91Ó°ÊÓ

To find the value of the constant \(c\), we need the total probability to equal 1. This means integrating the joint probability density function \(f_{XY}(x, y) = cxy\) over the specified range and setting the result to 1. The integral is:\[\int_{0}^{3} \int_{0}^{x} cxy \; dy \; dx = 1.\]

First, integrate with respect to \(y\) (inside integral):\[\int_{0}^{x} cxy \; dy = c \int_{0}^{x} xy \; dy = c \left[ \frac{1}{2}xy^2 \right]_{0}^{x} = c \cdot \frac{1}{2}x^3.\]

Now integrate with respect to \(x\):\[\int_{0}^{3} c \cdot \frac{1}{2}x^3 \; dx = c \cdot \frac{1}{2} \int_{0}^{3} x^3 \; dx = c \cdot \frac{1}{2} \cdot \left[ \frac{1}{4}x^4 \right]_{0}^{3} = c \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 81 = \frac{81c}{8}.\]

Set the result of the integral to 1 and solve for \(c\):\[\frac{81c}{8} = 1 \implies 81c = 8 \implies c = \frac{8}{81}.\]

The covariance, \(Cov(X, Y)\), is given by:\[Cov(X, Y) = E(XY) - E(X)E(Y).\]We will calculate \(E(XY)\), \(E(X)\), and \(E(Y)\) separately.

1. **Find \(E(XY)\):**\[E(XY) = \int_{0}^{3} \int_{0}^{x} xy \cdot \frac{8}{81} xy \; dy \; dx = \frac{8}{81} \int_{0}^{3} \int_{0}^{x} x^2y^2 \; dy \; dx.\]Calculate the inner integral:\[\int_{0}^{x} x^2y^2 \; dy = x^2 \cdot \left[ \frac{y^3}{3} \right]_{0}^{x} = \frac{x^5}{3}.\]Then,\[E(XY) = \frac{8}{81} \int_{0}^{3} \frac{x^5}{3} \; dx = \frac{8}{81} \cdot \frac{1}{3} \cdot \left[ \frac{x^6}{6} \right]_{0}^{3} = \frac{8}{81} \cdot \frac{1}{18} \cdot 729 = 4.\]2. **Find \(E(X)\) and \(E(Y)\):**- \(E(X) = \int_{0}^{3} x \left( \int_{0}^{x} \frac{8}{81} xy \; dy \right) dx.\) Computing this gives \(E(X) = 18/5.\)- \(E(Y) = \int_{0}^{3} \int_{0}^{x} y \cdot \frac{8}{81} xy \; dy \; dx.\) Computing this gives \(E(Y) = 6/5.\)

Substitute \(E(XY)\), \(E(X)\), and \(E(Y)\) into the covariance formula:\[Cov(X, Y) = 4 - \left( \frac{18}{5} \right) \left( \frac{6}{5} \right) = 4 - \frac{108}{25} = 4 - 4.32 = -0.32.\]

The correlation \( \rho(X, Y) \) is given by:\[\rho(X, Y) = \frac{Cov(X, Y)}{\sqrt{Var(X)} \sqrt{Var(Y)}}.\]First, find \(Var(X)\) and \(Var(Y)\).- Calculate \(Var(X) = E(X^2) - [E(X)]^2\), where \(E(X^2)\) is found similarly.- Calculate \(Var(Y)\) in a similar way.Finally, compute \(\rho(X, Y)\).

Compute the variances:- \(Var(X) = 117/5 - (18/5)^2 = 1539/25 - 324/25 = 1215/25 = 48.6\).- \(Var(Y) = 54/5 - (6/5)^2 = 270/25 - 36/25 = 234/25 = 9.36\).Calculate the correlation:\[\rho(X, Y) = \frac{-0.32}{\sqrt{48.6} \cdot \sqrt{9.36}} = \frac{-0.32}{\sqrt{48.6 \times 9.36}} \approx \frac{-0.32}{21.34} \approx -0.0149.\]

The constant \(c\) is \(\frac{8}{81}\), the covariance is \(-0.32\), and the correlation is approximately \(-0.0149\).