How can I find the correlation (rho) between $U$ and $V$, Where $U = X_1+X_2$ and $V= X_1-2X_2$ $X_1$ and $X_2$ are normally distributed with $\mu= 1$ and $\sigma= 2$.
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$\begingroup$ are $X_1$ and $X_2$ independent? $\endgroup$– MasoudCommented Mar 5, 2020 at 8:13
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2 Answers
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You can use the result that a linear combination of normally distributed random variables is also normal, in this case X is bivariate normal.
\begin{equation} \textbf{X}_{p=2} = \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} \in N_{p=2}( \bf{ \mu_x }, \bf{ \Sigma_x } ) = N\left( \begin{bmatrix} \mu_1 =1 \\ \mu_2 =1 \end{bmatrix} , \begin{bmatrix} 2 & \sigma_{12} \\ \sigma_{21} & 2 \end{bmatrix} \right) \end{equation} Then \begin{equation} \bf{A}\bf{X} = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} X_1 + X_2 \\ X_1 -2X_2 \end{bmatrix} = \begin{bmatrix} U \\ V \end{bmatrix} \in N_{p=2} \left(\bf{A}\bf{\mu_x}, \bf{A}\bf{\Sigma_x}\bf{A}' \right) \end{equation} Finally calculate $\rho = \sigma_{12}/(\sqrt{\sigma_{11}}\sqrt{\sigma_{22}})$ from the entries of $\bf{A}\bf{\Sigma}\bf{A}'$, which is the correlation between U and V. Note that $\sigma_{11},\sigma_{22}$ denotes the variances (just notation).
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Hint: If $X_1$ and $X_2$ are independent so $$cov(U,V)=cov(X_1+X_2,X_1-2X_2)$$ $$=cov(X_1,X_1)+cov(X_2,-2X_2)$$ $$=cov(X_1,X_1)-2cov(X_2,X_2)$$
