I understand that mean of Y is M'Y(0), whereas variance of Y is M''Y(0).

I can derive expressions through differentiation to get M'Y(0) and M''Y(0). The 2, however, have the expression Mx(0) in them.

How do I get Mx(0) from the fact that M'x(0)is μ and M''x(0) is σ^2?

  • 2
    $\begingroup$ Remember: the MGF is the expected value of $e^{tx}$, and if $t=0$, it becomes the expected value of $e^0$, i.e., $1$ - and the expected value of $1$ is also $1$, as distributions integrate to $1$. $\endgroup$ – jbowman Nov 9 '18 at 22:37

$M(0) = 1$ for all moment generating functions $M$. This can be seen from the definition of the mgf for a random variable X:

$$M_X(t) = E[e^{tX}]$$

Substituting $t=0$:

$$M_X(t) = E[e^{0X}] = E[e^0] = E[1] = 1$$


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