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The question is about the notion of variance in control theory, a college course I am taking.

Ever since I can remember, in every maths or physics class, whenever a professor started teaching something they just listed definitions from a textbook. For example: speed is equal to the ratio between distance and time. Thankfully, common sense tells me that speed just measures how fast I can go from x to y.

Keeping this in mind, variance, as my professor so dearly explained, is $$\frac1 {N+1} \sum_{t=-\frac N 2} ^ {\frac N 2} (u[t] - {\frac1 {N+1} \sum_{t=-\frac N 2} ^ {\frac N 2} (u[t])^2) }$$

Can someone be kind enough to tell me with English words what meaning this actually has? What is it that variance actually measures and how can I visualize it so I can actually understand it?

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  • $\begingroup$ I am fairly certain that the formula that you have so lovingly written out has some misplaced parentheses. Are you sure that your professor did not write $$\frac{1}{N+1}\sum_{t=-\frac N2}^{\frac N2} \left(u[t]- \left[\frac{1}{N+1} \sum_{t=-\frac N 2}^{\frac N2} u[t]\right]\,\, \right)^2??$$ What I have shown above is subtracting off the average of $N+1$ values of $u[t]$ (the quantity in square brackets) from each $u[t]$, squaring the difference, and averaging the sum of the squares. Physically, we are computing the moment of inertia about the center of mass of a set of point masses. $\endgroup$ – Dilip Sarwate Jan 26 '16 at 15:00
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Variance is a measure of how likely it is that a particular sample is very different from the average of all the samples.

Consider a sequence $\{5,5,5,5,5,5\}$. Its variance is zero because every sample is equal to the sequence average. Now consider the sequence $\{0,10,0,10,0,10\}$. Its mean is the same as the previous sequence (5), but now its variance is large, because samples are "far" from the mean.

The textbook formula can be interpreted as follows: it is the average of the square of the difference between every sample and the sequence average. The "inner" summation is the sequence average. The "outer" summation is the average of the difference between each sample and the sequence average.

The reason there is a square operation in the formula is that you don't want differences of equal magnitude but opposite sign to cancel each other out.

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  • $\begingroup$ Thank you! I appreciate your time. I still have one more question though. Unfortunately, there are many examples of such definitions and concepts that are increasingly more complex (or so they seem to me). I was wondering if there is any textbook / lecture / website that you can recommend that can explain such concepts pertaining to control theory the way you did. I don't know if this is appropriate to ask on stack exchange and so I apologize beforehand if it isn't. $\endgroup$ – Initiate Jan 25 '16 at 17:38
  • $\begingroup$ I don't know about control theory. I would ask your professor to provide more intuition, either in class or in his/her office hours. I've found the courses on ocw.mit.edu to be very good in general too. $\endgroup$ – MBaz Jan 25 '16 at 17:55
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    $\begingroup$ I think your explanation doesn't fit the formula that is shown in the question because of misplaced parentheses. What the OP's formula is doing is subtracting off the mean-square value of the $(x[t])^2$ from each $x[t]$ and averaging the result. The OP's formula simplifies to $$\left(\frac{1}{N+1}\sum x[t]\right) -\left( \frac{1}{N+1}\sum (x[t])^2\right),$$ that is, the average value of the $x[t]$ minus the average value of the $(x[t])^2$. Your explanation fits the formula that is shown in my comment on the question. $\endgroup$ – Dilip Sarwate Jan 26 '16 at 15:22
  • $\begingroup$ @DilipSarwate You're right, thanks for pointing it out. I assumed the formula posted was just the usual variance equation without checking it in detail. I'll assume it was a transcription error unless the OP replies to your comment indicating otherwise. $\endgroup$ – MBaz Jan 26 '16 at 20:31

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