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If $f(x) = 0.5 e^{-|x|}$ for $-\infty < x < \infty$, how would you find the moment generating function for this? Also how would you find the distribution of $Y = |X|$?

Attempt:

$E(e^{tX}) = \int_{-\infty}^\infty f(x) e^{tx} \; dx.$

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    The pdf doesn't integrate to one. I think there is a .5 in the exponential term.2012-02-28

1 Answers 1

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As Didier Piau stated: $ \begin{align} E(e^{tX}) & = \int_{-\infty}^\infty f(x) e^{tx} \ dx=\int_{-\infty}^0 f(x) e^{tx}\ dx+\int_0^\infty f(x) e^{tx}\ dx \\ \\ & =\int_{-\infty}^0 0.5e^x e^{tx}\ dx + \int_0^\infty 0.5 e^{-x} e^{tx}\ dx \\ \\ & = 0.5\left(\int_{-\infty}^0 e^{(t+1)x}\ dx+\int_{0}^\infty e^{(t-1)x} \ dx\right) = \frac{0.5}{t+1}-\frac{0.5}{t-1},\quad t\in (-1,1) \end{align} $

For the distribution of $Y=|X|$ we have the general rule for a transformation of the form $y=h(x)$ in our case $Y=|X|$ with $Y\in [0,+\infty)$: $g(y)=f(h^{-1}(y))|\frac{d(h^{-1}(y))}{dy}|$. Applying this rule to your problem will give you the distribution of Y.

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    One can simplify $E(\mathrm e^{tX})$ to $1/(1-t^2)$.2012-02-28