Timeline for Deconvolution of sum of two random variables
Current License: CC BY-SA 3.0
14 events
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| Feb 18, 2014 at 19:13 | comment | added | Philip | Sorry, i wasn't precise enough. What i wanted to say is that when $f$ is an arbitrary pdf then $1/c f(x/c)$ is not necessarily a pdf, too. In addition i am a bit puzzled by the linear system of equations: Since Fourier transform is complex, the log is ambiguous. So it cannot be transformed back to receive $g$? | |
| Feb 18, 2014 at 18:16 | comment | added | Robert | I was saying the if $g$ is a linear function, g cannot be a density (it cannot integrate to 1). | |
| Feb 18, 2014 at 17:38 | comment | added | Philip | I don't get it. I thought the 1/c was necessary to make g(x/c) a pdf? Because if $\int f(x) = 1$, then $\int g(x/c)$ will not be 1. | |
| Feb 18, 2014 at 16:55 | comment | added | Robert | No -- in fact, if g() is linear, it will not define a density at all. The above holds for a larger class of functions. You can easily reserve it (just write down the definition of Fourier). It holds for Lebesgue integrable functions. | |
| Feb 18, 2014 at 16:50 | comment | added | Philip | ${\cal F}g'(\omega)=1/c{\cal F}g(\omega/c)$ does only hold for linear functions, right? So this approach only works if g() is linear. | |
| Feb 16, 2014 at 20:24 | comment | added | Philip | c is known to me. g() and f() are continuous in principle. However, since i only have observations of Zi's i can only calculate a discrete pdf. | |
| Feb 16, 2014 at 16:42 | history | edited | Robert | CC BY-SA 3.0 |
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| Feb 15, 2014 at 1:41 | history | edited | Robert | CC BY-SA 3.0 |
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| Feb 15, 2014 at 0:40 | history | edited | Robert | CC BY-SA 3.0 |
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| Feb 15, 2014 at 0:36 | comment | added | Robert | I see I can add comments here -- I edited the above already in regards to your comment. | |
| Feb 15, 2014 at 0:34 | history | edited | Robert | CC BY-SA 3.0 |
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| Feb 14, 2014 at 19:31 | comment | added | Jochen Wengenroth | Forier transformation is of course a good idea. However, how do you solve the equation $c\hat{f}(x) =\hat{g}(x)\hat{g} (x/c)$ in an efficient way to calulate $g$ by means of the Fourier inversion formula? | |
| Feb 14, 2014 at 17:20 | history | edited | Robert | CC BY-SA 3.0 |
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| Feb 14, 2014 at 17:11 | history | answered | Robert | CC BY-SA 3.0 |