Question

Suppose \(X\) and \(Y\) are independent and identically distributed random variables that are distributed uniformly in the interval \([0,1]\). The probability that \(X \geq Y\) is \(\_\_\_\_\).

Hint:

For uniform distributions over a square, probabilities involving inequalities correspond to geometric areas. Use symmetry or simple integration for efficient calculation.

Answer 1

Step 1: Define the probability of interest.
The probability \(P(X \geq Y)\) is computed as: \[ P(X \geq Y) = \iint_{x \geq y} f_{X,Y}(x, y) \, dx \, dy, \] where \(f_{X,Y}(x, y)\) represents the joint probability density function. For a uniform distribution over the unit square \([0, 1] \times [0, 1]\), \(f_{X,Y}(x, y) = 1\). Step 2: Set up the integral over the valid region.
The region \(x \geq y\) in the unit square forms a triangular area with vertices at \((0, 0)\), \((1, 0)\), and \((1, 1)\). The integral becomes: \[ P(X \geq Y) = \int_0^1 \int_0^x 1 \, dy \, dx. \] Step 3: Evaluate the integral.
Integrate over \(y\): \[ \int_0^x 1 \, dy = x. \] Now, integrate over \(x\): \[ P(X \geq Y) = \int_0^1 x \, dx = \frac{x^2}{2} \bigg|_0^1 = \frac{1}{2}. \] Final Answer: \[\boxed{{0.50}}\]