Question

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Suppose that the joint probability density function for the random variables X and Y is f(x, y)= k exp(- x -y) ; x, y >0 .
Verify that f(x,y) is a valid probability mass function.

Answer 1

To verify that f(x, y) is a valid probability density function (PDF), we need to show that it satisfies the two properties of a PDF: non-negativity and total probability. By checking these properties, we can confirm that f(x, y) = exp(-x - y) is a valid PDF.

Step-by-step explanation:

To verify that f(x, y) is a valid probability density function (PDF), we need to show that it satisfies the two properties of a PDF:

1. Non-negativity: The PDF must be non-negative for all values of x and y. In this case, f(x, y) = k * exp(-x - y), where x and y are both greater than 0. Since the exponential function is always positive and k is a positive constant, f(x, y) must be non-negative.
2. Total probability: The integral of f(x, y) over the entire range of x and y must equal 1. In this case, we need to integrate f(x, y) over the range x > 0 and y > 0 to obtain the total probability. The integral of exp(-x - y) is -exp(-x - y), so the integral of f(x, y) is -k * exp(-x - y). We need to find the value of k that makes this integral equal to 1. Since the integral should be equal to 1, -k * exp(-x - y) should equal 1. Therefore, k = -1. So, the PDF is f(x, y) = exp(-x - y).

Therefore, f(x, y) = exp(-x - y) is a valid probability density function.