I have following characteristic function for certain random variable X:
$$\Phi (t) = \frac{\beta_1\beta_2}{\eta_1}\frac{\eta_1 - it}{(\beta_1 - it)(\beta_2 - it)}$$ where $\eta_1 > 0, \quad\beta_1 > 0, \quad\beta_2 > 0$.

For $a < b$, I want to use inverse formula to find the distribution of X:

$$\mu(a,b) + \frac{1}{2}\mu(\{a,b\}) = \lim_{T\to \infty}\frac{1}{2\pi}\int_{-T}^{T}\frac{e^{-ita}-e^{-itb}}{it}\Phi(t)\, dt$$


$$\mu(\{a\}) = \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^{T}e^{-ita}\Phi(t)\, dt$$

My question is how to compute:

$$\int_{-T}^{T}\frac{e^{-ita}-e^{-itb}}{it}\Phi(t)dt$$ and $$\int_{-T}^{T}e^{-ita}\Phi(t)\, dt$$

Is there a closed form solution? Thank you very much for your answers.

  • $\begingroup$ Got something from the answer below? $\endgroup$ – Did Aug 13 '14 at 11:57
  • $\begingroup$ Apparently not. $\endgroup$ – Did Mar 15 '18 at 14:53

A more direct approach is to note that $$\Phi(t)=\alpha\frac{\beta_1}{\beta_1-it}+(1-\alpha)\frac{\beta_2}{\beta_2-it}$$ for some suitable $\alpha$ hence the density $f_X$ of $X$ is $$f_X(x)=\left(\alpha\beta_1\mathrm e^{-\beta_1x}+(1-\alpha)\beta_2\mathrm e^{-\beta_2x}\right)\cdot\mathbf 1_{x\gt0}.$$


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