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My question has to do with integrating gaussian noise.

Let us assume we have samples of discrete gaussian white noise with mean $\mu = 0$ and variance $\sigma_{th}^2$. These noise samples are passed through the system shown in the Figure (a cascade of two integrators with outputs $y_1[n]$ and $y_2[n]$, respectively). enter image description here

What will be the mean and variance of $y_1$ and $y_2$ (let us say after $N$ cycles)?

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  • $\begingroup$ The output of your summer is a single random variable. It is a Gaussian random variable, and you don't need the Central Limit Theorem to assert this: the sum of independent Gaussian random variables is Gaussian: no asymptotics or limits needed.The rest of your question makes no sense because the output of the first summer is not a random process or white noise and so there is nothing except one random variable to pass on to the next summer. $\endgroup$ – Dilip Sarwate Mar 4 '16 at 3:34
  • $\begingroup$ I think perhaps a simpler way to the question is this - "IF white gaussian noise is passed through two integrators, what will the output of the second integrator look like?" I strongly suspect this has something to do with the Wiener Process, but I cant seem to figure it out. $\endgroup$ – Saqib Shah Mar 4 '16 at 11:44
  • $\begingroup$ Can you answer the question for a non-random sequence first? Go on! Take your best shot at writing down in mathematical notation and no verbal gobbledygook the value of the output of the first integrator at time $t=1$? the output of the second integrator at time $t=1$? Now, up the ante and try to do the same for $t=2$ and possibly even for $t=3$. If you cannot do these calculations by yourself, then any answer that can be given, no matter how ridiculous, will suffice for your needs since you cannot tell the difference between shinola. $\endgroup$ – Dilip Sarwate Mar 4 '16 at 13:45
  • $\begingroup$ In the case of a deterministic sequence, the outputs can easily be determined. assuming that x[n] is the sequence of inputs to the first integrator.At time t=0, both the integrators will have zero outputs (I am assuming delaying integrators$\frac{z−1}{1−z^{−1}}$). At $t=1$, integrator 1($y1$) will output $y1=x[0]$, integrator 2 ($y2=0$). At $t=2$, $y1=x[0]+x[1],y2=x[0]$. At $t=3$, $y1=x[0]+x[1]+x[2], y2=2×x[0]+x[1]$, so on and so forth. $\endgroup$ – Saqib Shah Mar 4 '16 at 14:54
  • $\begingroup$ Good! Now consider editing your question because it mentions only a $y$, not a $y_1$ (or $y_2$ for that matter). How does the time index come into play? How can you tell the difference between $y1 = x[0]+x[1]$ and $y1 = x[0]+x[1]+x[2]$?? and how are these things related to the $y$ in your displayed equation? $\endgroup$ – Dilip Sarwate Mar 4 '16 at 20:08
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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>>1) \end{equation}

Again, the mean will stay zero.

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