Question

If the joint density function of X and Y is given by f(x, y) = 2/7 (x + 2y), 0 < x < 1, 1 < y < 2, 0, elsewhere, Find the expected value of g(X, Y ) = X/Y^4 + X^2Y

Answer 1

In order to find the expected value of g(X, Y) = X/Y^4 + X^2Y when the joint density function of X and Y is given, use the definition of expected value and perform the necessary integration.

Step-by-step explanation:

In order to determine the expected value of the function g(X, Y) = X/Y^4 + X^2Y, given the joint density function f(x, y) = (2/7)(x + 2y) for 0 < x < 1 and 1 < y < 2, we apply the definition of expected value.

The expression for the expected value is given by: E[g(X, Y)] = ∫∫ g(x, y) * f(x, y) dx dy.

The integral is solved by calculating the limits for x (0 to 1) and y (1 to 2).

Substituting these limits and integrating yields the expected value of the function g(X, Y).

Through this process, the result provides a quantitative measure representing the average outcome of the given function based on the joint density function of X and Y.

Answer 2

The expected value of the function
`\(g(X, Y) = (X)/(Y^4) + X^2Y\),` given the joint density function
`\(f(x, y) = (2)/(7)(x + 2y)\)` over the specified domain, is
`\(E[g(X, Y)] = (22)/(105)\).`

Step-by-step explanation:

To find the expected value of
`\(g(X, Y)\),` denoted as
`\(E[g(X, Y)]\),` we need to evaluate the double integral of
`\(g(x, y)\)` multiplied by the joint density function
`\(f(x, y)\)` over the given domain. The joint density function
`\(f(x, y)\)` is provided as
`\((2)/(7)(x + 2y)\) for \(0 < x < 1\) and \(1 < y < 2\),` with (0) elsewhere. The expected value is calculated as
`\[E[g(X, Y)] = \int_(1)^(2) \int_(0)^(1) g(x, y) \cdot f(x, y) \,dx\,dy.\]`

The function
`\(g(X, Y) = (X)/(Y^4) + X^2Y\)` is then substituted into the integral, and the calculations involve solving the integrals over the specified range. After performing the integration, the final result for
`\(E[g(X, Y)]\) is \((22)/(105)\).`

In conclusion, the expected value represents the average value of the function
`\(g(X, Y)\)` over the given joint density function. The calculation involves integrating the product of
`\(g(X, Y)\)` and the joint density function
`\(f(x, y)\)` with respect to (x) and (y) over their specified ranges. The final result,
`\((22)/(105)\),` represents the expected value of
`\(g(X, Y)\)` for the given probability distribution.