# Simplifying correlation coefficient

1. Oct 11, 2014

### Dustinsfl

1. The problem statement, all variables and given/known data
The linear prediction of one random variable based on the outcome of another becomes more difficult if noise is present. We model noise as the addition of an uncorrelated random variable. Specifically, assume that we wish to predict $X$ based on observing $X + N$, where $N$ represents the noise. If $X$ and $N$ are both zero mean random variables that are uncorrelated with each other, determine the correlation coefficient between $W = X$ and $Z = X + N$. How does it depend on the power in $X$, which is defined as $E_X[X^2]$, and the power in $N$, also defined as $E_N[N^2]$?

2. Relevant equations
The correlation coefficient is define as
$$\rho_{W, Z} = \frac{cov(W, Z)}{\sqrt{var(W)var(Z)}}.$$
Let's note some standard identities which may be useful.
\begin{align*}
cov(X + Y, X) &= cov(X, X) + cov(Y, X)\\
cov(X, X) &= var(X)\\
var(X + Y) &= var(X) + var(Y) + 2cov(X, Y)
\end{align*}
Since $X$ and $N$ are uncorrelated, $cov(X, N) = 0$.

3. The attempt at a solution
I have reduced the coefficient down to
\begin{align*}
\rho_{W, Z} &= \frac{cov(W, Z)}{\sqrt{var(W)var(Z)}}\\
&= \frac{\sqrt{var(X)}}{\sqrt{var(X) + var(N)}}\\
&= \sqrt{\frac{E[X^2] - E^2[X]}
{E[X^2] - E^2[X] + E[N^2] - E^2[N]}}\\
&= \frac{1}{\sqrt{1 + \frac{E[N^2] - E^2[N]}
{E[X^2] - E^2[X]}}}
\end{align*}
but the answer can be reduced further to
$$\sqrt{\frac{\eta}{\eta + 1}}$$
where $\eta = \frac{E[X^2]}{E[N^2]}$.

I dont see how I can get to this expression.

2. Oct 12, 2014

### haruspex

You are given E[X] and E[N].

3. Oct 12, 2014

### Staff: Mentor

From this part (emphasis added) - "If X and N are both zero mean random variables that are uncorrelated with each other.."