Product of Two Uniform Random Variables

Editorials
1

[Jump to AI Interview]

Problem Statement

Two random variables X and Y represent independent uniform distributions between 0 and 1. What is the probability that their product is greater than 0.5?

Solution

This problem requires finding the probability of a product condition using double integration over the joint distribution.

Step 1: Define the Random Variables

Let:

  • X \sim \text{Uniform}(0, 1)

  • Y \sim \text{Uniform}(0, 1)

  • X and Y are independent

We want to find P(XY > 0.5).

Step 2: Set Up the Joint Distribution

Since X and Y are independent uniform random variables, their joint probability density function is:

f_{X,Y}(x, y) = f_X(x) \cdot f_Y(y) = 1 \cdot 1 = 1

for (x, y) \in [0, 1] \times [0, 1], and zero elsewhere.

Step 3: Identify the Favorable Region

We need to find the region in the unit square [0,1] \times [0,1] where xy > 0.5.

The boundary curve is xy = 0.5, which can be written as y = \frac{0.5}{x}.

This is a hyperbola. Let’s analyze the feasible region:

  • For x \in (0, 0.5]: We have \frac{0.5}{x} \geq 1. Since y \in [0, 1], the condition y > \frac{0.5}{x} would require y > 1, which is impossible. Therefore, no points in this region satisfy xy > 0.5.

  • For x \in (0.5, 1]: We have \frac{0.5}{x} < 1. The condition xy > 0.5 is equivalent to y > \frac{0.5}{x}, and since y \in [0, 1], we need y \in \left(\frac{0.5}{x}, 1\right].

Therefore, the favorable region is:

  • x \in (0.5, 1]

  • y \in \left(\frac{0.5}{x}, 1\right]

Step 4: Set Up the Double Integral

Using the definition of probability for continuous random variables:

P(XY > 0.5) = \int \int_{\{(x,y): xy > 0.5\}} f_{X,Y}(x, y) \, dy \, dx
P(XY > 0.5) = \int_{0.5}^1 \int_{0.5/x}^1 1 \, dy \, dx

Step 5: Evaluate the Inner Integral

For a fixed x \in (0.5, 1], we evaluate:

\int_{0.5/x}^1 1 \, dy = \left[y\right]_{0.5/x}^1 = 1 - \frac{0.5}{x} = 1 - \frac{1}{2x}

Step 6: Evaluate the Outer Integral

Now we integrate with respect to x:

P(XY > 0.5) = \int_{0.5}^1 \left(1 - \frac{1}{2x}\right) dx

Splitting the integral:

P(XY > 0.5) = \int_{0.5}^1 1 \, dx - \frac{1}{2} \int_{0.5}^1 \frac{1}{x} \, dx

Evaluating each term:

First term:

\int_{0.5}^1 1 \, dx = \left[x\right]_{0.5}^1 = 1 - 0.5 = 0.5

Second term:

\int_{0.5}^1 \frac{1}{x} \, dx = \left[\ln x\right]_{0.5}^1 = \ln 1 - \ln 0.5 = 0 - (-\ln 2) = \ln 2

Combining:

P(XY > 0.5) = 0.5 - \frac{1}{2} \cdot \ln 2 = \frac{1}{2} - \frac{\ln 2}{2} = \frac{1}{2}(1 - \ln 2)

Step 7: Numerical Evaluation

Using \ln 2 \approx 0.693:

P(XY > 0.5) = \frac{1}{2}(1 - 0.693) = \frac{1}{2} \cdot 0.307 = 0.1535

Alternative Verification: Geometric Interpretation

We can verify this result by computing the area under the curve y = \frac{0.5}{x} in the region x \in [0.5, 1].

The area below the curve (where xy < 0.5) is:

\int_{0.5}^1 \frac{0.5}{x} \, dx = 0.5 \left[\ln x\right]_{0.5}^1 = 0.5 \ln 2

The total area in the rectangle [0.5, 1] \times [0, 1] is 0.5 \times 1 = 0.5.

Therefore, the area above the curve (where xy > 0.5) is:

0.5 - 0.5 \ln 2 = \frac{1}{2}(1 - \ln 2)

This matches our previous result.

Final Answer

P(XY > 0.5) = \frac{1}{2}(1 - \ln 2) \approx 0.153

Generalization

For X, Y \sim \text{Uniform}(0, 1) independently, the probability that XY > c (where 0 < c < 1) is:

P(XY > c) = \int_c^1 \int_{c/x}^1 1 \, dy \, dx = \int_c^1 \left(1 - \frac{c}{x}\right) dx
= \left[x - c \ln x\right]_c^1 = (1 - 0) - (c - c \ln c) = 1 - c + c \ln c

For c = 0.5:

P(XY > 0.5) = 1 - 0.5 + 0.5 \ln 0.5 = 0.5 + 0.5(-\ln 2) = 0.5 - 0.5 \ln 2 = \frac{1}{2}(1 - \ln 2)