1

I would like to understand how I can show that the method to create two normally distributed random numbers given as an answer to this question is correct.

Given independent $X_1$ and $X_2$ normally distributed with mean zero and variance 1, one can get a randomvariable with correlation $\mathbb{E}[X_1 X_3] =\rho$ by setting $$ X_3=\rho X_1+\sqrt{1-\rho^2}X_2. $$

Community
  • 1
  • $X_1$ and $X_2$ should be independent, then you would have $\Bbb E[X_1X_2]=0$. – Tom-Tom Feb 21 '14 at 15:19
  • Yes, you are right. Question solved. Should I delete it? –  Feb 21 '14 at 15:21
  • In principle this is not a chat but a community, meaning that questions and answers are there to be seen by anyone and possibly commented or improved. However, as it was only a point you skipped in the formulation of your problem, this could be considered as not very relevant for other people. Up to you. And welcome in here, by the way., – Tom-Tom Feb 21 '14 at 15:24
  • @V.Rossetto thanks! –  Feb 21 '14 at 18:19

1 Answers1

1

It can be done like this: $$\mathbb{E}[X_1 X_3] = \mathbb{E}[X_1 (\rho X_1+\sqrt{1-\rho^2}X_2]$$ $$ = \mathbb{E}[\rho X_1^2 +\sqrt{1-\rho^2}X_1X_2] $$ $$ = \rho\mathbb{E}[ X_1^2] +\sqrt{1-\rho^2}\mathbb{E}[X_1X_2] $$ Here I can use the relation $\mathbb{E}(X^2)=\text{Var}(X)+(\mathbb{E}[X])^2$ to get $$=\rho + \sqrt{1-\rho^2}\mathbb{E}[X_1X_2].$$ Since $X_1$ and $X_2$ are independent we have $\mathbb{E}[X_1X_2]=0$, which leads to the desired result $$\mathbb{E}[X_1 X_3]=\rho$$

Community
  • 1