If $(Y_1, Y_2)$ are bivariate Normal with correlation $\rho$ and $Y_i$'s marginally $N(0,1)$, why is it true that we can express $Y_2 = \rho Y_1+(1-\rho^2)Z$ where $Z$ is a standard normal independent of $Y_1$?
If $(Y_1, Y_2)$ are bivariate Normal, why is it true that $Y_2 = \rho Y_1+(1-\rho^2)Z$ where $Z$ is a normal independent of $Y_1$?
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probability
probability-theory
1 Answers
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1) The normal distribution is stable: linear combinations of independent normal distributions are again normal. Proof: either explicit calculation of CDF, or just use characteristic functions.
2) A normal distribution is uniquely determined by it's mean and variance.
So, first verify that $Y_2$ is normally distributed like $N(0,1)$. Next verify that $\mbox{corr}(Y_1,Y_2)=\rho$. Then you're done.
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0Is there a standard way to derive this? The way I see it right now it seems to have popped out of thin air. – 2017-01-10
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0@user321627: Not really. It's more to do with the stability of the normal distribution. Another example is the Poisson distribution which can exhibit similar decompositions. – 2017-01-11