Marginal density function understanding

1. Dec 7, 2013

Given a plane with three points, (0, -5), (10,0), and (0, 5) with x-axis and y-axis connecting three points to make a triangle. Suppose this triangle represents the support for a joint continuous probability density

What can we say about the marginal density f_2(y)? Are they going to be increasing function, decreasing function, neither increasing nor decreasing function, or cannot be identified because of lack of information. Please explain you answer

2. Dec 7, 2013

jbunniii

You deleted the homework template and didn't show any attempt, so I can't offer much help, but consider this: is it possible for ANY probability density function to be monotonically increasing or decreasing?

Mod note from Mark44: Belatedly noted and addressed about the lack of an attempt.

Last edited by a moderator: Dec 11, 2013
3. Dec 7, 2013

No an example will be uniformly distribution function...this will give a constant function... With that said, the marginal density function f_2(y) could be increasing, decreasing, and constant. So we cannot classify whether marginal density function f_2(y) as increasing, decreasing, and constant without more information ....right?

4. Dec 7, 2013

jbunniii

No, it's not constant everywhere. Doesn't it have two jump discontinuities?
Well, a density function has to be nonnegative everywhere and it must integrate to 1. Therefore it must take on some positive value somewhere, say at $y_0$ we have $f_2(y_0) = c > 0$. Now if $f_2$ is increasing, what can you say about the values of $f_2$ for $y > y_0$? What does this imply about the integral of $f_2$?

5. Dec 7, 2013

Let change my understanding a little bit: Since we are not given the joint density function, we cannot conclude that the marginal density is increasing, or decreasing or neither. If our original joint density is constant, then we get increase up to y = 0, then decrease.

6. Dec 7, 2013

7. Dec 8, 2013

jbunniii

Forget about the joint density for a minute. Just consider a density function in one variable, say $f(y)$. Since $f$ is a density function, it must be nonnegative and it must integrate to 1. Therefore there must be some $y_0$ for which $f(y_0) > 0$. Now, suppose $f$ is monotonically increasing. This means that $f(y) \geq f(y_0)$ for all $y \geq y_0$. But this means that
$$\int_{-\infty}^{\infty} f(y) dy = \int_{-\infty}^{y_0} f(y) dy + \int_{y_0}^{\infty} f(y) dy \geq \int_{-\infty}^{y_0} f(y) dy + \int_{y_0}^{\infty} f(y_0) dy$$
Now what is the value of that last integral?

8. Dec 8, 2013

Where is this going? It seems like it is getting away from my understanding...

9. Dec 8, 2013

jbunniii

I am trying to demonstrate to you that the question as posed doesn't make much sense. No pdf can be an increasing function, or a decreasing function, or a constant function.

10. Dec 8, 2013

Just to clarify: you mean we cannot classify marginal density function f_ 2(y) as increasing, decreasing, and neither an increasing nor decreasing function, without more information, right?

Last edited: Dec 8, 2013
11. Dec 8, 2013

Ray Vickson

A probability density function can, indeed, be increasing, decreasing or constant within its natural domain (where it is > 0). Of course a probability distribution function (i.e., a cumulative distribution) must be monotone non-decreasing (and may be constant over some sub-regions). One problem is that the shorthand "pdf" can stand for either of the two, so ought to be defined before use.

12. Dec 8, 2013

The way I define marginal density f_2(y) here is f_2(y) = $$\int f(x,y) dx$$ ...if that makes the problem more clear

13. Dec 8, 2013

jbunniii

(My emphasis added.) Sure, that's true, but I did not see any mention in the question of this restriction. A probability density function of a real random variable is defined on all of $\mathbb{R}$, and it cannot be increasing, decreasing, or constant on all of $\mathbb{R}$.

14. Dec 8, 2013

Ray Vickson

In probability and statistics it is very common to say that a density is monotone (or not) on its natural domain, so we often speak of increasing densities, or decreasing densities, or U-shaped densities that first decrease, then increase, etc. On the whole line many of those behaviors are impossible, as you say, but that does not stop anybody from describing them that way. It is just one of the subject-specific conventions.

15. Dec 8, 2013

Could you check my answer to the original question : just to clarify my understanding after a long discussion, we cannot classify marginal density function f_ 2(y) as increasing, decreasing, and neither an increasing nor decreasing function, without more information, right?

16. Dec 8, 2013

jbunniii

Thanks for the info. I was unaware of this convention.

17. Dec 8, 2013

jbunniii

Well, this is a homework forum and you still haven't shown any attempt, so I can only offer a hint. Let's say you want to show that it is possible for $f_2$ to be increasing. Try choosing $f_2$ to be your favorite increasing marginal pdf with support $[-5,5]$. See if you can construct a joint pdf $f(x,y)$ whose support is the given triangle, such that $\int f(x,y) dx = f_2(y)$. Constraining $f(x,y)$ so it only depends on $y$ should make the math easier.

18. Dec 8, 2013

Go back to my original argument: is the following enought? Since we are not given the joint density function, we cannot conclude that the marginal density is increasing, or decreasing or neither. If our original joint density is constant, then it get increase up to y = 0, then decrease

19. Dec 9, 2013

jbunniii

OK, so that shows there is a joint density for which the marginal pdf is neither increasing nor decreasing. But it's plausible that EVERY joint pdf would result in a marginal pdf which is neither increasing nor decreasing. In order to show that this is not the case, you need to show that there exists a joint pdf which produces an increasing $f_2$. (It suffices to focus on the increasing case, because if you can find such a joint pdf, then you can then immediately modify it by replacing $f(x,y)$ with $f(x,-y)$ to yield a decreasing $f_2$.)

20. Dec 9, 2013

Ray Vickson

I would say it is not enough; what you seem to be saying is that you cannot see any reason why something can happen. That is very different from demonstrating that it really can happen. In other words, I would only give full marks on the question if you either prove that one or more of the behaviors cannot happen, or else give actual examples of each of the listed behaviors.