If the variance of the random variable X exists, show that $\mathbf{E}(X^2)≥[\mathbf{E}(X)]^2$. I don't know how to start, so any hint is appreciated. Thank you in advance.
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Hint: $(X - \mathbf{E}X) ^ 2 = $ ? – Tlön Uqbar Orbis Tertius Mar 03 '15 at 12:04
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$(X - \mathbf{E}X) ^ 2 = X^2 - 2X \mathbf{E}X + \mathbf{E}X^2$. This quantity is nonnegative. Taking expectations yeld $$0 \leq \mathbf{E}\big[(X - \mathbf{E}X)^2 \big] = \mathbf{E}[X^2] - 2\mathbf{E}X \mathbf{E}X + (\mathbf{E}X)^2 = \mathbf{E}[X^2] - (\mathbf{E}X)^2$$ wich is what you wanted.
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