Is there a simpler way to calculate correlation? 
Let's consider that a variable y constructed from x
$x_i ∈ \left\{1,3,5,7,8\right\}$
$f(x_i)=2x_i+1$
$y_i=f(x_i) + ε_i, ∀i∈ \left\{1;...;5\right\} $ 
where $ε_i$ is a identically and independantly distributed random variable which follows a normal law $\mathcal{N(0,2)}$

calculate the correlation coefficient of $x$ and $y$. Is it still valid as an informer of dependance?


given $\sigma_{f(x)}$ which is about $5.66$ from the last exercise (which can be found again easily with a bit of calculation)
Let's try to find $σ_y$ from its formula:
$$σ_y=\sqrt[2]{Var(y)}$$
$$<=>σ_y=\sqrt[2]{\frac{\sum\limits_{i=1}^5 = (y_i-E(Y))²}{5}}$$
And that the expectetation $E(Y)$ of Y is
$$E(Y)={\frac{\sum\limits_{i=1}^5 = (y_i)}{5}}$$
Using $E(Y)=E(f(x_i))+E(ε_i)$?
then
$$=>E(Y)=E(f(x_i))$$
then
$$<=>σ_x=\sqrt[2]{\frac{\sum\limits_{i=1}^5 (f(x_i)+\frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/(2\sigma^2)})-E(Y))²}{5}}$$
But here I'm not sure to be going in the right direction... It give unbelievable complex calculation to cope with...

Hints: 
Actually $\sigma^2_x=8$ so $\sigma_x \approx 2.83$.
It is $\sigma_{f(x)}$ which is about $5.66$.
You should then be able to calculate $\sigma^2_y$ (an integer) since it is $f(x)+\epsilon$, assuming the $x$ and $\epsilon$ are independent.  That gives you $\sigma_y$.
The covariance $\sigma_{f(x) y}$ is equal to the variance of $f(x)$, again assuming independence, and this is twice the covariance $\sigma_{xy}$, so you can easily calculate the correlation coefficient $\rho_{x y}= \dfrac{ \sigma_{x y} }{\sigma_x \sigma_y}$. This is a theoretical result informing you about the dependence of $y$ on $x$. 
