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I want to prove the fact that $\forall$ random variable $X$ the following condition holds: $D\left[X\right] = E\left[X^2\right] - E\left[X\right]^2$.

Where, $D\left[X\right]$ denotes dispersion of random variable $X$ and $E\left[X\right]$ is an Expected value of $X$.

I've tried some calculations, but unfortunately i got stuck at some point:

From the definition $D\left[X\right] = E\left[(X- E\left[X\right])^2\right] = E\left[X^2- 2\cdot X \cdot E\left[X\right] + E\left[X\right]^2\right] = E\left[X^2\right] - 2 \left[X \cdot E\left[X\right]\right] + E\left[E\left[X\right]^2\right]$

False Promise
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1 Answers1

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From the definition $D\left[X\right] $ $= E\left[(X- E\left[X\right])^2\right] \\= E\left[X^2- 2\cdot X \cdot E\left[X\right] + E\left[X\right]^2\right] \\= E\left[X^2\right] - 2 \left[X \cdot E\left[X\right]\right] + E\left[E\left[X\right]^2\right]$

You dropped an $\mathsf E$ (typo?).   To continue on, recall that $\mathsf E[X]$ is a constant, and that for any constant $a$, then $\mathsf E(aX)=a~\mathsf E(X)$ and $\mathsf E(a^2)=a^2$. $$\begin{align}\mathsf D\left[X\right] &= \mathsf E\left[X^2\right] - 2~\mathsf E\Bigl[X \cdot \mathsf E\bigl[X\bigr]\Bigr] + \mathsf E\Bigl[\mathsf E\bigl[X\bigr]^2\Bigr] \\[0.75ex] & = \mathsf E\left[X^2\right] - 2~\mathsf E\left[X \right]\cdot \mathsf E\left[X\right] + \mathsf E\left[X\right]^2 \\[1ex] & = \mathsf E[X^2]-\mathsf E[X]^2\end{align}$$

And... well, that's it.

Graham Kemp
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