Deriving cdf of ricean distribution + help

AI Thread Summary
The discussion focuses on deriving the cumulative distribution function (CDF) of the Rice distribution using the generalized Marcum Q function and the probability density function (PDF) of the Rice distribution. The user attempts to express the CDF as F_{r}(r) = 1 - Q_{M}(A/σ, r_{min}/σ) and seeks clarification on their approach involving substitutions for variables. They express uncertainty about the correctness of their integration method, particularly in relation to the constants involved. The conversation highlights the challenges in integrating the PDF to obtain the CDF while maintaining the integrity of the parameters. Overall, the user is looking for guidance on their mathematical approach to derive the Rice CDF accurately.
JamesGoh
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Im aware that the generalised form of the Marcum Q function, which is

Q_{M}(\alpha,\beta)=1/(\alpha)^{M-1}\int_{\beta}x^{M}.exp(-x^{2} +\alpha^{2})/2.I_{M-1}(\alphax)dx

and the pdf of the amplitude in rice distribution is

f_{r}(r)=r/\sigma^{2}exp( (-r^{2}-A^{2})/2\sigma^{2} )I_{0}(rA/\sigma^{2})

where I_{0}(x) is a modified bessel function of first kind, zero order

and the cdf of the rice distribution is

F_{r}(r) =1-Q_{M}(A/\sigma,r_{min}/\sigma)

Using the formula for Qm and the rice pdf, I have tried to get the rice cdf, however I have not had much success. I have tried the following

Let x=r/\sigma, \alpha=A/\sigma and \beta=0

Q_{1}(\alpha,\beta)=\int_{0}^{r_{min}}=(r/\sigma)exp( (-r^{2}-A^{2})/2\sigma^{2} )I_{0}(r/\sigmaA/\sigma)d(r/\sigma)

Im aware that the cdf is the integral of the pdf and \sigma is a constant (which means it cannot change), so is my approach correct ?
 
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Im aware that the generalised form of the Marcum Q function, which is

Q_{M}(\alpha,\beta)=1/(\alpha)^{M-1}\int_{\beta}x^{M}.exp(-x^{2} +\alpha^{2})/2.I_{M-1}(\alphax)dx

and the pdf of the amplitude in rice distribution is

f_{r}(r)=r/\sigma^{2}exp( (-r^{2}-A^{2})/2\sigma^{2} )I_{0}(rA/\sigma^{2})

where I_{0}(x) is a modified bessel function of first kind, zero order

and the cdf of the rice distribution is

F_{r}(r) =1-Q_{M}(A/\sigma,r_{min}/\sigma)

Using the formula for Qm and the rice pdf, I have tried to get the rice cdf, however I have not had much success. I have tried the following

Let x=r/\sigma, \alpha=A/\sigma and \beta=0

Q_{1}(\alpha,\beta)=\int_{0}^{r_{min}}=(r/\sigma)exp( (-r^{2}-A^{2})/2\sigma^{2} )I_{0}(r/\sigmaA/\sigma)d(r/\sigma)

Im aware that the cdf is the integral of the pdf and \sigma is a constant (which means it cannot change), so is my approach correct ?
 
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