# Marginal probability distribution

• _joey
In summary, the conversation discusses finding the marginal density distribution for two random variables X and Y, using the probability density function f(x,y)= 6x^2y with constraints 0<x<y and x+y<2. The calculations for the marginal density of X are correct, but there are errors in the marginal density for Y. After correcting the errors, integrating the marginal density over the appropriate bounds gives a volume of 1, confirming that the marginal density is normalized. The conversation also includes a question about finding the bounds for the conditional probability f(y|x), which is clarified as being the same as the bounds for f_X(x).
_joey
I have a question I need to answer.

$$f\left(x,y\right)= 6x^2y$$ such that $$0<x<y$$ and $$x+y<2$$ where $$f\left(x,y\right)$$ is a probability density function for two random variables: $$X,\;Y$$

I need to find marginal density distribution for the random variables $$X$$ and $$Y$$

It appears to be a straight forward question except that when I integrate twice, in instance with the marginal density for $$Y,\;f_Y$$, it does not integrate to 1.

Here are my calculations:

$$0&<&x<y$$ and $$x+y<2 \implies 0<x<1$$ and $$x<2-y$$. Hence, marginal density for $$X$$ is

$$f_x\left(x,y\right)=\int\limits_{x}^{2-x}6x^2y\,dy =$\left. 3{{x}^{2}}{{y}^{2}} \right|_{x}^{2-x}$=12x^2-12x^3,\;0<x<1$$

If I integrate the above function again over $$\left(0,1\right)$$I will obtain 1. This is a property of marginal density distribution (?!)

Things start falling apart with marginal density for $$Y$$ variable

$$f_y\left(x,y\right)=\int\limits_{0}^{1}6x^2y\,dx =$\left. 2{{x}^{3}}{{y}} \right|_{0}^{1}$=2y,\;x<y<2-x$$

If I integrate $$f_Y(y)=2y$$ again over $$x<y<2-x$$ I will obtain $$4-4x$$.

Any help and suggestions will be much appreciated.

Thanks!

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_joey said:
Things start falling apart with marginal density for $$Y$$ variable

$$f_y\left(x,y\right)=\int\limits_{0}^{1}6x^2y\,dx =$\left. 2{{x}^{3}}{{y}} \right|_{0}^{1}$=2y,\;x<y<2-x$$

Shouldn't you integrate x rather from 0 to y instead of from 0 to 1?
Also note that for arbitrary value of y, x can range from 0 to 2 so your first integral for the marginal density of X is also not correct.

DrDu said:
Shouldn't you integrate x rather from 0 to y instead of from 0 to 1?
Also note that for arbitrary value of y, x can range from 0 to 2 so your first integral for the marginal density of X is also not correct.

We have a set of inequalities:

0<x<y and x+y<2. If you solve the inequalities, you will get 0<x<1.
0<x+x<y+(2-y)

If set inequality constraints on a plane, you will see a triangle with points (0,0) (1,0) and (0, 2). x is bounded above by 1

If we integrate from 0 to y, as you suggest, and then from x to 2-x with respect to y (y is bounded)? The volume will not be equal to 1. This is a property of marginal density

Ok, I was to quick. Note that the second corner of the triangle must be (1,1), not (1,0) as x must be smaller than y.
The marginal density for y is then $$\int_0^y dx 6x^2y$$ for y<1 and $$\int_0^{2-y} dx 6x^2y$$ for $$y \ge 1$$.
I'm in a hurry, hope I didn't make an error this time.

DrDu said:
Ok, I was to quick. Note that the second corner of the triangle must be (1,1), not (1,0) as x must be smaller than y.
The marginal density for y is then $$\int_0^y dx 6x^2y$$ for y<1 and $$\int_0^{2-y} dx 6x^2y$$ for $$y \ge 1$$.
I'm in a hurry, hope I didn't make an error this time.

Yes, point (1,1), it was a typo.

