Timeline for Distribution of ratio between complex Gaussian and Chi-square R.V.s
Current License: CC BY-SA 3.0
43 events
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| Feb 5, 2019 at 12:52 | comment | added | Felipe Augusto de Figueiredo | thanks a lot for the notebook. | |
| Feb 5, 2019 at 0:29 | comment | added | Iosif Pinelis | Here is a link to my calculation of $f_U(u)$: my.pcloud.com/publink/… | |
| Feb 4, 2019 at 20:32 | comment | added | Felipe Augusto de Figueiredo | Simple question, how did you find $f_{\text{U}}(\text{u})$ with Mathematica? I'm trying to reproduce the result and the equation get is totally different: my.pcloud.com/publink/… | |
| Jan 22, 2019 at 8:32 | comment | added | Felipe Augusto de Figueiredo | Dear, thanks for your comment. In fact I have done that already a couple of days ago. Please, have a look at this link: mathoverflow.net/questions/321344/… | |
| Jan 22, 2019 at 1:08 | comment | added | Iosif Pinelis | I have already answered many of your questions, in addition to your original question. If you have more questions yet, you should ask them in separate posts. | |
| Jan 20, 2019 at 11:25 | comment | added | Felipe Augusto de Figueiredo | Do you think it is possible to find the moments of $U$ as it was done for $R$? I've tried Mathematica's "MomentGeneratingFunction" but it returns the p.d.f. of $U$. I have never used Mathematica before, so I might be doing something wrong. | |
| Jan 20, 2019 at 9:46 | comment | added | Felipe Augusto de Figueiredo | thanks, in fact I figured out the problem with the notebook and now I able to plot for n = 60 with a larger expansion, however, it takes too long to get an expansion for values of n greater than 60. I think I will stick to n = 60, as it already proves the point. | |
| Jan 20, 2019 at 0:42 | comment | added | Iosif Pinelis | I think for large $n$, like $n=100$, the computational implementation of the expansion needs extra care, since some of the terms there (such as $16n^5$) then become very large. Concerning your Mathematica notebook, there seems to be some coding error there, looking at the message "Set: Tag Times [...] is Protected". | |
| Jan 18, 2019 at 22:58 | comment | added | Felipe Augusto de Figueiredo | Dear, sorry to bother you. I'm trying to plot the distribution for n=100, however, I get the following plot pasteboard.co/HX4TUP8.png. This was plotted using the expansion 9.7.1 in people.math.sfu.ca/~cbm/aands/page_377.htm, to which you provided a solution in one of the comments above. I also tried to reproduce your results my.pcloud.com/publink/… but it was to no avail. I couldn't find the same expansion as you. My notebook and the expansion I found are: my.pcloud.com/publink/…. | |
| Jan 9, 2018 at 23:12 | comment | added | Iosif Pinelis | Here is the link to the pdf image of the Mathematica notebook illstrating my last comment: my.pcloud.com/publink/… | |
| Jan 9, 2018 at 22:46 | comment | added | Iosif Pinelis | If $|u|$ is small enough, then the $O(u^6)$ summand will be much smaller than the other summands, proportional to $u^0,u^2,u^4$. So, for such small enough $|u|$, you can just neglect the $O(u^6)$ term. That will give you an approximation to $f_U$ in a small enough neighborhood of $0$. | |
| Jan 9, 2018 at 18:05 | comment | added | Felipe Augusto de Figueiredo | Sorry for the inconvenience, but I still don't understand how to apply that ($O(.)$) on the equation you gave me to calculate $f_{u}(u)$ around 0. | |
| Jan 9, 2018 at 14:02 | comment | added | Iosif Pinelis | It's standard notation: $A=O(u^6)$ as $u\to0$ means that, for some constant $C$ not depending on $u$ and for all small enough $|u|$, we have $|A|\le Cu^6$. | |
| Jan 9, 2018 at 11:14 | vote | accept | Felipe Augusto de Figueiredo | ||
| Jan 9, 2018 at 11:13 | comment | added | Felipe Augusto de Figueiredo | Many thanks for your answer! I've got one last question, what would be $O(.)$, i.e., the last term in your equation? | |
| Jan 9, 2018 at 6:45 | comment | added | Iosif Pinelis | I don't know Matlab. However, you can get the asymptotics of $f_U$ near $0$ by using the asymptotic expansion 9.7.1 in people.math.sfu.ca/~cbm/aands/page_377.htm , which yields, in particular, $f_U(u)=\frac1{2\sqrt{2 \pi }}\,[4 n-2+(-8 n^3+12 n^2+2 n-3) u^2+(16 n^5-40 n^3+9 n) u^4+O(u^6)]$ for small $u$. So, for $n=10$ you get $f_U(0)=\frac{19}{\sqrt{2 \pi }}\approx7.58$, which is what you can see in the picture. | |
| Jan 9, 2018 at 1:33 | comment | added | Felipe Augusto de Figueiredo | Now I was able to plot it almost correctly. The only exception are the points around 0 more specifically from -0.01 up to 0.01, for these points I have $f_{u}(u) = NaN$. Do you have any suggestion on how to plot the points around 0? It seems Mathematica is better on that sense. | |
