# What does variance actually mean in control theory?

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?

• 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. – Dilip Sarwate Jan 26 '16 at 15:00

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.
• 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. – Dilip Sarwate Jan 26 '16 at 15:22