We appreciate your visit to 1 point Let T be the region inside the triangle with vertices 0 0 2 0 and 2 3 and let f z y be. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
The value of E(XY) is 16384/3. To find E(XY), we need to calculate the expected value of the product of the random variables X and Y.
Here, X and Y represent the coordinates of a point (x, y) within the region T. First, let's find the joint probability density function (PDF) of X and Y within T. Since the function f(x, y) is defined only within T, we can determine the PDF by normalizing f(x, y) over the region T.
To find the normalization constant, we integrate f(x, y) over T:
∫∫[T] f(x, y) dA = ∫∫[T] (66 + 156x + y) dA
Here, dA represents the differential area element.
To integrate over the triangle T, we can split it into two regions: T1 and T2.
T1 is the triangle bounded by the points (0, 0), (2, 0), and (2, 2), and T2 is the triangle bounded by the points (0, 0), (2, 2), and (2, 3).
Calculating the integral over T1:
∫∫[T1] (66 + 156x + y) dA = ∫[0 to 2] ∫[0 to x] (66 + 156x + y) dy dx
After integrating with respect to y, we get:
∫[0 to 2] [66y + 78xy + (1/2)y²] | [0 to x] dx
Simplifying this, we have:
∫[0 to 2] (66x + 39x² + (1/2)x³) dx
Evaluating this integral, we get:
[33x²+ 13x³ + (1/8)x⁴] | [0 to 2]
= 33(2)² + 13(2)³ + (1/8)(2)⁴ - (1/8)(0)⁴- 33(0)²- 13(0)³
= 132 + 104 + 8 - 0 - 0 - 0
= 244.
Similarly, calculating the integral over T2:
∫∫[T2] (66 + 156x + y) dA = ∫[0 to 1] ∫[x to 2] (66 + 156x + y) dy dx
After integrating with respect to y, we get:
∫[0 to 1] [(66 + 156x)y + (1/2)y²] | [x to 2] dx
Simplifying this, we have:
∫[0 to 1] (66 + 156x - 66x + (1/2) - (1/2)x² - (1/2)x²) dx
= ∫[0 to 1] (66 - 66x + 156x - x^2) dx
= ∫[0 to 1] (156x - x²) dx
= [78x² - (1/3)x³] | [0 to 1]
= 78(1)² - (1/3)(1)³ - 78(0)² - (1/3)(0)³
= 78 - (1/3) - 0 - 0
= 77/3.
Now, to find E(XY), we multiply the calculated integrals by X and Y respectively and integrate over the region T:
E(XY) = ∫∫[T] XY f(x, y) dA
= ∫∫[T] (XY)(66 + 156x + y) dA
= ∫∫[T1] (XY)(66 + 156x + y) dA + ∫∫[T2] (XY)(66 + 156x + y) dA
= (244)(77/3) + (77/3)(244)
= 2(244)(77/3)
= 16384/3.
Therefore, E(XY) = 16384/3.
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