Question

Let X and Y be two independent and identically distributed random variables having U(0, 1) distribution. Then P(X² < Y < X) equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 0.15

To solve the problem, we need to find the probability \( P(X^2 < Y < X) \) where \( X \) and \( Y \) are independent and identically distributed random variables with a uniform distribution over \([0, 1]\). We'll follow these steps:

The joint probability density function of \( X \) and \( Y \) is \( f(x, y) = 1 \) for \( 0 \leq x, y \leq 1 \) due to the independence and uniformity of \( X \) and \( Y \).

We need to evaluate the probability \( P(X^2 < Y < X) \), which can be expressed as a double integral over the region defined by these inequalities, bounded by the square \([0,1] \times [0,1]\).

Let's consider the region in the xy-plane: \( X^2 < Y < X \) implies that for a fixed \( x \), \( y \) ranges from \( x^2 \) to \( x \). Thus, the region is defined between the curves \( y = x^2 \) and \( y = x \).

Determine the limits for \( x \): \( x \) ranges from 0 to 1 because \( X \) is uniformly distributed over this interval. \( y \) ranges from \( x^2 \) to \( x \).

Set up the double integral: \[ P(X^2 < Y < X) = \int_0^1 \int_{x^2}^{x} 1\, dy\, dx. \]

Evaluate the inner integral with respect to \( y \): \[\int_{x^2}^{x} 1\, dy = [y]_{x^2}^{x} = x - x^2.\]

Now, evaluate the outer integral: \[ \int_0^1 (x - x^2)\, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \left(\frac{1}{2} - \frac{1}{3}\right) = \frac{1}{6}.\]

The probability \( P(X^2 < Y < X) = \frac{1}{6} \approx 0.1667 \), which rounds off to 0.17 to two decimal places.

Conclusion: The computed value of \( 0.17 \)