Timeline for Asymptotic scaling of mean and variance for non-central chi distribution
Current License: CC BY-SA 4.0
13 events
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| Sep 4, 2023 at 15:12 | vote | accept | user1172131 | ||
| Aug 31, 2023 at 14:40 | comment | added | Carlo Beenakker | happy to see your results now agree with mine; it's no secret that I am a physicist, not a mathematician, so my analytical skills lack your rigor. | |
| Aug 31, 2023 at 13:50 | comment | added | Iosif Pinelis | @CarloBeenakker : This now fixed. Again, hopefully you can make your answer rigorous and detailed. | |
| Aug 31, 2023 at 13:49 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 31, 2023 at 13:05 | comment | added | Iosif Pinelis | @CarloBeenakker : You are right, the cross term is missing. I will fix this. Meanwhile, hopefully you can make your answer rigorous and detailed. | |
| Aug 31, 2023 at 12:59 | comment | added | Carlo Beenakker | So I would think that in your definition of $Y$ a sum over $2(X_i -\mu_i)\mu_i/\sigma_i^2$ is missing under the square root. | |
| Aug 31, 2023 at 12:37 | comment | added | Carlo Beenakker | Apologies for being slow to understand: if you subtract $\mu$ from $X$ to obtain a central variable, shouldn’t the square also contain the cross term $X\mu$ ? | |
| Aug 31, 2023 at 12:20 | comment | added | Iosif Pinelis | Previous comment continued: As for Mathematica, it sometimes does strangest things when dealing with special functions. To minimize that "hallucination" factor, I used the formula $Y=\sqrt{V_k+\lambda^2}$, where, as in my answer, $V_k$ is a random variable with the central chi-squared distribution with $k$ degrees of freedom and hence with an elementary pdf. | |
| Aug 31, 2023 at 12:10 | comment | added | Iosif Pinelis | @CarloBeenakker : My answer contains a complete, transparent, and simple proof. Have you found any issues with this proof? I did find a number of issues with your answer, and you are not responding to the comments of mine on those issues. In addition to the mentioned problems in your answer, I specifically don't understand how you got $M^{1/2}-\tfrac{1}{8}VM^{-3/2}$ for $EY$. | |
| Aug 31, 2023 at 5:33 | comment | added | Carlo Beenakker |
to check; the Mathematica code for the variance is var = k + k*L^2 - (Pi/2)*LaguerreL[1/2, k/2 - 1, -k*L^2/2]^2 If I plot this as a function of $k$ for $L=1$ it converges to 3/4, not to 1/4.
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| Aug 31, 2023 at 4:58 | comment | added | Carlo Beenakker | your answer for $Var Y$ for $L=1$ ($\mu_i=\sigma_i=1$) is 1/4; mine is 3/4; I may have made a mistake, but the exact result computed from the noncentral chi distribution does seem to converge to 3/4 (see lower right plot in my answer) | |
| Aug 31, 2023 at 2:30 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 30, 2023 at 22:04 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |