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I have a question about conditional probability.

Let's say

$y = x + e$

where $x$ ~ $U[0,1]$ and $e$ ~ $N(0,1)$ and both are independent.

What is $p_x(x | y = 10)$ ?

In particular why isn't it the same as $p_e(10 - x)$, i.e. $N(10-x, 1)$? I'm trying to see where the constraint of x being in the range $[0,1]$ comes into play here. I'm likely missing something fundamental, would appreciate any pointers on this.

user3882729
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1 Answers1

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In particular why isn't it the same as $p_e(10−x)$, i.e. $N(10−x,1)$? I'm trying to see where the constraint of $x$ being in the range $[0,1]$ comes into play here.

$p_x(x\mid y=10)$ is a probability density function, so its integral over the support of $[0,1]$ must equal $1$. This is not the case for a normal distributed random variable, so it is not that.

$$\int_0^1 p_x(x\mid y=10)\,\mathrm d x=1\\\int_0^1 p_e(10-x)\,\mathrm d x\neq 1$$

Use Bayes' Rule to find what actually is the conditional distribution.

$$\begin{align} p_x(x\mid y=10) &=\dfrac{p_x(x)\,p_e(10-x)}{p_y(10)}\\[1ex]&=\dfrac{p_x(x)\,p_e(10-x)}{\displaystyle\int_0^1p_x(s)\,p_e(10-s)\,\mathrm d s} \end{align}$$

Graham Kemp
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  • Thanks @Graham, using Bayes theorem as above makes sense. Could you please explain though why $y = x + e$ cannot be re-written as $x = y - e$ and therefore making $p_x(x|y)$ normally distributed. IOW, what is the argument that prevents this math (aside from the integral over support not being 1) ? – user3882729 Aug 23 '23 at 17:52
  • $x$ is equal to $y-e$. However, this will not make $x$ conditionally normal, no more than $e=y-x$ makes $e$ conditionally uniform. – Graham Kemp Aug 23 '23 at 23:40