If we integrate the marginal density you obtained ($$12y^3-24y^2+16y$$) for 2nd time to see if it is valid (and the joint distribution from which density was obtained is valid), we don't get volume equal to 1. Unless the bounds are mixed in the 2nd integration.

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This problem is giving me a headache.:(

I did the integrals and got 1. The integral over y from 0 to 1 gives 2/5 and the second one over y from 1 to 2 gives 1-2/5.

Could you please elaborate on how and why you are choosing these bounds or refer me to relevant information.

DrDu,

What do you think of my solution for marginal density of Y that is, for f(x,y) over (0, 1) and then over (x, 2-x)

The ranges of integration are easy to find. You already identified the relevant triangle.
To get the marginal distribution of y, you have to integrate over x with y hold fixed. If y is smaller 1, x can range from 0 t0 y. If y is greater than 1, you are on the other side of the triangel and x can range from 0 to 1-y.
Hence f(y)=2y^4 for y<1 and 2(2-y)^3 y for y>1.
To confirm that the marginal density is normalized, you integrate over y. The first integral with y ranging from 0 to 1 is easy and yields 2/5, the second integral with y ranging from 1 to 2 becomes easy after introduction of z=2-y as a new variable of integration. It yields 1-2/5.
So the total integral is 1.

DrDu said:
The ranges of integration are easy to find. You already identified the relevant triangle.
To get the marginal distribution of y, you have to integrate over x with y hold fixed. If y is smaller 1, x can range from 0 t0 y. If y is greater than 1, you are on the other side of the triangel and x can range from 0 to 1-y.
Hence f(y)=2y^4 for y<1 and 2(2-y)^3 y for y>1.
To confirm that the marginal density is normalized, you integrate over y. The first integral with y ranging from 0 to 1 is easy and yields 2/5, the second integral with y ranging from 1 to 2 becomes easy after introduction of z=2-y as a new variable of integration. It yields 1-2/5.
So the total integral is 1.

Another question on bounds. It is a simple question but I lost confidence in setting up the bounds.

If I need to find conditional probability f(y|x). Then, $$f(y|x) = \frac{f(x,y)}{f_X(x)}$$. In our case, $$f_X(x)}=\int\limits_{x}^{2-x}6x^2y dy = 12x^2-12x^3, 0<x<1$$ and $$f(y|x)=\frac{6x^2y}{12x^2-12x^3}=\frac{2y}{2-2x}$$

Question: the bounds for $$f(y|x)$$ are the same as for $$f_X(x)$$? That is, $$0<x<1$$ and $$x<y<2-x$$?

Thanks

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## 1. What is a marginal probability distribution?

A marginal probability distribution is a type of probability distribution that shows the probabilities of each possible value of a single random variable. It is calculated by summing or integrating the joint probability distribution over all other variables, leaving only the variable of interest.

## 2. How is a marginal probability distribution different from a joint probability distribution?

A marginal probability distribution shows the probabilities of a single variable, while a joint probability distribution shows the probabilities of multiple variables occurring together. Marginal probability distributions are calculated by summing or integrating over all other variables, while joint probability distributions are calculated by multiplying the probabilities of each individual variable.

## 3. Why is the marginal probability distribution important in statistics?

The marginal probability distribution is important in statistics because it allows us to analyze and understand the behavior of individual variables in a larger system. It can also be used to calculate important statistics such as means, variances, and correlations.

## 4. How is the marginal probability distribution related to the cumulative distribution function?

The marginal probability distribution is related to the cumulative distribution function (CDF) because the CDF is the integral of the marginal probability distribution. In other words, the CDF shows the probability of a variable being less than or equal to a certain value, while the marginal probability distribution shows the probability of a variable taking on that specific value.

## 5. Can a marginal probability distribution be used to make predictions about future events?

Yes, a marginal probability distribution can be used to make predictions about future events. By analyzing the probabilities of different values of a variable, we can estimate the likelihood of certain outcomes occurring. However, it is important to note that the accuracy of these predictions depends on the quality of the data and assumptions used to create the marginal probability distribution.

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