{1}{2}(y+1)^2} + e^{-\frac{1}{2}(y-1)^2}}$$ $$= \frac{e^{-\frac{1}{2}(y-1)^2}}{e^{-\frac{1}{2}(y+1)^2} + e^{-\frac{1}{2}(y-1)^2}}\cdot\frac{e^{\frac{1}{2}(y-1)^2}}{e^{\frac{1}{2}(y-1)^2}}$$ $$= \frac{1}{e^{-\frac{1}{2}(y+1)^2}e^{\frac{1}{2}(y-1)^2} + 1}$$ $$= \frac{1}{e^{\frac{1}{2}((y-1)^2-(y+1)^2)} + 1}$$ $$= \frac{1}{e^{\frac{1}{2}(-4y)} + 1}$$ $$= \frac{1}{e^{-2y} + 1}$$ Done.

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