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I need to find the expression for the following equation, where $m>0$

Probability[y≤m,y\[Distributed]TransformedDistribution[x^2,x\[Distributed]RiceDistribution[v,Sqrt[α/2]]]]

But, MMA does not give me what I want. any help. Am I using the right syntax?

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  • $\begingroup$ 1) I has a special meaning in Mathematica as the imaginary unit, so you shouldn't use it as your variable; more in general, try to avoid uppercase variable names altogether. 2) when you specify the distribution of y, I think you should use the transformed distribution directly, not its PDF. $\endgroup$ – MarcoB Apr 26 '16 at 12:32
  • $\begingroup$ Related: mathematica.stackexchange.com/q/113617/27951 $\endgroup$ – MarcoB Apr 26 '16 at 12:52
  • $\begingroup$ @MarcoB, thanks for your comment. I have tried with your suggestion. It still does not work...... $\endgroup$ – Srestha Narayanan Apr 26 '16 at 12:53
  • $\begingroup$ Have you tried writing the integral yet? Are you familiar without how to do that? Use PDF to get the PDF of the Rice Distribution. Integrate[( 2 E^(-((t^2 + v^2)/[Alpha])) t BesselI[0, (2 t v)/[Alpha]])/[Alpha] t^2, {t, 0, m}] $\endgroup$ – Searke Apr 26 '16 at 13:27
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    $\begingroup$ There's no reason to believe there's a solution to this problem. You may have to use an approximation instead. $\endgroup$ – Searke Apr 26 '16 at 13:42
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@wolfies essentially gave you the answer in your original question where he wrote

PDF[TransformedDistribution[x^2, x \[Distributed] RiceDistribution[v, Sqrt[α/2]]], y]

Just change PDF to CDF (and avoid the use of I as @MarcoB recommended):

CDF[TransformedDistribution[x^2, x \[Distributed] RiceDistribution[v, Sqrt[α/2]]], y]

Cumulative distribution function

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  • $\begingroup$ Huh. I assumed Integrate would have found that when I wrote it out. I wonder if I did something wrong or if Integrate didn't find it. $\endgroup$ – Searke Apr 26 '16 at 16:00

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