2.5 Correlation Matrix

The population correlation matrix is defined in a manner similar to its sample counter part.
Let us denote the correlation coefficient between the \(i^{\text {th}}\) and \(j^{\text {th}}\) as \(\rho _{ij}\) \[\rho _{ij} = \frac {\sigma _{ij}}{\sigma _i\sigma _j}\]

The matrix \(P = (\rho _{ij})\), with \(\rho _{ij} = 1 \) is called the population correlation matrix.

Taking \(D=Diag (\sigma _i)\) we have that \[P = D^{-1}\Sigma D^{-1}\] The matrix \(P\geq 0\) because \(\Sigma \geq 0\) and \(D\) is symmetric.

1.

\[E(\underline {X}) = \begin {pmatrix} E(x_1)\\ E(x_2)\\ \end {pmatrix} , \underline {X}= \begin {pmatrix} x_1\\ x_2\\ \end {pmatrix} \]

\begin {align*} E(x_1) & = \int _{x_1,x_2} x_1f(x_1,x_2)dx_1dx_2\\\\ & = \int ^1_0\int ^1_0 x_1(x_1+x_2)dx_2dx_1\\\\ & = \int ^1_0\big (x^2_1x_2 +\frac {x_1x_2^2}{2}\Big |^1_0\big )dx_1\\\\ & = \int ^1_0\big (x^2_1 + \frac {x_1}{2}\big )dx_1=\big (\frac {x^3_1}{3}+ \frac {x^2_1}{4}\Big |^1_0=\frac {1}{3}+\frac {1}{4}\\\\ & = \frac {7}{12}\\ \end {align*}

\[E(x_2) = \int ^1_0\int ^1_0 x_2(x_1+x_2)dx_2= E(x_1) = \frac {7}{12}\]

So the mean vector is \(\displaystyle {E(\underline {X})= \begin {pmatrix} 7/12\\ 7/12\\ \end {pmatrix} }\)

2.

\[\Sigma = \begin {pmatrix} \sigma _{11} & \sigma _{12}\\ \sigma _{21} & \sigma _{22}\\ \end {pmatrix} \]

\[\sigma _{11} = E(x_1)^2 - (E(x_1))^2\]

\begin {align*} E(x^2_1) & = \int ^1_0\int ^1_0 x^2_1(x_1+x_2)dx_1dx_2\\\\ & = \int ^1_0\int ^1_0 \big (x^3_1+x_1^2x_2\big )dx_2dx_1\\\\ & = \int ^1_0 \big (x^3_1 + \frac {x^2_1}{2}\big ) dx_1\\ & = \frac {1}{4}+ \frac {1}{6}\\\\ & = \frac {5}{6}\\ \end {align*}

\begin {align*} var(x_1) = \sigma _{11} & = E(x_1^2) - (E(x_1))^2\\ & = \frac {5}{12}-\Big (\frac {7}{12}\Big )^2\\ & = \frac {5}{12}-\frac {49}{144}=\frac {60-49}{144}\\\\ & = \frac {11}{144}\\ \end {align*}

\begin {align*} E(x_1,x_2) & = E(g(\underline {X}))\\ & = \int ^1_0 \int ^1_0 x_1x_2 f(x_1,x_2)dx_1dx_2\\\\ & = \int ^1_0\int ^1_0\big (x^2_1x_2 + x^2_2x_1\big )dx_1dx_2\\\\ & = \frac {1}{3}\\ \end {align*}

\begin {align*} Cov(x_1,x_2) & = E(x_1 x_2) - E(x_1)E(x_2)\\ & = \frac {1}{3}-\Big (\frac {7}{12}\Big )^2\\ & = \frac {48-49}{144}\\\\ & = -\frac {1}{144}\\ \end {align*}

\begin {align*} \text {Hence}\quad var(\underline {X}) & = E(\underline {X}-\underline {\mu })(\underline {X}-\underline {\mu })'= E[G(\underline {X})]\\\\ & = \begin {pmatrix} var(x_1) & Cov(x_1,x_2)\\ Cov(x_1,x_2) & var(x_2)\\ \end {pmatrix}\\\\ & = \begin {pmatrix} \frac {11}{144} & -\frac {1}{144}\\\\ -\frac {1}{144} & \frac {11}{144}\\ \end {pmatrix}\\ \end {align*}

3.

(a)

\begin {align*} f_1(x_1) & = \int ^1_0f(\underline {X})dx_2\\ & = \int ^1_0 (x_1+x_2) dx_2\\\\ & = \begin {cases} x_1 + \frac {1}{2} , & 0<x_1<1\\\\ 0, & \text {otherwise}\\ \end {cases}\\ \end {align*}

(b)

\[f_2(x_2)= \begin {cases} x_2 + \frac {1}{2}, & 0< x_2 < 1\\\\ 0, & \text {Otherwise}\\ \end {cases} \]

(a)

\(\displaystyle {f(x_2/x_1) = \frac {f(x_1,x_2)}{f_1(x_1)}= \begin {cases} \frac {2(x_1+x_2)}{2x_1+1}, & 0<x_1<1\\\\ 0, & \text {Otherwise}\\ \end {cases} }\)

(b)

\begin {align*} E(x_2/x_1=0.5) & = \int ^1_0 x_2 \frac {2(0.5+x_2)}{2(0.5)+1}dx_2\\\\ & = \int ^1_0 x_2 \frac {1+ 2x_2}{2}dx_2\\\\ & = \frac {1}{2}\Bigg (\frac {1}{2}+\frac {2}{3}\Bigg )\\\\ & = \frac {7}{12}\\\\ \end {align*}

(c)

(Exercise) Find \(var(x_2/x_1=0.5)\)

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