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given f(x) = 2e^-2x for x > 0 and 0 elsewhere. determine Mx(t) for exponential distribution (theta = 1/2)
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M(t) is the expected value of e^tx. \[2\int\limits_{0}^{\infty}e ^{tx}e ^{-2x}dx=2\int\limits_{0}^{\infty}e ^{(t-2)x}dx\] evaluate that integral looking at when t is less than 2. you should find that M(t) is : \[M(t)=1/(1-(\theta)t)\]
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