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The random vector (X, Y) is said to be uniformly distributed over a region R in the plane if, for some constant c, its joint density is f(x, y) = {c if (x, y) R 0 else Show that 1/c = area of region R. Suppose that (X, Y) is uniformly distributed over the square centered at (0, 0) and with sides of length 2. Show that X and Y are independent, each being distributed uniformly over (-1, 1). What is the probability that (X, Y) lies in the circle of radius 1 centered at the origin? That is, find P

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Final answer:

To show that 1/c is the area of region R, integrate f(x, y) over R and solve for c. X and Y are independent and uniformly distributed over (-1, 1). To find the probability (X, Y) lies in a circle of radius 1 centered at the origin, find the area of the circle and divide by the area of the square.

Step-by-step explanation:

A random vector (X, Y) is said to be uniformly distributed over a region R in the plane if its joint density is given by f(x, y) = c if (x, y) is in R, and 0 otherwise. In order to show that 1/c is equal to the area of region R, we need to find the value of c that makes the integral of f(x, y) over region R equal to 1. This can be done by integrating f(x, y) over the region R and setting it equal to 1, then solving for c. Once we find the value of c, we can use it to find the probability that (X, Y) lies in a certain region, such as a circle of radius 1 centered at the origin.

In the case where (X, Y) is uniformly distributed over the square centered at (0, 0) with sides of length 2, we can show that X and Y are independent by showing that their joint density function can be written as a product of their marginal density functions. Since the joint density function for (X, Y) is constant over the entire square, it means that X and Y are each uniformly distributed over the interval (-1, 1). To find the probability that (X, Y) lies in the circle of radius 1 centered at the origin, we can find the area of the circle and divide it by the area of the square.

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