Proof

This can be proven by the property : \(E[X^{k}]= [\frac{d^{k}}{dt^{k}}M_{X}(t)]_{t=0}\).

Actually, if we differentiate \(M_{X}(t)=E[\mathrm{e}^{Xt}]\) respect to \(t\), we have

$$M’_{X}(t)=E[X\mathrm{e}^{Xt}]$$

and if we put \(t=0\), we obtain

$$M’_{X}(0)=E[X]$$

The second derivative of \(M_{X}(t)\) is given by

$$M”_{X}(t)=E[X^{2}\mathrm{e}^{Xt}]$$

and and if we put \(t=0\), we obtain

$$M”_{X}(0)=E[X^{2}]$$

Thus, we get

$$V[X]=E[X^{2}]-{E[X]}^{2}=M”_{X}(0)-M’_{X}(0)^{2}$$