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I have 2 standard normal, bivariate correlated random variables, $corr \ (X_1, X_2)=\rho$.

I want to generate two independent standard normal random variables from these 2.

I tried to use what I learned from my previous thread: How does the formula for generating correlated random variables work?

I got the following:

$0=cov(\alpha X_1+\beta X_2, X_1)=\alpha\cdot cov(X_1X_2)+\beta\cdot\rho=\alpha+\beta\rho$

and

$\alpha^2+\beta^2=1$

But then I can't proceed. Can someone help me with that?

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    $\begingroup$ So what's the problem? You can derive a formula for $\beta$ (as a function of $\alpha$) from the first equation, put it into the second one, solve it and get the value for $\alpha$ by substituting for $\beta$. $\endgroup$ – Artem Sobolev Mar 13 '15 at 0:00
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    $\begingroup$ Presumably you intend that $X_1$ and $X_2$ are bivariate normal. It's possible for $X_1$ and $X_2$ to be normal and for them to be correlated, but for their joint distribution to be non-normal. $\endgroup$ – Glen_b Mar 13 '15 at 7:20
  • $\begingroup$ I agree with @Glen_b . Lanza, please indicate whether you mean that $X_1$ and $X_2$ are bivariate normal. If the answer is no, then I should edit my answer accordingly. Thank you! $\endgroup$ – JellicleCat Mar 13 '15 at 22:41
  • $\begingroup$ @JellicleCat Sorry, they are bivariate, I edited my post. $\endgroup$ – Lanza Mar 14 '15 at 12:26
  • $\begingroup$ math.stackexchange.com/questions/446093/… $\endgroup$ – jameselmore Mar 14 '15 at 13:44
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You say you need solve $$0=\alpha+\beta\rho\ (1)$$ $$\alpha^2 + \beta^2=1\ (2)$$

However, I think (2) is improper. You want to get two independent standard normal random variables, which are $X_1$ and $Y=\alpha X_1 + \beta X_2$. $X_1$ is made to follow standard normal. How about $Y$? $Y$ follows normal distribution, the mean of $Y$ is apparently 0, and the variance of $Y$ should be 1, so you need $1=var(Y)=\alpha^2 var(X_1)+\beta^2 var(X_2)+2\alpha\beta cov(X_1,X_2)=\alpha^2 + \beta^2 + 2\alpha\beta\rho$, which means $$1=\alpha^2 + \beta^2 + 2\alpha\beta\rho \ (3)$$ instead of (2).

Therefore, you need solve $$0=\alpha+\beta\rho\ (4)$$ $$\alpha^2 + \beta^2 + 2\alpha\beta\rho = 1\ (5)$$

From (4) you get $$\alpha=-\beta\rho\ (6)$$ after squaring you have $$\alpha^2=\beta^2\rho^2\ (7)$$ Put (6) and (7) into (5) $$\beta^2\rho^2+\beta^2-2\beta^2\rho^2=1\ (8)$$ $$\beta^2-\beta^2\rho^2=1\ (9)$$ Therefore $$\beta=\pm\sqrt\frac{1}{1-\rho^2}\ (10)$$ Using (6) and (10), you get $$\alpha=\sqrt\frac{\rho^2}{1-\rho^2},\ \beta=-\sqrt\frac{1}{1-\rho^2}$$ or $$\alpha=-\sqrt\frac{\rho^2}{1-\rho^2},\ \beta=\sqrt\frac{1}{1-\rho^2}$$

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