1.
The joint and marginal distributions of \(X\) and \(Y\) are given below. Find \(\Cov(X, Y)\) and \(\rho_{X, Y}\text{.}\)
| \(Y = 1\) | \(Y = 2\) | \(Y = 3\) | |
|---|---|---|---|
| \(X = 0\) | \(0.1\) | \(0.15\) | \(0.1\) |
| \(X = 1\) | \(0.15\) | \(0.2\) | \(0.3\) |
| \(x\) | \(0\) | \(1\) |
|---|---|---|
| \(\Pr(X = x)\) | \(0.35\) | \(0.65\) |
| \(y\) | \(1\) | \(2\) | \(2\) |
|---|---|---|---|
| \(\Pr(Y = y)\) | \(0.25\) | \(0.35\) | \(0.4\) |
Solution.
\(\Cov(X, Y) = \E(XY) - \E(X)\E(Y)\text{.}\) We have:
\begin{align*}
\E(X) \amp = 0(0.35) + 1(0.65) = 0.65 \\
\E(Y) \amp = 1(0.25) + 2(0.35) + 3(0.4) = 2.15 \\
\E(XY) \amp = (0)(1)(0.1) + (0)(2)(0.15) + (0)(3)(0.1) + \\
\amp \quad\, (1)(1)(0.15) + (1)(2)(0.2) + (1)(3)(0.3) = 1.45 \\ \\
\Cov(X, Y) \amp = 1.45 - (0.65)(2.15) = \boxed{0.0525}
\end{align*}
Since \(X\) is an indicator random variable, we have \(\Var(X) = (0.65)(0.35) = 0.2275\text{.}\) For \(\Var(Y)\text{:}\)
\begin{align*}
\E(Y^2) \amp = 1^2(0.25) + 2^2(0.35) + 3^2(0.4) = 5.25 \\
\Var(Y) \amp = \E(Y^2) - (\E(Y))^2 = 5.25 - (2.15)^2 = 0.6275
\end{align*}
So the correlation is:
\begin{align*}
\rho_{X, Y} \amp = \frac{\Cov(X, Y)}{\sqrt{\Var(X)\Var(Y)}} = \frac{0.0525}{\sqrt{(0.2275)(0.6275)}} \approx \boxed{0.139}.
\end{align*}
