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I'm working on some game theory models that involve modeling players' beliefs, and am not sure how to figure out this probability problem.

Suppose I choose some number $x$ between 0 and 1 inclusive (i.e., $0 \leq x \leq 1)$. You do not see my choice, but instead get a signal $S$ normally distributed around my choice. That is, $S$ is a draw from the normal distribution:

$$Z \sim N(x, \sigma^2)$$

You know $S$ (a draw from $Z$), $\sigma^2$, and that $0 \leq x \leq 1$. What are your beliefs about $x$? That is, what do you believe $P(x < y \mid S)$ for any $y$?

I think you might need to have some prior distribution over $x$ in order to solve this. If you do, I think the best prior to use is the uniform distribution $U(0,1)$, since the choice could be anywhere on that interval.

Thanks for any help!

Note: This question is similar to my previous one about two overlapping distributions, except that in this case there are infinitely many possible distributions, not just two.

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Using Bayes' Theorem we have that:

$$ P(\mu|S) = \frac{P(S|\mu)P(\mu)}{P(S)} $$

rewriting $P(S)$.

$$ P(\mu|S) = \frac{P(S|\mu)P(\mu)}{\int_{0}^{1}{d\mu'}{P(S|\mu')P(\mu')}} $$

Using that $P(S|\mu)$ is a normal distribution and that $P(\mu)$ is uniform.

$$ P(\mu|S) = \sqrt{\frac{2}{\pi}} \frac{e^{\frac{-(S-\mu)^2}{2\sigma^2}}}{\mathrm{erf}\left(\frac{S}{\sqrt{2}\sigma}\right)- \mathrm{erf}\left(\frac{S-1}{\sqrt{2}\sigma}\right)} $$

Here is how the distribution looks for $\sigma=0.1$ and $S = 0.8$.*

Example case

Here is how the distribution looks for $\sigma=0.4$ and $S = 0.2$.*

Example case

(*) In the plot y is S

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  • $\begingroup$ This is great, thank you! So to get the probability I want, I can just integrate over that density function from 0 to $x$. Is there any way I can get the result of that integral (the CDF) symbolically? Basically, allowing $x$ to be a variable, I want to solve the CDF for $S$, but the $S$'s are trapped inside those error functions. Any idea of how I might do that? $\endgroup$ – sundance Mar 4 '16 at 3:30
  • $\begingroup$ @sundance. I'm not aware of any analytical expresion for erf. Regards $\endgroup$ – ALucero Mar 4 '16 at 4:13

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