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I have a prior distribution $f(x)=\pi cos(\pi x) $ where $x$ is the probability of getting tails in a coin toss. Should a coin toss result in tails, how would this be reflected in the posterior distribution?

Would the posterior distribution simply be $f(x)=\pi cos(\pi/2) $, as $P(x)=1/2$?

Thank you!

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    $\begingroup$ Is $x$ in $[0,0.5]$? $\endgroup$
    – gunes
    May 11 '19 at 19:29
  • $\begingroup$ Hi gunes, $x$ is in [0,1] $\endgroup$
    – Sarina
    May 11 '19 at 19:30
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    $\begingroup$ Then, $f(x)$ is not a valid PDF, since it is negative after $0.5$. $\endgroup$
    – gunes
    May 11 '19 at 19:31
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If your prior is made valid, e.g. as in your comment, $f(x)=\pi/2\sin(\pi x)$, you can find the posterior via Bayes Rule ($D$ denotes your experiment): $$f(x|D)=\frac{p(D|x)f(x)}{p(D)}\propto p(D|x)f(x)=x\pi/2\sin(\pi x)$$ $p(D|x)=x$ because we have only one toss and it is tails. Finally, we need to normalize this expression by calculating the following integral and dividing the expression by it: $$Z=\pi/2\int_0^1 x\sin(\pi x)dx\rightarrow f(x|D)=\frac{1}{Z}x\pi/2\sin(\pi x)$$

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  • $\begingroup$ Hi gunes, thank you so much! If this were two tails, how would the $p(D|x)$ function change? $\endgroup$
    – Sarina
    May 11 '19 at 20:11
  • $\begingroup$ Thank you! Can I ask you one more question regarding this.. in order to calculate posterior probability, would the function need to be integrated at a certain value? $\endgroup$
    – Sarina
    May 11 '19 at 20:24
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    $\begingroup$ You should just calculate $Z$ above. I didn't get what you mean by integrating at a certain value. $\endgroup$
    – gunes
    May 11 '19 at 20:26
  • $\begingroup$ thank you for your help! $\endgroup$
    – Sarina
    May 11 '19 at 20:43
  • $\begingroup$ @Sarina if it were two tails: $p(D|x)=x^2$ $\endgroup$
    – gunes
    May 12 '19 at 4:13
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If $f(x)\propto x\,\sin \pi x\Bbb 1_{(0,1)}(x)$, then $$\int_0^1 x\,\sin \pi x\,\text{d}x=\pi^{-2}\int_0^\pi x\,\sin x\,\text{d}x=\pi^{-2}\underbrace{\left[-x\cos x\right]_0^\pi}_{=\pi-0}+\pi^{-2}\underbrace{\int_0^\pi \cos x\,\text{d}x}_{=0}=\pi^{-1}$$ therefore $$f(x)=\pi x \sin \pi x$$

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