| Jan 8, 2018 at 23:27 | comment | added | Iosif Pinelis | I think in your code you need a parenthesis before 4 in 1/4*u(u_idx).^2 and the corresponding closing parenthesis. Without those parentheses, I also get a (wrong) picture similar to yours. Please try to correct your code. | |
| Jan 8, 2018 at 23:13 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 8, 2018 at 23:11 | comment | added | Iosif Pinelis | I have added the Mathematica code (right above the graph of $f_U$), which implements the obtained expression for $f_U$. | |
| Jan 8, 2018 at 18:33 | comment | added | Felipe Augusto de Figueiredo | Thanks for your update! I'm trying to simulate and compare the histogram of $z$ and you p.d.f. function, however, when I plot your function it gives me a totally different distribution. After plotting your pdf I get this (n=M=10 and -0.3 <= u <= 0.3): pasteboard.co/H20F8oI.png Could you check if your equation is correct, please? The script I'm using to plot can be found at this link: pastebin.com/YP1C8aC4 | |
| Jan 8, 2018 at 14:16 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 20:31 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 20:17 | comment | added | Iosif Pinelis | I now have an explicit formula for the pdf of $\Re z$, in terms of the modified Bessel function of the first kind. | |
| Jan 7, 2018 at 20:16 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 19:51 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 19:44 | comment | added | Iosif Pinelis | I have added the expression for the pdf of $\Re z$, with its bell-shaped graph. | |
| Jan 7, 2018 at 19:43 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 19:01 | comment | added | Felipe Augusto de Figueiredo | I see that $R = |z|^2$, however, what I need is the p.d.f of $z$. How can I find that based on $f_{R}(r)$? | |
| Jan 7, 2018 at 18:59 | comment | added | Felipe Augusto de Figueiredo | As requested, I have reverted $z$ to its original definition. The figures don't need the captions as I have plotted them using $z$ and not $Mz$. | |
| Jan 7, 2018 at 18:54 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 18:51 | comment | added | Iosif Pinelis | The graph of the pdf of $R$ is different from the graph of the pdf of $\Re z$ -- because $R$ is not $\Re z$. Rather, $R$ equals $|z|^2$ in distribution, as I noted. | |
| Jan 7, 2018 at 18:48 | comment | added | Felipe Augusto de Figueiredo | Yes, I'll change to the original definition. | |
| Jan 7, 2018 at 18:47 | comment | added | Felipe Augusto de Figueiredo | Regarding your answer, the figures added to my question show that the resulting distribution is bell-shaped, which is not in line with the p.d.f $f_{R}(r)$ | |
| Jan 7, 2018 at 18:46 | comment | added | Iosif Pinelis | Can you reverse the change in the definition of $z$, and then accordingly change the captions to the pictures, as I suggested? | |
| Jan 7, 2018 at 18:45 | comment | added | Felipe Augusto de Figueiredo | Sorry for the change, but I have only added the constant $M$ as it is in line to what I really want to know, the other things (figures and script) are only to show what I have already said since the original post. | |
| Jan 7, 2018 at 17:18 | comment | added | Iosif Pinelis | The distribution is not Student's $t$ either; in particular, it is not Cauchy -- even after rescaling. But to prove stuff such as something is not something else may be pretty nasty business. Also, your changing the question a number of times makes it hard to keep up with you. | |
| Jan 7, 2018 at 17:13 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 15:54 | comment | added | Felipe Augusto de Figueiredo | I have just updated the question as I forgot to add the term $M$ to $z$ ratio. This error was pointed out by user "ofer zeitouni". | |
| Jan 7, 2018 at 12:55 | comment | added | Felipe Augusto de Figueiredo | I followed your answer and it seems OK, however I've run some simulations in Matlab and the distribution indeed is bell-shaped. It might not be a Gaussian distribution but it has a bell like distribution. Could it be a Cauchy or student's-t distribution? I will try to post the histogram of my simulation and the script I have. BTW, your assumption is correct, x_{i} are i.i.d R.V.s | |
| Jan 7, 2018 at 4:20 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 4:08 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 7, 2018 at 2:59 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |
