Why does $ \operatorname{Var}(X) = E[X^2] - (E[X])^2 $

Some times ago, a professor showed me this right triangle:

The formula you reported can be seen as the application of the Phytagora's theorem:

$$P = \mathbb{E}[X^2] = \text{Var}[X] + \mathbb{E}^2[X].$$

Here, $P = \mathbb{E}^2[X]$ (which is the second uncentered moment of $X$) is read as "the power" of $X$. Indeed, there is a physical explanation.

In physics, energy and power are related to the "square" of some quantity (i.e. $X$ can be velocity for kinetic energy, current for Joule law, etc.).

Suppose that these quantities are random (indeed, $X$ is a random variable). Then, the power $P$ is the sum of two contribution:

1. The square of the expected value of $X$;
2. Its variance (i.e. how much it varies from the expected value).

It is clear that, if $X$ is not random, then $\text{Var}[X] = 0$ and $\mathbb{E}^2[X] = X^2$, so that:

$$P = X^2,$$

which is a typical physical definition of energy/power. When randomness is present, the we must use the whole formula

$$P = \mathbb{E}[X^2] = \text{Var}[X] + \mathbb{E}^2[X]$$

to evaluate the power of the signal.

As a final remark, the power of $X$ can be seen as the length of the vector which components corresponds to the square of its expected value plus its variability.

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P.S. A further clarification... the values $P$, $\text{Var}[X]$ and $\mathbb{E}^2[X]$ represent the squares of the sides of the triangle, not their length...

Easy! Expand by the definition. Variance is the mean squared deviation, i.e., $V(X) = E((X-\mu)^2).$ Now:

$$ (X-\mu)^2 = X^2 - 2X \mu + \mu^2$$

and use the fact that $E(\cdot)$ is a linear function and that $\mu$ (the mean) is a constant.

The shortcut computes the same thing, but counts the difference in the mean of squares and the square of the mean.

The other formula tells you exactly the same thing as the one that you have given with $x,x^2$ $\&$ $n$. You say you understand this formula so I assume that you also get that variance is just the average of all the deviations squared.

Now, $\mathbb{E}(X)$ is just the average of of all $x’_is$, which is to say that it is the mean of all $x’_is$.

Let us now define a deviation using the expectation operator. $$Deviation = D = (X-\mathbb{E}(X))$$ And Deviation squared is, $$D^2 = (X-\mathbb{E}(X))^2$$

Now that we have deviation let’s find the variance. Using the above mentioned definition of variance, you should be able to see that

$$Variance = \mathbb{E}(D^2)$$ Since $\mathbb{E}(X)$ is the average value of $X$,The above equation is just the average of deviations squared.

Putting the value of $D^2$, we get, $$Var(X) = \mathbb{E}(X-\mathbb{E}(X))^2 = \mathbb{E}(X^2+\mathbb{E}(X)^2-2X*\mathbb{E}(X)) = \mathbb{E}(X^2)+\mathbb{E}(X)^2-2\mathbb{E}(X)^2 = \mathbb{E}(X^2)-\mathbb{E}(X)^2$$ Hope this helps.