This figure below describes the joint PDF of the random variables X and Y. These random variables take values over the unit disk. Over the upper half of the disk (i.e., when y>0), the value of the joint PDF is 3a, and over the lower half of the disk (i.e., when y<0), the value of the joint PDF is a (for some positive constant a which we will solve for in the problem).
In each part below, express your answer using standard notation . Enter 'pi' for π.
Helpful fact: the area of a circle of radius r is πr2.
1)Find a.
2)Compute E[XY].
Hint: once you setup your integral in terms of dx and dy, you may want to use polar coordinates (take x=rcos(θ), y=rsin(θ), and dxdy=rdrdθ). Alternatively, you can find the solution with very little calculation through a symmetry argument.
3)Compute the marginal PDFs fX(x) and fY(y).
If −1<x<1, fX(x)=
If −1<y<0, fY(y)=
If 0<y<1, fY(y)=
4)Compute the expected values E[X] and E[Y].
Hint: In evaluating integrals, it may be useful to write 2ydy as d(y2) and make a change of variables.
5)Compute E[X|Y>0] and var(X|Y>0).
question 2 and the first part of 4 and 5 =0
question 3:
If −1<x<1, fX(x)= 2/(pi) * sqrt(1-x^2)
If −1<y<0, fY(y)= 1/pi * sqrt(1 - y^2)
If 0<y<1, fY(y)= 3/pi * sqrt(1 - y^2)
so what is the answer for 1) and 4b) + 5b)?Thanks for helping out
1: a=0.1592
4b: 0.2122
5b is a pain. Did not find the correct answer so far.
For part 5 (though this only affects 5b), ensure that you have normalized the conditional PDF.
5b: 1/4 (one fourth) or 0.25 if you want
1) To find the constant "a", we need to set up and solve an equation using the given information about the joint PDF.
Since the joint PDF is constant over the upper half of the disk (y > 0) with a value of 3a, and constant over the lower half (y < 0) with a value of a, we can use the fact that the total probability over the entire unit disk must be 1.
The area of the upper half of the disk is half the area of the whole disk, which is (π * 1^2) / 2 = π/2. The area of the lower half of the disk is also (π * 1^2) / 2 = π/2.
So, we can set up the equation:
(3a * π/2) + (a * π/2) = 1
To solve for "a", we can simplify and solve the equation:
(3a/2) + (a/2) = 1
(4a/2) = 1
2a = 1
a = 1/2
Therefore, "a" is equal to 1/2.
2) To compute E[XY], we need to set up and evaluate the integral using the joint PDF.
In this case, we can use polar coordinates since the random variables take values over the unit disk. We have x = r * cos(θ) and y = r * sin(θ), and the range of integration for r is from 0 to 1 and for θ is from 0 to 2π.
The integral for E[XY] can be set up as follows:
E[XY] = ∫∫xy * f(x, y) * dx * dy
= ∫∫(r * cos(θ)) * (r * sin(θ)) * f(x, y) * r * dr * dθ
= ∫∫r^2 * cos(θ) * sin(θ) * f(x, y) * dr * dθ
Since the joint PDF is a constant over the upper and lower halves of the disk, we can also use a symmetry argument. The product of cos(θ) and sin(θ) is an odd function over the interval [0, π], so the integral of this product over the upper and lower halves will cancel each other out.
Therefore, E[XY] = 0.
3) To compute the marginal PDFs fX(x) and fY(y), we need to integrate the joint PDF over the appropriate ranges.
For fX(x), we integrate the joint PDF over y:
fX(x) = ∫f(x, y) * dy
Since the joint PDF is constant over the upper and lower halves of the disk, we can set up the integral accordingly:
If -1 < x < 1, then:
fX(x) = ∫[(3a * 1/2) + (a * 1/2)] * dy
= ∫2a * dy
= 2a * y
If -1 < y < 0, then:
fY(y) = a
If 0 < y < 1, then:
fY(y) = 3a
4) To compute the expected values E[X] and E[Y], we need to integrate the respective random variables multiplied by their corresponding marginal PDFs.
For E[X], we evaluate:
E[X] = ∫x * fX(x) * dx
Using the marginal PDF we derived earlier:
E[X] = ∫x * 2a * y * dx
= 2a * ∫x * y * dx
To simplify the integral, we can use the fact that 2y * dy is equivalent to d(y^2) and make a change of variables:
E[X] = 2a * ∫x * d(y^2)
= 2a * ∫x * d(y^2/dx) * dx
= 2a * ∫d(x * y^2) / dx * dx
= 2a * ∫d(x * y^2)
= 2a * (x * y^2) + C
For E[Y], we can use the derived marginal PDF:
E[Y] = ∫y * fY(y) * dy
If -1 < y < 0, then:
E[Y] = ∫y * a * dy
= a * (1/2) * y^2 + C
If 0 < y < 1, then:
E[Y] = ∫y * 3a * dy
= 3a * (1/2) * y^2 + C
5) To compute E[X|Y > 0] and var(X|Y > 0), we need to condition on the event Y > 0 and use conditional expectations and variances.
To compute E[X|Y > 0], we need to find the conditional PDF fX|Y > 0(x) and then find the expectation:
E[X|Y > 0] = ∫x * fX|Y > 0(x) * dx
To compute var(X|Y > 0), we need to find the conditional PDF fX|Y > 0(x) and then find the variance:
var(X|Y > 0) = ∫(x - E[X|Y > 0])^2 * fX|Y > 0(x) * dx