Joint Distributions and Correlation

By independence, \((X, Y)\) is uniform on the rectangle \([1, 3] \times [2, 4]\), whose area is 4, so a probability is the favourable area divided by 4.

Since \(Y\) is never below 2, \(X \ge Y\) can only happen when \(X \ge 2\). The favourable region is the triangle \(2 \le Y \le X \le 3\), with both legs of length 1:

\[ \text{area} = \int_{2}^{3} (x - 2)\,dx = \frac{1}{2}, \qquad P(X \ge Y) = \frac{1/2}{4} = \frac{1}{8} = 0.125 \]

Answer: 0.125

The marks went to all because the printed function is not a valid density: over its region it integrates to \(\int_{0}^{2}\int_{0}^{x} 2xy\,dy\,dx = \int_{0}^{2} x^{3}\,dx = 4\), not 1.

The conditional expectation can still be found as printed, because a constant factor cancels in a conditional density. For a fixed \(x\),

\[ f_{Y|X}(y|x) = \frac{2xy}{\int_{0}^{x} 2xy\,dy} = \frac{2xy}{x^{3}} = \frac{2y}{x^{2}}, \quad 0 < y < x \]

\[ E[Y|X = x] = \int_{0}^{x} y \cdot \frac{2y}{x^{2}}\,dy = \frac{2x}{3}, \qquad E[Y|X = 1.5] = \frac{2 \times 1.5}{3} = 1 \]

Answer: 1

Use \(\operatorname{Cov}(X, Y) = E[XY] - E[X]\,E[Y]\). The four outcomes HH, HT, TH, TT are equally likely.

• \(X = 1\) only for HH, so \(E[X] = \frac{1}{4}\).
• \(Y = 1\) for every outcome except TT, so \(E[Y] = \frac{3}{4}\).
• Whenever \(X = 1\), also \(Y = 1\), so \(XY = X\) and \(E[XY] = \frac{1}{4}\).

\[ \operatorname{Cov}(X, Y) = \frac{1}{4} - \frac{1}{4} \cdot \frac{3}{4} = \frac{1}{16} = 0.0625 \]

The covariance is positive because two heads guarantee at least one head. As 0.0625 lies exactly halfway, the key accepts both 0.062 and 0.063.

Answer: 0.063