Probability P(X ≥ Y) Calculation

We are given two random variables, X and Y, which are independent and identically distributed (i.i.d.). Both follow a uniform distribution on the interval $[0, 1]$, denoted as X, Y ~ U[0, 1]. We need to find the probability P(X ≥ Y).

Geometric Interpretation

The joint distribution of X and Y is uniform over the unit square defined by $0 \le x \le 1$ and $0 \le y \le 1$. The total area of this sample space is $1 \times 1 = 1$.

The condition X ≥ Y corresponds to the region within the unit square where the x-coordinate is greater than or equal to the y-coordinate. This region is a triangle with vertices at (0, 0), (1, 0), and (1, 1).

Calculating the Probability

The area of this favorable region (the triangle) is calculated as:

Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.

Since the joint distribution is uniform, the probability of the event X ≥ Y is the ratio of the favorable area to the total area of the sample space:

P(X ≥ Y) = $\frac{\text{Area where } X \ge Y}{\text{Total Area}} = \frac{1/2}{1} = \frac{1}{2}$.

Alternatively, due to symmetry (since X and Y are i.i.d. from the same distribution), the probability P(X > Y) must be equal to P(Y > X). Since P(X=Y) = 0 for continuous distributions, we have P(X ≥ Y) = P(X > Y). Also, P(X > Y) + P(Y > X) + P(X=Y) = 1. This simplifies to 2 * P(X > Y) = 1, so P(X > Y) = 1/2. Therefore, P(X ≥ Y) = 1/2.