# Correlation between two (correlated) normally distributed variables

In an electrical circuit I wish to find the noise correlation between two signal paths (which in the end are added). Let me also say, that I am quite useless at probability theory (but trying to become better!). I setup my problem as follows:

Given $$X = X_1 + c_1\cdot Z$$ and $$Y = X_2 + c_2 \cdot Z$$, where $$X_1 \sim \mathcal{N}(0,\sigma^2 _1)$$, $$X_2 \sim \mathcal{N}(0,\sigma^2 _2)$$, and $$Z \sim \mathcal{N}(0,\sigma^2 _Z)$$. $$X$$, $$Y$$, and $$Z$$ are not correlated (independent). Further, $$c_1\in \mathbb{C}$$ and $$c_2\in \mathbb{C}$$. What is the correlation coefficient between $$X$$ and $$Y$$?

As far as I have understood,

$$corr(X,Y)=\frac{cov(X,Y)}{(\sigma^2_X+\sigma^2_Y)^{0.5}}$$,

where $$\sigma^2_X=\sigma^2_1+c_1^2\sigma^2_Z$$ and $$\sigma^2_Y=\sigma^2_2+c_2^2\sigma^2_Z$$ - is this true with complex constants? The covariance is defined by

$$cov(X,Y)=E[XY]-E[X]E[Y]$$,

where $$E[X]=0$$ and $$E[Y]=0$$. How do I find $$E[XY]$$?

Thanks for the help in advance.

## 1 Answer

With the use of the following:

$$Cov(X,Z+Y)=Cov(X,Z)+Cov(X,Y)$$

$$Cov(aX+b,Y)=aCov(X,Y)$$

$$Cov(Y,Y)=Var(Y)$$

$$Cov(X,Y)=0$$ if $$X,Y$$ are independent

we have

$$Cov(X,Y)=Cov(X_1 + c_1\cdot Z, X_2 + c_2 \cdot Z)=Cov(X_1,X_2)+Cov(X_1,c_2Z)+Cov(X_2,c_1Z)+Cov(c_1Z,c_2Z)=c_1\cdot c_2 Cov(Z,Z)=c_1\cdot c_2 Var(Z)=c_1\cdot c_2 \cdot \sigma_z^2$$