An intuitive understanding of the equation $Var(X)=E(X^2)-E(X)^2$ I know the equation $Var(X)=E(X^2)-E(X)^2$ and its proof. After reading the textbook of mine, I found that this equation has been used in a lot of place. I want to know whether there is an intuitive understanding of this equation?
If you add $E[X]^2$ to both sides it looks like the Pythagorean theorem. In fact, $|\mathrm{E}[x]|$ and $\sqrt{\mathrm{Var}[X]}$ are the lengths of two legs of a right triangle.

The right triangle exists in an abstract space called a Hilbert Space. The Hilbert space in question is the set of all variables with finite second moment ($\mathrm{E}[X^2]$) on a probability space. To be specific, let the space be $\mathcal{H}= \{X: \int_\Omega X^2 \ d\mathbb{P} &lt; \infty\}$ and let us extend the notion of length on $\mathbb{R}^2$ to $\mathcal{H}$ by defining the norm of $X\in\mathcal{H}$ by $\|X\|=\left(\int_\Omega X^2 \ d\mathbb{P}\right)^{1/2}$. We can also extend our notion of orthogonality in $\mathbb{R}^2$ by saying that $X$ is orthogonal to $Y$ in $\mathcal{H}$ if $\int_\Omega X  Y \ d\mathbb{P}=0$. 

Although this space is very abstract, there are theorems that allow us to extend our geometric intuition about the Euclidean plane to $\mathcal{H}$. 
The important theorem here is Parseval's Identity which generalizes the Pythagorean Theorem. If $X$ and $Y$ are orthogonal in $\mathcal{H}$ and $Z=X+Y$, then $\|X\|^2+\|Y\|^2 = \|Z\|^2$. 

In the context of the question, if we define $Y=X-E[X]$ (more about this choice of $Y$ at the end) and $Z=E[X]$, that is, $Z$ is the function on $\Omega$ that maps every element to $E[X]$, then the norm of $Y$ is
\begin{align}
 \|Y\| = \left(\int_\Omega (X-E[X])(X-E[X]) \ d\mathbb{P}\right)^{1/2} = \sqrt{\mathrm{Var}[X]},
\end{align} (which by assumption is finite and verifies that $Y\in\mathcal{H}$). If it exists, then the norm of $Z$ is 
\begin{align}
  \|Z\|=\left(\int_\Omega E[X] \ d\mathbb{P}\right)^{1/2},
\end{align}
and since $E[X]$ is a constant, 
\begin{align}
  \|Z\|&amp;=\left(E[X]^2 \ \int_\Omega 1_\Omega \ d\mathbb{P} \right)^{1/2} \\
   &amp;= |E[X]|\cdot 1 \\
   &amp;= |E[X]|.
\end{align}
Thus, $\|Z\|=|E[X]|$ and $Z\in\mathcal{H}$. 

In this case $Y+Z=X$, so if we can verify that $Y$ and $Z$ are orthogonal then we can apply Parseval's Theorem. The calculation is simple:
\begin{align}
   \int_\Omega Y Z \ d\mathbb{P} &amp;= 
        \int_\Omega (X-E[X])E[X] \ d\mathbb{P} \\
          &amp;= E[X]\int_\Omega X \ d\mathbb{P} - E[X]^2\int_\Omega \ d\mathbb{P} \\
         &amp;= E[X]^2-E[X]^2 \\
         &amp;= 0.
\end{align}

Plugging these in yields 
\begin{align*}
   \|X\|^2 &amp;= \|Y\|^2 + \|Z\|^2 \\
   \mathrm{E}[X^2] &amp;= \mathrm{Var}[X] + \mathrm{E}[X]^2.
\end{align*}

To add a little more intuition to the choice of $Y=X-E[X]$, $Y$ can be thought of as a projection of $X$ onto the subspace of $\mathcal{H}$ consisting of all variables with 0 mean. It is analogous to projecting a point in $\mathbb{R}^2$ onto a line. In Hilbert Space this line is more like a very high dimensional plane cutting through $\mathcal{H}$.  

So in summary, there is an intuitive understanding and it comes from thinking of the set of variables as a geometric space.
This has a physical interpretation if $X$ is the speed of a moving particle parametrized by the interval $[0,1]$.  Here the expected value of $X$ is the length of the trajectory while the expected value of $X^2$ is the energy.

IF the trajectory of a uniformly moving particle is reparametrized, its length will remain the same but the energy will increase.  The difference is measured in terms of the variance of the changing speed.