Finding marginal distribution of 2d of probability density function

[Master1022]

Homework Statement

Given the following probability distribution $f(x, y)$, find $p(x)$, the marginal distribution.

Relevant Equations

Marginal probability

Hi,

I have question about finding marginal distributions from 2d marginal pdfs that lead to the probabilities being greater than 1.

Question:
If we have the joint probability distribution $f(x, y) = k \text{ for} |x| \leq 0.5 , |y| \leq 0.5$ and 0 otherwise. I have tried to define a square with side lengths 1 which is centered at the origin.
1) Find the constant k
2) Find p(x), the marginal probability distribution

Attempt:
1) Using the law of total probability:
\iint f(x, y) dx dy = 1 \rightarrow k(1)(1) = 1 \rightarrow k = 1

2) Now when we find the marginal distribution p(x). Given the symmetry of the distribution, I thought this could also be done by folding the square over the x-axis and combining the probabilities. However, if we do this, then the distribution becomes a uniform distribution with p(x) = 2 for

-0.5 \leq x \leq 0.5. This doesn’t seem to be correct.

Would someone be able to point me the right direction of how to proceed? Thank you in advance

 

[haruspex]

I thought this could also be done by folding the square over the x-axis and combining the probabilities.

That would still give a joint distribution, but with a halved range for y.
You need to eliminate y by integrating across its range, keeping x constant.
In general, the bounds of y might depend on x, but not here.

 

[FactChecker]

2) Now when we find the marginal distribution p(x). Given the symmetry of the distribution, I thought this could also be done by folding the square over the x-axis and combining the probabilities.

Ok. But you still have to integrate this over the "folded" values of $0 \lt y \lt 0.5$, so what does that give you? In fact, you didn't get any benefit from "folding" the original joint PDF. You could have just integrated $f(x,y)=1$ over the values of $-0.5 \lt y \lt 0.5$.

However, if we do this, then the distribution becomes a uniform distribution with p(x) = 2 for

-0.5 \leq x \leq 0.5. This doesn’t seem to be correct.

It is not correct.

 

[Master1022]

Thank you very much @haruspex and @FactChecker for your replies! I just have some follow up questions. I do see that I ought to have integrated.

That would still give a joint distribution, but with a halved range for y.

Okay that makes more sense - compressing that y range from 0 to 0.5 would result in a uniform distribution of $p(x) = 1$ from $-0.5 \leq x \leq 0.5$.

You could have just integrated $f(x,y)=1$ over the values of ##-0.5 \lt y \lt 0.5

You need to eliminate y by integrating across its range, keeping x constant.

So if I integrate then $p(x) = \int f(x, y) dy = \int_{-0.5}^{0.5} (1) dy = 1$. This agrees with the answer above of $p(x) = 1$ from $-0.5 \leq x \leq 0.5$

In general, the bounds of y might depend on x, but not here.

Okay thank you. So if we had a similar situation, but instead we had a circle (still centered) at the origin with radius 1/\sqrt{\pi} and we wanted to find the marginal distribution $p(x)$, then would have an integral that looked something like?
p(x) = \int f(x, y) dy = \int_{-\sqrt{\frac{1}{\pi} - x^2}}^{\sqrt{\frac{1}{\pi} - x^2}} (1) dy = 2 \sqrt{\frac{1}{\pi} - x^2}

However, this does seem to suggest a probability greater than $1$ for $x = 0$...

 

[haruspex]

this does seem to suggest a probability greater than 1 for x=0...

Why is that a problem? Probability densities can be as high as you like.

 

[FactChecker]

However, this does seem to suggest a probability greater than $1$ for $x = 0$...

Not probability, which can never be greater than 1, but probability density, which can be as great as you like as long as its integral does not exceed 1.

 

[Master1022]

Why is that a problem? Probability densities can be as high as you like.

Not probability, which can never be greater than 1, but probability density, which can be as great as you like as long as its integral does not exceed 1.

That is true. Thank you very much for the clarification!