I have a question about the moment generating function of 2X. If the MGF of X is $M_X(t)$, then why is the MGF of 2X $M_X(2t)$? Why is it not $[M_X(t)]^2$?

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    $\begingroup$ What reason do you have to believe that the moment generating function of $2X$ is the square of $M_X$? By definition $M_X(t) = \mathbb{E}[e^{tX}]$, hence $$M_{2X}(t) = \mathbb{E}[e^{t(2x)}] = \mathbb{E}[e^{(2t)X}] = M_X(2t).$$ $\endgroup$ – Cavents Apr 10 '16 at 19:46

By definition, $$M_X(t) = \operatorname{E}[e^{tX}].$$ So if $Y = 2X$, then $$M_Y(t) = \operatorname{E}[e^{tY}] = \operatorname{E}[e^{2tX}] = M_X(2t).$$ On the other hand, if $X_1$ and $X_2$ are IID and distributed as $X$, then their sum $W = X_1 + X_2$ has MGF $$M_W(t) = \operatorname{E}[e^{tW}] = \operatorname{E}[e^{t(X_1 + X_2)}] = \operatorname{E}[e^{tX_1} e^{tX_2}] \overset{\text{ind}}{=} \operatorname{E}[e^{tX_1}] \operatorname{E}[e^{tX_2}] = (M_X(t))^2.$$ So the MGF of $2X$ is $M_X(2t)$, but $(M_X(t))^2$ is the MGF of the sum of two IID variables, each distributed as $X$. These expressions are different because $2X$ is twice the value of $X$, but in the sum $X_1 + X_2$, $X_1$ need not equal $X_2$.

  • $\begingroup$ This is so helpful. Thank you so much! $\endgroup$ – user2817869 Apr 10 '16 at 20:26

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