Timeline for Why can a polynomial of degree $>2$ not be a cumulant generating function?
Current License: CC BY-SA 3.0
13 events
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| Nov 8, 2012 at 18:21 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
explained relation to char. function
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| Nov 8, 2012 at 18:07 | comment | added | whuber♦ | Perhaps. I think my concerns began when an earlier formulation of the property was edited; I was reacting to the change. What I would like to suggest is that if your characterization of the ridge property is not in the same form found in your reference, then it would help future readers to quote the reference verbatim and then restate it in the form you find suitable, so that they can see the connection. | |
| Nov 8, 2012 at 18:06 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
made formulation of ridge property more precise
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| Nov 8, 2012 at 18:04 | comment | added | Arnold Neumaier | maybe I should have written ''logarithmic ridge property'' to make it clear that it has not the same form? | |
| Nov 8, 2012 at 18:02 | comment | added | Arnold Neumaier | @whuber: in my first comment: latter = char.function. I didn't notice that my formulation was ambiguous. - In my answer, the ridge property as formulated there (with $\Re$ rather than absolute values) applies to the CGF, consistent with the quadratic case. Whereas the correct ridge property of the exponential must have an absolute value. | |
| Nov 8, 2012 at 17:55 | comment | added | whuber♦ | Right, no quarrels there! But in your answer you claim the ridge property applies to the CGF. It looks like you meant to write that it applies to the CF, that's all. I hope you won't take it wrong if I suggest you reread the last sentence of your answer very carefully, because perhaps you just aren't seeing anymore what is actually written there. | |
| Nov 8, 2012 at 17:40 | comment | added | Arnold Neumaier | @whuber: en.wikipedia.org/wiki/Cumulant defines the cumulant generating function to be $f(z)=\log E(e^{zX})$. This is the usage that I am used to, and which I implied in my question and answer. Lucacz states the ridge property for the characteristic function, which is the analytically continued $e^{f(iz)}$; there the absolute value figures. | |
| Nov 8, 2012 at 17:34 | comment | added | whuber♦ | I think something is missing, Arnold: your answer asserts that the CGF must satisfy the ridge property, not the CF. Your comment, on the contrary, states that the CF must satisfy the ridge property. Which is correct? | |
| Nov 8, 2012 at 17:32 | comment | added | Arnold Neumaier | @whuber: $f(z)$ is the log of the analytically continued characteristic function, hence the absolute value in the ridge property for the latter translates into that for the real part of $f$. If $f(z)=z^2$ then $\Re f(x+it)=x^2-t^2\le x^2=f(x)$, as stated. | |
| Nov 8, 2012 at 16:15 | comment | added | whuber♦ | (+1) However, I believe the "ridge property" compares the absolute value of $f(z)$ to $f(\mathbb{Re}\ z)$ and that the comparison is enforced only for $\mathbb{Re}\ z \gt 0$. Otherwise, even quadratic polynomials would be ruled out. | |
| Nov 8, 2012 at 16:07 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
added qualification real
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| Nov 8, 2012 at 14:43 | vote | accept | Arnold Neumaier | ||
| Nov 8, 2012 at 14:41 | history | answered | Arnold Neumaier | CC BY-SA 3.0 |