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Suppose that $Y_1$ and $Y_2$ follow a bivariate normal distribution with parameters $\mu(Y_1)= \mu(Y_2)= 0, {\sigma^2}(Y_1)= 1, {\sigma^2}(Y_2)= 2$, and $\rho = 1/\sqrt 2$. Find a linear transformation $X_1 = a_{11} Y_1 + a_{12} Y_2, X_2 = a_{21} Y_1 + a_{22} Y_2$ such that $X_1$ and $X_2$ are independent standard normal random variables.

My work so far:

$$ f(y_1,y_2)=\frac{1}{2\pi\sigma_{Y_1}\sigma_{y_2}\sqrt{1-\rho^2}}exp{\left [ -\frac{1}{2(1-\rho^2)}\left( \frac{(y_1-\mu_{Y_1})^2}{\sigma_{y_1}^2}-\frac{2\rho(y_1-\mu_{Y_1})(y_2-\mu_{Y_2})}{\sigma_{Y_1}\sigma_{Y_2}}+ \frac{(y_2-\mu_{Y_2})^2}{\sigma_{y_2}^2}\right)\right]} $$ IF $(Y_1,Y_2)\sim N(\mu_{Y_1},\mu_{Y_2},\sigma_{Y_1}^2,\sigma_{Y_2}^2,\rho)$ then $aY_1+bY_2 \sim N(a\mu_{Y_1}+b\mu_{Y_2},a^2\sigma_{Y_1}^2+b^2\sigma_{Y_2}^2+2ab\rho\sigma_{Y_1}\sigma_{Y_2})$

$$ J(y_1,y_2)=\begin{Vmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} \\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} \end{Vmatrix} =\begin{Vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{Vmatrix}=a_{11}a_{22}-a_{12}a_{21} $$ \begin{aligned} y_1&=\frac{a_{12}x_2-a_{22}x_1}{a_{12}a_{21}-a_{22}a_{11}}\\ y_2&=\frac{a_{11}x_2-a_{21}x_1}{a_{22}a_{11}-a_{21}a_{12}}\\ f_{X_1X_2}(x_1,x_2)&=J^{-1}(y_1,y_2)\cdot f_{Y_1Y_2}(\frac{a_{12}x_2-a_{22}x_1}{a_{12}a_{21}-a_{22}a_{11}},\frac{a_{11}x_2-a_{21}x_1}{a_{22}a_{11}-a_{21}a_{12}})\\ &=\frac{1}{a_{11}a_{22}-a_{12}a_{21}}\cdot\frac{1}{2\pi}exp{\left [-\left( \frac{(y_1)^2}{1}-\frac{2\frac{1}{\sqrt2} y_1 y_2}{1\sqrt2}+ \frac{(y_2)^2}{2}\right)\right]}\\ &=\frac{1}{a_{11}a_{22}-a_{12}a_{21}}\cdot\frac{1}{2\pi}exp{\left [-\left( (\frac{a_{12}x_2-a_{22}x_1}{a_{12}a_{21}-a_{22}a_{11}})^2- (\frac{a_{12}x_2-a_{22}x_1}{a_{12}a_{21}-a_{22}a_{11}})(\frac{a_{11}x_2-a_{21}x_1}{a_{22}a_{11}-a_{21}a_{12}})+ \frac{(\frac{a_{11}x_2-a_{21}x_1}{a_{22}a_{11}-a_{21}a_{12}})^2}{2}\right)\right]}\\ \end{aligned} $$ \begin{aligned} f_{X_1}&=\frac1{\sqrt{2\pi}\sqrt{a_{11}^2\sigma_{Y_1}^2+a_{12}^2\sigma_{Y_2}^2+2a_{11}a_{12}\rho\sigma_{Y_1}\sigma_{Y_2}}}exp\left[ -\frac{x_1-(a_{11}\mu_{Y_1}+a_{12}\mu_{Y_2})}{2(a_{11}^2\sigma_{Y_1}^2+a_{12}^2\sigma_{Y_2}^2+2a_{11}a_{12}\rho\sigma_{Y_1}\sigma_{Y_2})}\right]\\ &=\frac1{\sqrt{2\pi}\sqrt{a_{11}^2+2a_{12}^2+2a_{11}a_{12}}}exp\left[ -\frac{x_1}{2(a_{11}^2+a_{12}^2\cdot2+2a_{11}a_{12})}\right]\\ f_{X_2}&=\frac1{\sqrt{2\pi}\sqrt{a_{21}^2\sigma_{Y_1}^2+a_{22}^2\sigma_{Y_2}^2+2a_{21}a_{22}\rho\sigma_{Y_1}\sigma_{Y_2}}}exp\left[ -\frac{x_2-(a_{21}\mu_{Y_1}+a_{22}\mu_{Y_2})}{2(a_{21}^2\sigma_{Y_1}^2+a_{22}^2\sigma_{Y_2}^2+2a_{21}a_{22}\rho\sigma_{Y_1}\sigma_{Y_2})}\right]\\ &=\frac1{\sqrt{2\pi}\sqrt{a_{21}^2+2a_{22}^2+2a_{21}a_{22}}}exp\left[ -\frac{x_2}{2(a_{21}^2+a_{22}^2\cdot2+2a_{21}a_{22})}\right]\\ f_{X1}f_{X_2}&=\frac1{2\pi\sqrt{a_{11}^2+2a_{12}^2+2a_{11}a_{12}}\sqrt{a_{21}^2+2a_{22}^2+2a_{21}a_{22}}}exp\left[ -\frac{x_1}{2(a_{11}^2+a_{12}^2\cdot2+2a_{11}a_{12})}-\frac{x_2}{2(a_{21}^2+a_{22}^2\cdot2+2a_{21}a_{22})}\right] \end{aligned} $$ $$f_{X_1X_2}(x_1,x_2)=f_{X_1}(x_1)f_{X_2}(x_2) $$ ---Thank you StubbornAtom,I overlooked the key word "standard "

