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Suppose that $X_1$ and $X_2$ randon variables are independent with normal distribution $(0,1)$ . Let $Y_1=X_1+3X_2-2$ and $Y_2=X_1-2X_2+1$. Determine the distribution of $Y=(Y_1,Y_2)$

I know that

  • $Var(Y_1)=Var(X_1+3X_2-2)=Var(X_1)+9Var(X_2)+6Cov(X_1,X_2)=1+9+0=10$
  • $Var(Y_2)=Var(X_1-2X_2+1)=Var(X_1)+4Var(X_2)-4Cov(X_1,X_2)=1+4+0=5$
  • And that $\mu=(-2,1)$

Then $\Sigma$=\begin{bmatrix} 10 & 0 \\0 & 5\end{bmatrix}

Since $Cov(Y_1,Y_1)=0$, but the answer in the book is $\Sigma$=\begin{bmatrix} 10 & -5 \\-5 & 5\end{bmatrix}

Am i doing something wrong with the covariance between the variables?

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1 Answer 1

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$\newcommand{\Cov}{\mathrm{Cov}}$ $\newcommand{\Var}{\mathrm{Var}}$ The off-diagonal elements represent $\Cov(Y_1, Y_2)$, which it doesn't look like you calculated. $$\begin{align}\Cov(Y_1, Y_2) &= \Cov(X_1 + 3X_2 + 2, X_1 - 2X_2 + 1) \\ &= \Cov(X_1+3X_2, X_1 - 2X_2) \\ &= \Cov(X_1,X_1)+\Cov(3X_2,X_1)+\Cov(X_1,-2X_2)+\Cov(3X_2,-2X_2) \\ &= \Var(X_1)+0+0-6\Var(X_2) \\ &= -5 \end{align}$$ So, $\Cov(Y_1, Y_2) = -5$

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  • $\begingroup$ You're right, i was doing the covariance of $Y_1$ and $Y_2$ wrong. Thank you! $\endgroup$
    – Emma Wool
    Commented May 24, 2017 at 14:56

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