$Y_i=f(x_i)+e_i$ where $e_i \sim N(0,1)$ : what is the variance of $Y$? I have $n$ $Y$s: $Y=(Y_1,\ldots,Y_n)$. What is the variance of $Y$ when $e_i$ are the only random variables and they are independent. I can for sure say that the mean is $0$, but what about the variance? is that 1 or does $f(x)$ have an impact
 A: If $x_i$ is fixed, i.e. not random, then $\operatorname{var} (f(x_i) + e_i) = \operatorname{var}(e_i).$
(But if, for example, you randomly choose an index $i,$ then $x_i$ becomes random, even if each of $x_1,x_2,x_3,\ldots$ is fixed and not random.)
PS: Perhaps I overlooked that you want the variance of the vector as a whole. By the definition in Feller's celebrated book on probability, if $Y$ is an $n\times 1$ column vector then $$\operatorname{var}(Y) = \operatorname{E}((Y-\operatorname{E}Y)(Y-\operatorname{E}Y)^T)$$
is an $n\times n$ matrix. The entries are the covariances. You have told us nothing about the joint distribution of $e_1,\ldots,e_n,$ so we don't know the covariances. If they are all $0$ (which would be the case if we could say that $e_1,\ldots,e_n$ are independent) then we have
$$
\operatorname{var}(Y) = \begin{bmatrix} \operatorname{var}(e_1) & 0 & 0 & \cdots & 0 \\ 0 & \operatorname{var}(e_2) & 0 & \cdots & 0 \\ 0 & 0 & \operatorname{var}(e_3) & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & \operatorname{var}(e_n) \end{bmatrix}.
$$
