Marginal densities of a Probability

In summary, the joint probability density function for Y1 and Y2 is given by f(y1,y2) = 8y1y22 for 0<=y1<=1, 0<=y2<=1, and y12<=y2, and 0 elsewhere. The marginal functions are f1(y1) = 8/3y1 - 8/3y17 and f2(y2) = 4y23. The limits of integration for both are from 0 to 1, and the expected values for Y1 and Y2 should be calculated using these limits.
  • #1
RET80
15
0

Homework Statement


Y1 and Y2 have a joint probability density function given by:
f(y1,y2) = {8y1y22, 0<=y1<=1, 0<=y2<=1, y12<=y2
0, Elsewhere

Homework Equations


f1(y1) =ʃ f(y1,y2) dy2
f2(y2) =ʃ f(y1,y2) dy1

For E(Y) (later, discussed in part 3):
E(Y1) = ʃ y1f(y1,y2) dy1
E(Y2) = ʃ y1f(y1,y2) dy2

All integrals are set to -infinity to +infinity, which are then adjusted to the boundaries of the density function

The Attempt at a Solution


I attempted both marginal functions and set the limits of integration as follows:
for f1(y1), limits of integration were: y1 to 1
for f2(y2), limits of integration were: sqrt(y2) to 0

With those limits of integration set up, I then solved for both separately and received the answers:
f1(y1) = 8/3y1 - 8/3y17
f2(y2) = 4y23

Now my question is, are my limits of integration setup correctly for this kind of question/equation. I don't think they are because after this, I solve for E(Y1) and E(Y2) and if I'm not mistaken, those must be whole numbers and not numbers with variables and if I use the same limits of integration that I used here, it will not work (or it will and just look awful). Not too sure here.
 
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  • #2
RET80 said:

Homework Statement


Y1 and Y2 have a joint probability density function given by:
f(y1,y2) = {8y1y22, 0<=y1<=1, 0<=y2<=1, y12<=y2
0, Elsewhere

Homework Equations


f1(y1) =ʃ f(y1,y2) dy2
f2(y2) =ʃ f(y1,y2) dy1

For E(Y) (later, discussed in part 3):
E(Y1) = ʃ y1f(y1,y2) dy1
E(Y2) = ʃ y1f(y1,y2) dy2

All integrals are set to -infinity to +infinity, which are then adjusted to the boundaries of the density function

The Attempt at a Solution


I attempted both marginal functions and set the limits of integration as follows:
for f1(y1), limits of integration were: y1 to 1
for f2(y2), limits of integration were: sqrt(y2) to 0

With those limits of integration set up, I then solved for both separately and received the answers:
f1(y1) = 8/3y1 - 8/3y17
f2(y2) = 4y23

Now my question is, are my limits of integration setup correctly for this kind of question/equation. I don't think they are because after this, I solve for E(Y1) and E(Y2) and if I'm not mistaken, those must be whole numbers and not numbers with variables and if I use the same limits of integration that I used here, it will not work (or it will and just look awful). Not too sure here.

You have them both correct. But remember, they are both valid for their variable from 0 to 1, and 0 elsewhere. So when you calculate expected values your integrals should go from 0 to 1; no variables in the limits or answers.
 

1. What is a marginal density?

A marginal density is a probability density function that describes the probability of a subset of random variables occurring, given that other variables have taken on specific values. It is obtained by summing or integrating a joint probability distribution over the values of the variables that are not of interest.

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

A joint probability distribution describes the probability of multiple variables occurring together, while a marginal density focuses on the probability of a subset of those variables occurring. It is obtained by "marginalizing" or removing the other variables from the joint probability distribution.

3. Can a marginal density be used to calculate individual probabilities?

Yes, a marginal density can be used to calculate probabilities for individual variables within a joint probability distribution. This is especially useful when dealing with a large number of variables, as it allows for simpler and more focused calculations.

4. Are marginal densities only applicable to continuous variables?

No, marginal densities can be used for both continuous and discrete variables. For continuous variables, the marginal density is a probability density function, while for discrete variables it is a probability mass function.

5. How do you interpret a marginal density graph?

A marginal density graph shows the relationship between the values of one variable and the probability of that variable occurring. The area under the curve represents the total probability for that variable. The shape of the curve can also provide information about the distribution of the variable. For example, a bell-shaped curve indicates a normally distributed variable.

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