Continuous joint probability density functions

[das1]

Consider the following joint probability distribution function of (X , Y):

a(x + y^2) {0<=x<=2, 0<=y<=2}
0 otherwise

Calculate the value of the constant a that makes this a legitimate probability distribution. (Round your answer to four decimal places as appropriate.)

And then,
For the distribution above, determine the marginal probability density function of Y and use it to calculate the probability that 0.25 < Y < 0.75. (Round your answer to the fourth decimal place as appropriate.)

 

[Jameson]

What have you tried so far? What properties of joint probability distribution functions do you know? There is one property that was used in a thread you made yesterday which is useful here, just extended to two variables.

 

[das1]

OK so this is what I've done:
I set up the integral from 0 to 2 of (x+y^2)dx dy = 1
I've never done an integral with 2 variables before but I plugged it into this calculator:
Integral Calculator - Symbolab
and it gave me 28/3 ? Set that = 1 and divide and get 3/28?

That's my best guess so far, but I could be way off.

 

[Jameson]

You've never worked with double integrals before? Do you know how to solve something like this?

\(\displaystyle \int_{0}^{1} \int_{1}^{5} x+ydydx\)

If not you haven't been given the tools you need to solve this problem so it's totally understandable why this is difficult! :)

You are on the right track though in your reasoning. The double integral of the joint pdf should equal one:

\(\displaystyle \int_{0}^{2} \int_{0}^{2} a(x+y^2)dydx=1\)

 

[CellCoree]

I would first like to clarify that the given joint probability density function is not continuous as it has a jump discontinuity at x = 2 and y = 2. It can be considered as a piecewise continuous function.

To calculate the value of the constant a that makes this a legitimate probability distribution, we need to ensure that the total probability over the entire domain is equal to 1. Therefore, we can set up the following equation:

1 = ∫∫ a(x + y^2) dxdy

Using the given limits of integration, we can solve this integral as follows:

1 = a∫∫ (x + y^2) dxdy
= a∫(0 to 2)∫(0 to 2) (x + y^2) dxdy
= a∫(0 to 2) (x^2/2 + xy^2) from 0 to 2 dy
= a∫(0 to 2) (4 + 2y^2) dy
= a(4y + (2y^3)/3) from 0 to 2
= a(8 + 16/3)
= (8a + 16/3)

Equating this to 1, we get:

1 = 8a + 16/3
=> 8a = 1 - 16/3
=> 8a = -13/3
=> a = (-13/3)/8
=> a = -13/24

Therefore, the value of the constant a that makes this a legitimate probability distribution is -13/24.

Moving on to the second part, to determine the marginal probability density function of Y, we need to integrate the joint probability density function over all possible values of X. This can be expressed as follows:

fY(y) = ∫∫ a(x + y^2) dxdy
= a∫(0 to 2)∫(0 to 2) (x + y^2) dxdy
= a∫(0 to 2) (x^2/2 + xy^2) from 0 to 2 dy
= a∫(0 to 2) (4 + 2y^2) dy
= a(4y + (2y^3)/3)