$$ \begin{aligned} a_{11}a_{22}-a_{12}a_{21}=1\\ a_{11}^2+2a_{12}^2+2a_{11}a_{12}=1\\a_{21}^2+2a_{22}^2+2a_{21}a_{22}=1\\ 2a_{12}a_{22}+a_{12}a_{21}+a_{11}a_{22}+a_{11}a_{21}=0 \end{aligned} $$

$$a_{11}=1 ,a_{12}=0 ,a_{21}=-1 a_{22}=1$$ In my opinion,There is no sinlger answer for my question.

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  • $\begingroup$ Since $(X_1,X_2)$ is jointly normal, they are independent iff they are uncorrelated. Find the constants from this condition and from the means and variances of $X_1,X_2$. $\endgroup$ – StubbornAtom Apr 18 at 8:33
  • $\begingroup$ The answer in the book: $x_1=y_1,x_2=-y_1+y_2 $ in other words $a_{11}=1,a_{12}=0,a_{21}=-1,a_{22}=1$ $\endgroup$ – liuzhiwei Apr 19 at 4:08
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A standard way of solving that is by using the $U\Sigma U^\dagger$ decomposition. The Jordan decomposition of your covariance matrix (with unitary matrices) is given by \begin{align*} \begin{bmatrix} 1&1\\ 1&2 \end{bmatrix} &= \begin{bmatrix} \frac{-1-\sqrt 5}{\sqrt{6}}&\frac{-1+\sqrt 5}{\sqrt{6}}\\ \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2} \end{bmatrix}\cdot\begin{bmatrix} \frac{3-\sqrt 5}{2}&0\\ 0&\frac{3+\sqrt 5}{2} \end{bmatrix}\cdot\begin{bmatrix} \frac{-1-\sqrt 5}{\sqrt{6}}&\frac{\sqrt{2}}{2}\\ \frac{-1+\sqrt 5}{\sqrt{6}}&\frac{\sqrt{2}}{2} \end{bmatrix} \end{align*}

Hence if \begin{align*} A=\begin{bmatrix} \frac{-1-\sqrt 5}{\sqrt{6}}&\frac{-1+\sqrt 5}{\sqrt{6}}\\ \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2} \end{bmatrix}^{-1} \end{align*}

Then $\begin{bmatrix} X_1\\X_2 \end{bmatrix}=A \begin{bmatrix} X_1\\X_2 \end{bmatrix}\sim \mathcal N\left( \begin{bmatrix} 0\\0 \end{bmatrix}, \begin{bmatrix} \frac{3-\sqrt 5}{2}&0\\ 0&\frac{3+\sqrt 5}{2} \end{bmatrix} \right)$

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  • $\begingroup$ $X_1,X_2$ are supposed to be standard normal. $\endgroup$ – StubbornAtom Apr 20 at 14:32
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The covariance matrix of $A \mathbf Y$ is $A \Sigma A^t = I$, therefore we need to solve the equation $A^t A = \Sigma^{-1}$. We can look for an upper-triangular $A$ (the Cholesky decomposition), which gives $$\begin{pmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{pmatrix}^t \begin{pmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{pmatrix} = \begin{pmatrix} a_{11}^2 & a_{11} a_{12} \\ a_{11} a_{12} & a_{12}^2 + a_{22}^2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix},$$ from which we find $a_{11}$, then $a_{12}$, then $a_{22}$.

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