The random variables X and Y are distributed according to the joint PDF:
It has the value of a*x^2 if 1 <= x <= 2 and 0 <= y <= x, and 0 otherwise. What's the marginal fY(y) in two scenarios:
1. 0 <= y <= 1
2. 1 <= y <= 2
Just give me useful hints please
1. To find fY(y) for 0 <= y <= 1, you need to integrate the joint PDF over x for this range of y. In other words, you need to find:
fY(y) = ∫fXY(x,y)dx, where the limits of integration are 1 for the lower limit and y for the upper limit.
2. To find fY(y) for 1 <= y <= 2, you need to integrate the joint PDF over x for this range of y. In other words, you need to find:
fY(y) = ∫fXY(x,y)dx, where the limits of integration are y for the lower limit and 2 for the upper limit.
1. The result is 28/15
2. The result is 4/15*(8-y^3)
The result for the marginal fY(y) for 0 <= y <= 1 is correct, which is fY(y) = (28/15)y^3.
However, the result for the marginal fY(y) for 1 <= y <= 2 needs some clarification. Using the same formula, we have:
fY(y) = ∫ fXY(x,y) dx, with the limits being y <= x <= 2 because the joint PDF is defined for y <= x <= 2.
So we have:
fY(y) = ∫[0 to 2] a*x^2 dx if y <= x <= 2
fY(y) = a/3 (2^3 - y^3) if y <= 2,
where a is some constant that we need to determine by integrating the joint PDF over its entire domain, that is:
∫∫ fXY(x,y) dx dy = 1
Integrating fXY(x,y) over its support, we get:
∫[0 to 1] ∫[y to 2] a*x^2 dx dy = 1
This integral evaluates to:
a/3 (2^3 - 1) = 1
Therefore,
a = 3/7
Substituting this value of a in the expression for fY(y) above, we obtain:
fY(y) = (3/7) * (8 - y^3) if 1 <= y <= 2
So the correct expression for the marginal fY(y) for 1 <= y <= 2 is fY(y) = (3/7) * (8 - y^3).
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You're welcome! I'm glad I could help. If you have any more questions, feel free to ask!
To find the marginal PDF, we need to integrate the joint PDF with respect to the other variable.
1. For scenario 1 (0 <= y <= 1):
- To find fY(y), we need to integrate the joint PDF with respect to x.
- Since 0 <= y <= x, the limits for x integration will depend on the value of y.
- When y is between 0 and 1, the lower limit for x integration will be y.
- The upper limit for x integration will be 2, as given in the joint PDF.
- Hence, you need to integrate the joint PDF fXY(x, y) = a*x^2 with respect to x, from y to 2.
2. For scenario 2 (1 <= y <= 2):
- Similar to scenario 1, to find fY(y), we need to integrate the joint PDF with respect to x.
- Since 0 <= y <= x, again, the limits for x integration will depend on the value of y.
- When y is between 1 and 2, the lower limit for x integration will be y.
- The upper limit for x integration will still be 2.
- Hence, you need to integrate the joint PDF fXY(x, y) = a*x^2 with respect to x, from y to 2.
Remember to normalize the resulting marginal PDFs by dividing them by their respective normalization constants to ensure they integrate to 1.
To find the marginal PDF of Y, you need to integrate the joint PDF over the range of X for each scenario.
For scenario 1 (0 <= y <= 1):
1. Set the bounds of integration for X based on the given range of y (0 <= y <= x).
2. Integrate the joint PDF over the bounds: ∫(0 to x) a*x^2 dx.
3. Apply the bounds for y (0 <= y <= 1) to evaluate the integral.
4. Normalize the PDF by dividing by the total probability.
For scenario 2 (1 <= y <= 2):
1. Set the bounds of integration for X based on the given range of y (0 <= y <= x).
2. Integrate the joint PDF over the bounds: ∫(0 to x) a*x^2 dx.
3. Apply the bounds for y (1 <= y <= x) to evaluate the integral.
4. Normalize the PDF by dividing by the total probability.
By following these steps, you should be able to find the marginal PDF of Y for both scenarios.