#StackBounty: #self-study #bayesian #continuous-data #uniform #analytical Two dependent uniformly distributed continuous variables and …

Bounty: 50

I am trying to solve the following exercise from Judea Pearl’s Probabilistic Reasoning in Intelligent Systems: Networks of Plausible Inference.

2.2. A billiard table has unit length, measured from left to right. A ball is rolled on this table, and when it stops, a partition is placed at its stopping position, a distance $x$ from the left end of the table. A second ball is now rolled between the left end of the table and the partition, and its stopping position, $y$, is measured.

a. Answer qualitatively: How does knowledge of $y$ affect our belief about $x$? Is $x$ more likely to be near $y$ , far from $y$, or near the midpoint between $y$ and 1?

b. Justify your answer for (a) by quantitative analysis. Assume the stopping position is uniformly distributed over the feasible range.

For b., I clearly need to use Bayes’ theorem:

$$
P(X|Y) = dfrac{P(Y|X)P(X)}{P(Y)}
$$

where I expressed

$$
P(X) sim U[0,1] =
begin{cases}
1, text{where } 0 leq x leq 1\
0, text{else}
end{cases}
\
P(Y|X) sim U[0,x] =
begin{cases}
1/x, text{where } 0 leq y leq x\
0, text{else}
end{cases}
$$

I tried getting $P(Y)$ by integrating the numerator over $X$.

$$
int_{-infty}^{infty} P(Y|X)P(X)dx = int_{0}^{1}P(Y|X)cdot 1 dx = int_{0}^{1}dfrac{1}{x} dx
$$

But the integral doesn’t converge.

I also tried to figure out the numerator itself, but I don’t see how $frac{1}{x}$ can represent $P(X|Y)$.

Where did I go wrong?


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