Marginal density function understanding

[Askhwhelp]

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

 

[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.

 

[Askhwhelp]

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?

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?

 

[jbunniii]

No an example will be uniformly distribution function...this will give a constant function...

No, it's not constant everywhere. Doesn't it have two jump discontinuities?

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?

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 $f_2$?

 

[Askhwhelp]

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 $f_2$?

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.
What is about this?

 

[Askhwhelp]

Please check my understanding

 

[jbunniii]

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.
What is about this?

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?

 

[Askhwhelp]

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?

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

 

[jbunniii]

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

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.

 

[Askhwhelp]

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.

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?

 

[Ray Vickson]

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.

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.

 

[Askhwhelp]

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.

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

 

[jbunniii]

A probability density function can, indeed, be increasing, decreasing or constant within its natural domain (where it is > 0).

(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 $\mathbb{R}$.

 

[Ray Vickson]

(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 $\mathbb{R}$.

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.

 

[Askhwhelp]

.

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?

 

[jbunniii]

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.

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

 

[jbunniii]

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?

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.

 

[Askhwhelp]

Well, this is a homework forum and you still haven't shown any attempt,

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

 

[jbunniii]

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

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 $f_2$.)

 

[Ray Vickson]

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

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.

 

[Askhwhelp]

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 $f_2$.)

My new understanding is that
If we have a joint (continuous) probability density function = 0 such that F(y)=\begin{cases} \int_0^{2y+10} 0 dx,& -5\le x \le 0\\ \int_0^{10-2y} 0 dx, & 0\le x \le 5\end{cases} = F(y)=\begin{cases} 0, & -5\le x \le 0\\ 0 ,& 0\le x \le 5\end{cases}.

so this is a constant function.

But if we have a joint (continuous) probability density function = 1 such that F(y)=\begin{cases} \int_0^{2y+10} 1 dx,& -5\le x \le 0\\ \int_0^{10-2y} 1 dx, & 0\le x \le 5\end{cases} = F(y)=\begin{cases} 2y+10, & -5\le x \le 0\\ 10-2y ,& 0\le x \le 5\end{cases}

so this is a increasing and then decreasing function.

Different joint (continuous) probability density gives different characteristics of the functions, increasing, decreasing, and constant...Therefore, without more information, we cannot classify whether it is increasing, decreasing or constant.

Is this right? Is this enough?

 

[jbunniii]

My new understanding is that
If we have a joint (continuous) probability density function = 0 such that F(y)=\begin{cases} \int_0^{2y+10} 0 dx,& -5\le x \le 0\\ \int_0^{10-2y} 0 dx, & 0\le x \le 5\end{cases} = F(y)=\begin{cases} 0, & -5\le x \le 0\\ 0 ,& 0\le x \le 5\end{cases}.

so this is a constant function.

It is a constant function, but it is not a valid probability density function, because it doesn't integrate to 1. Same problem with the joint pdf. Your reasoning is valid, though. If you can find a valid joint pdf which results in a valid constant pdf, then the conclusion is that there are at least two possibilities: (1) constant pdf, (2) increasing then decreasing pdf.

That doesn't answer your original question, however, which asked about these four possibilities: (1) increasing pdf, (2) decreasing pdf, (3) neither increasing nor decreasing; (4) not enough information. Your two examples both fall into category (3).

 

[Askhwhelp]

It is a constant function, but it is not a valid probability density function, because it doesn't integrate to 1. Same problem with the joint pdf. Your reasoning is valid, though. If you can find a valid joint pdf which results in a valid constant pdf, then the conclusion is that there are at least two possibilities: (1) constant pdf, (2) increasing then decreasing pdf.

That doesn't answer your original question, however, which asked about these four possibilities: (1) increasing pdf, (2) decreasing pdf, (3) neither increasing nor decreasing; (4) not enough information. Your two examples both fall into category (3).

Could you show me an example where the probability density function integrate 1?

 

[jbunniii]

Could you show me an example where the probability density function integrate 1?

Actually, I just noticed that your other example doesn't integrate to 1, either. The one where you set the joint pdf equal to 1. But that one can easily be fixed, by dividing by an appropriate constant (the area of the triangular region). If you do that, then that gives you an example that integrates to 1.

 

[Askhwhelp]

Actually, I just noticed that your other example doesn't integrate to 1, either. The one where you set the joint pdf equal to 1. But that one can easily be fixed, by dividing by an appropriate constant (the area of the triangular region). If you do that, then that gives you an example that integrates to 1.

To make the problem easier, let make the triangle be three points at (0,1), (0,-1), and (2,0)...
In this case... a joint (continuous) probability density function = 1/2 such that F(y)=\begin{cases} \int_0^{2y+2} 1/2 dx,& -1\le x \le 0\\ \int_0^{2-2y} 1/2 dx, & 0\le x \le 1\end{cases} ==> F(y)=\begin{cases} y+1, & -1\le x \le 0\\ 1-y ,& 0\le x \le 1\end{cases}

so this is a increasing and then decreasing function. This example is right?

As you say, this falls into category (3) neither an increasing nor decreasing function...so I guess my next step is to find a function that is increasing function or decreasing function. So that I can pick answer (4) for cannot classify them...now I am wondering do they exist ...even they exist, I also have to be able to integrate the function to 1 which is very difficult?...If they exist, can you think of a good way to explain it other than finding an example
, assuming finding a increasing function example is difficult? If not, do you have any idea or suggestion how to find an increasing or decreasing function?

 

[Askhwhelp]

Just realize something ... When the marginal density function begins at y=-1, it must increase at some point before y=1 since the PDF has to be able to integrate to 1 ..on the other than when it got to y=1 ...it must decrease back to 0...so the marginal density function f(y) could not be increasing or decreasing function right...therefore, the answer would be neither increasing nor decreasing