Let $X \sim \operatorname{Exp}(\lambda)$
How do we find $E\cos(tX)$ and $E\sin(tX)$?
I understand that a characteristic function can be derived as the following:
$\phi(t) = E(e^{itX}) = E\cos(tX) + iE(\sin(tX))$
Using characteristic functions to find $E\cos(tX)$ and $E\sin(tX)$
1
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probability
probability-theory
characteristic-functions
moment-generating-functions
exponential-distribution
1 Answers
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I'm going to guess you mean the distribution is $e^{-\lambda x}(\lambda\,dx)$ for $x\ge0$ and not $e^{-x/\lambda}(dx/\lambda)$ for $x\ge0.$
\begin{align} & \operatorname{E}(\cos(tX) + i\sin(tX)) = \operatorname{E}(e^{itX}) = \int_0^\infty e^{itx} e^{-\lambda x}(\lambda\,dx) = \lambda \int_0^\infty e^{(it-\lambda)x} \,dx \\[10pt] = {} & \left.\lambda \frac{e^{it-\lambda}}{it-\lambda} \right|_{x\,:=\,0}^{x\,:=\,\infty} = \frac \lambda {it-\lambda} \text{ since } \lambda >0\\[10pt] = {} &\frac{-\lambda(it+\lambda)}{(\lambda-it)(\lambda+it)} = \frac{-\lambda^2 - \lambda it}{\lambda^2-t^2} = \frac{\lambda^2}{t^2-\lambda^2} + i\frac{\lambda t}{t^2-\lambda^2}. \end{align} The real part of this last expression ie $\operatorname{E}(\cos(tX))$ and the imaginary part is $\operatorname{E}(\sin(tX)).$