Double Integrating Gaussian Noise
Thanks to Dilip, I managed to figure out the answer to the question.
$y_1[n]$ will  be a normal random variable, with its variance given by the following equation:

\begin{equation}
\sigma_{y_1}^2=N\sigma_{th}^2
\end{equation}

This can be explained by using the fact that the variances will just add up. The mean will stay zero.$\\$
A similar approach can be used to find the variance of $y_2[n]$ ($\sigma_{y_2}$), which begins by writing the equation for $y_2[N-1]$ ($y_2[n]$ after $N$ cycles):

\begin{equation}
y_2[N-1]=(N-2)x[0] + (N-3)x[1] + (N-4)x[2]...+x[N-3]
\end{equation}

Taking variances on both sides:

\begin{equation}
\sigma_{y_2}^2=\sigma_{th}^2(1+2^2+3^2+...(N-2)^2)=\sigma_{th}^2\frac{(N-2)(N-1‌​)(2N-3)}{6}\approx \sigma_{th}^2\frac{N^3}{3} (N&gt;&gt;1)
\end{equation}

Again, the mean will stay zero.