What is the characteristic function of the Dirac delta function?
Is it $e^{i*0}=1$?
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1Your question is already answered in the article you link to. – Glen_b Nov 12 '14 at 12:52
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@glen_b You are quite correct. Please close the answer if you think is the best. – Hansel Nov 12 '14 at 12:54
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I imagine it would be better if it were answered. If one doesn't appear soon I may put one up. – Glen_b Nov 12 '14 at 12:56
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@Glen_b I will accept it if you post it soon. It doesn't have to be too elaborate. – Hansel Nov 12 '14 at 12:59
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So the Dirac delta function has pmf:
$f(x;k_0) = \delta(x-k_0)\,,\quad -\infty<x<\infty,\: -\infty<k_0<\infty$
and cdf
$F(x;k_0)=\left\{\begin{matrix} 1, & \mbox{if }x\ge k_0 \\ 0, & \mbox{if }x<k_0 \end{matrix}\right.$
It is a discrete distribution with all probability at a single point.
Consider the definition of characteristic functions:
$\varphi_X(t) = \operatorname{E}\left[e^{itX}\right] = \int_{\mathbf{R}} e^{itx}\,dF_X(x)$
where the integral may be taken to be a Reimann-Stieltjes integral.
Now $F$ has a jump at $k_0$, so the integral is simply $e^{ik_0t}$.
[Alternatively, if you need a simpler argument, we could just use the definition of expectation for a discrete distribution and arrive at the same result.]
See also here.
You can also investigate the Fourier transform of the Dirac delta function from which the characteristic function immediately follows.
Glen_b
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Thanks. Just an additional comment, this result can also be proved by using the theory of approximate identities and generalized functions. – Hansel Nov 12 '14 at 13:37
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1An interesting result with respect to Fourier transforms is that the Fourier transform of the Dirac delta has value $1$ regardless of whether the Fourier transform is defined as $$G(\omega) = \int_{\mathbb R} e^{-i\omega x}f(x)\,\mathrm dx$$ or as $$H(t) = \int_{\mathbb R} e^{-i2\pi t x}f(x)\,\mathrm dx$$ – Dilip Sarwate Nov 12 '14 at 15:01