# Thread: How can we condition on events with probability 0?

1. ## How can we condition on events with probability 0?

We are given that the joint density of X and Y is

$\displaystyle f(x,y) = \begin{cases} & .5 \; \text{if } \; 0 \leq x \leq .5 \; \;\text{and}\; 0 \leq y \leq .5 \\ & 1.25 \; \text{if } \; 0 \leq x \leq .5 \; \;\text{and}\; .5 \leq y \leq 1 \\ & 1.5 \; \text{if} \; .5 \leq x \leq 1 \; \;\text{and}\; 0 \leq y \leq .5 \\ & .75 \; \text{if} \; .5 \leq x \leq 1 \; \;\text{and}\; .5 \leq y \leq 1 \\ \end{cases}$

And are asked to find the conditional distribution of Y given X = .75

the numerator $\displaystyle f(.75,y)$ is
1.5 if $\displaystyle 0 \leq y \leq 5$
.75 if $\displaystyle .5 \leq y \leq 1$

The thing that troubles me is calculating the marginal density of X at .75
$\displaystyle f_x(.75) = \int_{0}^{.5} 1.5 \;\text{d}y \;+\; \int_{.5}^{1} .75 \;\text{d}y \; = 1.125$

We have the event X=.75, which should have probability 0, but when we calculate it's marginal density at .75 we get a non zero number which is (even worse) greater than 1.

Is this something that just happens when we condition on events with probability 0? In general, how do conditional distributions even work in the continous case when we coondition on an event probability 0?

2. ## Re: How can we condition on events with probability 0?

Originally Posted by Jame
We are given that the joint density of X and Y is

$\displaystyle f(x,y) = \begin{cases} & .5 \; \text{if } \; 0 \leq x \leq .5 \; \;\text{and}\; 0 \leq y \leq .5 \\ & 1.25 \; \text{if } \; 0 \leq x \leq .5 \; \;\text{and}\; .5 \leq y \leq 1 \\ & 1.5 \; \text{if} \; .5 \leq x \leq 1 \; \;\text{and}\; 0 \leq y \leq .5 \\ & .75 \; \text{if} \; .5 \leq x \leq 1 \; \;\text{and}\; .5 \leq y \leq 1 \\ \end{cases}$

And are asked to find the conditional distribution of Y given X = .75

the numerator $\displaystyle f(.75,y)$ is
1.5 if $\displaystyle 0 \leq y \leq 5$
.75 if $\displaystyle .5 \leq y \leq 1$

The thing that troubles me is calculating the marginal density of X at .75
$\displaystyle f_x(.75) = \int_{0}^{.5} 1.5 \;\text{d}y \;+\; \int_{.5}^{1} .75 \;\text{d}y \; = 1.125$

We have the event X=.75, which should have probability 0, but when we calculate it's marginal density at .75 we get a non zero number which is (even worse) greater than 1.

Is this something that just happens when we condition on events with probability 0? In general, how do conditional distributions even work in the continous case when we coondition on an event probability 0?
It is a density. The probability of being in the interval [x-dx/2,x+dx/2] is approximately f(x)dx and it goes to zero as dx goes to zero. Which is what you want, the probability of x is zero, but the density is non-zero.

The requirement for being a density is that its integral over the relevant space be 1, it does not itself have to be less than 1.

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