Probablity Question - Joint PDF Expectation/Variance

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To find the expected values E(X) and E(Y) from a joint probability density function (PDF) for X and Y with the constraint Y > X ≥ 0, the correct approach involves integrating over the appropriate limits. Specifically, E(X) is calculated using the integral of x multiplied by the marginal PDF of X, f_X(x), which is derived from integrating f(x,y) with respect to y from x to infinity. Similarly, E(Y) is found by integrating y multiplied by the marginal PDF of Y, f_Y(y), which is obtained by integrating f(x,y) with respect to x from 0 to y. The limits for these integrals should reflect the constraints of the region defined by the joint PDF, ensuring that x ranges from 0 to infinity. Properly distinguishing between the marginal distributions is crucial for accurate calculations.
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Homework Statement


I have been given a joint PDF for X and Y, with ranges Y>X≥0.

I need to find the E(x) and E(y).

Homework Equations


I know E(x) = ∫(x)*(f(x,y) dx and E(y) = ∫(y)*(f(x,y)) dy

The Attempt at a Solution


For ∫x*f(x,y) dx, i used the limits = x to ∞

For ∫y*f(x,y) dy, i used the limits = 0 to y

Is this correct?
 
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MCooltA said:

Homework Statement


I have been given a joint PDF for X and Y, with ranges Y>X≥0.

I need to find the E(x) and E(y).

Homework Equations


I know E(x) = ∫(x)*(f(x,y) dx and E(y) = ∫(y)*(f(x,y)) dy


The Attempt at a Solution


For ∫x*f(x,y) dx, i used the limits = x to ∞

For ∫y*f(x,y) dy, i used the limits = 0 to y

Is this correct?

These are both wrong. EX = int x*f(x,y) dx dy, etc.

RGV
 
Im confused by what you meant above, but i meant to say;

I have found f(x) by ∫f(x,y) dy, with the limits x to ∞.

To then find the E(x) do i ∫x * f(x) dx, with the limits 0 to y?
 
MCooltA said:
Im confused by what you meant above, but i meant to say;

I have found f(x) by ∫f(x,y) dy, with the limits x to ∞.

To then find the E(x) do i ∫x * f(x) dx, with the limits 0 to y?

OK, this is equivalent to what I wrote. You really should use different symbols for the different distributions, such as g(x) = int f(x,y) dy and h(y) = int f(x,y) dx, or use f_X(x) instead of g(x) and f_Y(y) instead of h(y).

As to your second question: the region in (x,y) space is {0 <= x <= y}, so yes, for any given x, y goes from x to infinity. However, once y has been "integrated out" it is no longer present, so NO, in EX = int x*f_X(x) dx, x does NOT go from 0 to y---there is no y now!. The variable x goes from 0 to infinity: when x was 5, y went from 5 to infinity, when x was 10 million, y went from 10 million to infinity, etc.

RGV
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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