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Let $x\sim U[0,1]$ be drawn from a uniform distribution. We receive an imperfect signal $s$ about $x$ of the following form: $s=x$ with probability $p$, while with probability $1-p$ the signal $s$ is drawn from an (independent) uniform distribution.

After observing the signal $s$, the conditional distribution of $x$ is simply $$E(x|s)=s\cdot p+ (1-p)Eu=s\cdot p+ \frac{1}{2}(1-p).$$

Now, what happens if we replace $p$ with a function $p(x)$? Somehow, I cannot see how the Bayes rule would work here. What would be the distribution of $E(x|s)$, given some known $p(x)$?

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    Your statement of the conditional distribution of $x|s$ is not right. Perhaps you meant the conditional mean $E(x|s)$? – 2017-02-05
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    Yes, I corrected that. Thanks! – 2017-02-06
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    Still mixing the two concepts. The conditional expected value should be $sp + \frac{1}{2}(1-p).$ See answer. – 2017-02-06
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    Oh God, this is embarassing! ;-) You're right again and I should probably get more sleep. – 2017-02-06
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    It's okay, I also found this oddly confusing for such a seemingly simple problem – 2017-02-06

1 Answers 1

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The conditional distribution of $s$ given $x$ is $x$ with probabilty $p(x)$ and $U(0,1)$ with probability $1-p(x).$ We can write this out formally as $$ f(s|x) = \delta(s-x)p(x) + (1-p(x)) $$ for $0\le s\le 1$ where the $\delta(s-x)$ is an atom at $x.$

We can then get the unconditional distribution of $s,$ which is $$ \int f(s|x)f(x)dx = \int_0^1 (\delta(s-x)p(x) + (1-p(x)))dx = p(s) +1-\int_0^1p(x)dx$$ where we used that since $x$ is $U[0,1],$ its pdf is $f(x) = 1$ if $x\in[0,1]$ and $f(x)=0$ if $x\notin [0,1]$ and also used that $\int \delta(x-s)g(x)dx = g(s).$

Then, we can apply Bayes' rule and write $$ f(x|s) = \frac{f(s|x)f(x)}{f(s)} = \frac{\delta(s-x)p(x) + (1-p(x))}{p(s) +1-\int_0^1p(x)dx}.$$

In english this means that, conditional on $s,$ $x$ takes the value $s$ with probability $$ \frac{p(s)}{p(s) +1-\int_0^1p(x)dx}$$ and otherwise is distributed with a PDF proporitional to $1-p(x)$ on $[0,1].$ As a check, in the case that $p(x)=p$ is constant, this means that $x$ takes the value $s$ with probability $p$ and is uniform on $[0,1]$ otherwise, which is true.

Taking the conditional expectation, $$ E(x|s) = \int xf(x|s) = \frac{1}{p(s) +1-\int_0^1p(x)dx}\int_0^1 (x \delta(s-x)p(x) + x(1-p(x)))dx \\= \frac{sp(s)+\frac{1}{2}-\int_0^1 xp(x)dx}{p(s) +1-\int_0^1p(x)dx}.$$

As a check, when p(x) = p is constant, this reduces to $$E(x|s) = sp + \frac{1}{2}(1-p)$$ which makes sense since there's probabiliy $p$ that $x=s$ in which case the expected value should be $s$ and probability $1-p$ that $s$ gives us no info about $x$ so that $x$ is uniform and has expected value $1/2.$

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    Thanks a lot, everything is clear now! :) – 2017-02-06