Discussion

The moment generating function of a random variable X is given by \(Mx(t) = \exp[2e^{t} - 2]\). Then \(E[X] =\)

Answer: C

\(E[X] = \) 2.

\(\exp[2e^{t} - 2] = \exp[2(e^{t} - 1)]\) is the moment generating function of a Poisson distribution with \(\lambda = 2\), whose mean is 2.

Directly, \(E[X] = M'(0)\):

\[ \begin{aligned} M'(t) &= 2e^{t}\exp[2e^{t} - 2] \\ M'(0) &= 2 \cdot 1 \cdot e^{0} = 2 \end{aligned} \]

0 and 1 come from evaluating the exponent rather than differentiating, and the mean is finite because the MGF exists near 0.

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