# Deriving cdf of ricean distribution + help

• JamesGoh
In summary, the conversation discusses the generalised form of the Marcum Q function, which is used to calculate the cdf of the amplitude in rice distribution. The formula for Qm and the rice pdf are also mentioned, and the person has attempted to calculate the rice cdf using these formulas but has not had much success. They have tried substituting variables and are aware that the cdf is the integral of the pdf with a constant value for \sigma. They question if their approach is correct.

#### JamesGoh

Im aware that the generalised form of the Marcum Q function, which is

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

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 $$\beta=0$$

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

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 ?

Im aware that the generalised form of the Marcum Q function, which is

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

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 $$\beta=0$$

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

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 ?

## 1. How do you derive the cdf of a Ricean distribution?

The cdf (cumulative distribution function) of a Ricean distribution can be derived using the characteristic function of the distribution. This involves taking the Fourier transform of the pdf (probability density function) and simplifying it using trigonometric identities. The resulting expression can then be used to calculate the cdf for any value of the random variable.

## 2. What is the purpose of deriving the cdf of a Ricean distribution?

The cdf is an important tool in probability and statistics as it gives the probability that a random variable takes on a value less than or equal to a given value. By deriving the cdf of a Ricean distribution, we can better understand the behavior of the distribution and make predictions about the likelihood of certain values occurring.

## 3. Are there any limitations to deriving the cdf of a Ricean distribution?

One limitation is that the cdf can only be derived for a specific type of Ricean distribution, known as the complex-valued Ricean distribution. This distribution assumes that the amplitude and phase components of a signal are independent and follow different distributions. Other types of Ricean distributions may not have a known cdf.

## 4. Can you provide an example of calculating the cdf of a Ricean distribution?

Sure, let's say we have a complex-valued Ricean distribution with a mean of 1 and a scale parameter of 2. To calculate the cdf at a value of 3, we can use the formula: cdf(x) = 1 - exp(-x^2/8) * (1 + x^2/4), where x = 3. Plugging in the values, we get cdf(3) = 0.7887. This means that there is a 78.87% chance that the random variable will take on a value less than or equal to 3.

## 5. Are there any resources available to help with deriving the cdf of a Ricean distribution?

Yes, there are many online resources and textbooks that provide step-by-step explanations and examples of how to derive the cdf of a Ricean distribution. Some resources also offer tools and calculators to make the process easier. Additionally, consulting with other experts in the field or attending workshops or seminars can also be helpful in understanding and deriving cdf's for various distributions.