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I am given $Z_1$ and $Z_2$ independent standard normal variables, I have to find two random variables $X_1$ and $X_2$ with correlation $\operatorname{corr}(X_1,X_2) = p$, where $p\in (−1, 1)$.

Any help as to how I can get to the answer or which property or formula I should look into to get to this answer..


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The formula you were given is $X_2=pZ_1+\sqrt{1-p^2}Z_2$, please correct your post. – Did Mar 13 '12 at 6:09
Still incorrect. – Did Mar 13 '12 at 6:36

We use the amended suggestion $X_1=Z_1$, $X_2=p Z_1+\sqrt{1-p^2}Z_2$.
The correlation coefficient of $X_1$ and $X_2$ is $$\frac{E(X_1X_2)}{\sqrt{\text{Var}(X_1)\text{Var}(X_2)}}.\qquad\qquad(\ast)$$ We first calculate the numerator of $(\ast)$. We have $$E(X_1X_2)=E\left(Z_1\left(p Z_1+\sqrt{1-p^2}Z_2\right)\right)=p E(Z_1^2) +\sqrt{1-p^2}E(Z_1Z_2).$$ Since $Z_1$ is standard normal, it has mean $0$ and variance $1$, so $E(pZ_1^2)=p$. By independence, $E(Z_1Z_2)=E(Z_1)E(Z_2)=0$. So $E(X_1X_2)=p$.

We now deal with the variances in the denominator of $(\ast)$. The variance of $X_1$ is $1$. For the variance of $X_2$, use the fact that $X_2=p Z_1+\sqrt{1-p^2}Z_2$, a linear combination of independent normals with variance $1$. So $\text{Var}(X_2)=p^2+(\sqrt{1-p^2})^2=1$.

It follows that the correlation coefficient of $X_1$ and $X_2$ is equal to $p$.

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+1. Definitely simpler than the first version. – Did Mar 13 '12 at 7:26
Hello Andre, In this question we dont have to prove $\operatorname{corr}(X_1,X_2) = p$ is equal to $X_1=Z_1$ and $X_2=p\cdot Z_1+ \sqrt{1-p^2}\cdot Z_2$. We have to find 2 random variables X1 and X2. I have editied the question for further clarity...any hint will be appreciated... Thanks – newbietolinalg Mar 14 '12 at 5:45
The answer I gave proved that if $X_1=Z_1$, $X_2=pZ_1+\sqrt{1-p^2} Z_2$, then the correlation coefficient turns out to be $p$. The proof used the basic formula $(\ast)$ for the correlation coefficient, plus some useful facts about the normal, in particular the formula for the variance of a linear combination of independent normals. There are other ways to produce $X_1$ and $X_2$ that have correlation coefficient $p$, but this one is probably the simplest. – André Nicolas Mar 14 '12 at 5:52

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