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In continuation of Lower bounds of laplace transform of characteristic functions.

My question is:

Can anyone point out where i'm going wrong in the derivation below.

It's been a while since I've used transforms and i know I'm missing something as the product of the derivation does not agree with direct calculation.

Thanks in advance.


We wish to solve $K = f(x;a)$ where $K$ is some complex number and \begin{equation} f(x;a) = \int_{-\infty}^\infty h(x-\omega;a)g(\omega) \text{d}\omega \end{equation} with $g$ some probability distribution, $a,x$ are real numbers and \begin{equation} h(x;a) = \frac{1}{a+ i x},\quad a>0 \end{equation}

We let $\bar{\varphi}(t)= \hat{g}(t)$ and calculate the Fourier transform of $h$ \begin{equation} \hat{h}(t) = 2\pi e^{-at}\Theta(t), \end{equation} where $\Theta(t)$ is the Heaviside step function. Thus $\mathcal{F}[K](t) = \mathcal{F}[f](t) $ gives \begin{equation} 2\pi K\delta(t) =2\pi e^{-at}\Theta(t)\bar{\varphi}(t) \implies \frac{K}{\bar{\varphi}(t)}\delta(t)=e^{-at}\Theta(t) \end{equation} Taking the IFT we get \begin{equation} \frac{K}{\bar{\varphi}(0)} = \int_0^\infty e^{-(a-ix)t}\text{d}t \end{equation} Which gives $\frac{1}{a-ix} = \frac{K}{\bar{\varphi(0)}}$ thus $$a+ix = \frac{\varphi(0)}{\bar{K}}$$

However this does no agree with direct calculation. Take the Cauchy distribution with median $0$ $g(\omega;\sigma) = \frac{\sigma}{\pi(\sigma^2+\omega^2)}$ and its analytic continuation $G(\omega;\sigma) =\frac{1}{\pi(\sigma-i \omega)}$ $$f(x;a) =i\int_{-\infty}^\infty \frac{g(z)}{z - i(a +ix)}dz = \pi G(ia-x)$$ by the Schwarz Integral formula. Which implies $f(x;a) =\frac{1}{x+ia + \sigma}$. Thus if $f(x;a) = K$, then $$ x+ia = \frac{1}{K} - \sigma$$ and from the section above, this is not equal to $x+ia = \frac{\varphi(0)}{\bar{K}} = \frac{1}{\bar{K}}$

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Then you pass to $$ 2\pi K\delta(t) =2\pi e^{-at}\Theta(t)\bar{\varphi}(t) $$ it means that you are solving equation $$ K\equiv \int_{-\infty}^\infty h(x-\omega;a)g(\omega) \text{d}\omega $$ and it's not that is required. The Fourier transform is not applicable in this way since the lhs is not given, $K$ being just a value of rhs at some point $x$.

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  • $\begingroup$ I don't understand your answer, i do want to solve that equation. $\endgroup$ – MrSlunk May 27 '13 at 6:07
  • $\begingroup$ @MrSlunk One thing is to solve equation $F(x)=K$ for some x. Another is to solve $F(x)\equiv K$. For example, trying to solve $e^{-x^2}=1$ one cannot take the Fourier transform from both sides. Equality F[e^{-x^2}](ξ)=F[1](ξ) for all ξ would imply that $e^{-x^2}\equiv1$. $\endgroup$ – Andrew May 27 '13 at 6:26
  • $\begingroup$ Thanks, I understand. Considering i arrived in the ballpark of the answer i'm looking for, Do you have any suggestions of alternate approaches? $\endgroup$ – MrSlunk May 27 '13 at 7:05
  • $\begingroup$ @MrSlunk Since $g$ is not specified the resulting function $f$ is rather arbitrary. So the question is a general one: how to find inverse function $f^{-1}$ for a given $f$. It hasn't an explicit general answer. The fact that $f$ is an integral transform of some probability distribution doesn't seem to be of much help. $\endgroup$ – Andrew May 27 '13 at 7:32
  • $\begingroup$ yes, i was hoping that $f$ being an integral transform and $g$ a probability distribution(symmetric and zero mean can be assumed) was enough to provide some general results on the inverse. $\endgroup$ – MrSlunk May 27 '13 at 8:23

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