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.


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

  • $\begingroup$ 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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