# Proof of CDF of nakagami-m distribution?

I want to derive the CDF of Nakagami-m distribution. Here is what i had tried:

My approach: We know that to find CDF, we have to integrate the PDF. Hence, first writing the PDF of nakagami random variable (X) as $$f_X(x)=\frac{2}{\Gamma(m)}\left(\frac{m}{\Omega}\right)^mx^{(2m-1)}e^{-\left(\frac{m}{\Omega}x^2\right)}$$-------(1).

Next, i assume $$\alpha = m$$ and $$\beta = \frac{m}{\Omega}$$.

Thus (1) becomes: $$f_X(x)=\frac{2}{\Gamma(\alpha)}\left(\beta\right)^{\alpha}x^{(2\alpha-1)}e^{-\beta x^2}$$-------(2).

Integrating (2) we get CDF as $$F_{X}(x) =$$ $$\int_0^u\frac{2}{\Gamma(\alpha)}\left(\beta\right)^{\alpha}x^{(2\alpha-1)}e^{-\beta x^2}\text{dx}$$

Thus, $$F_{X}(x) = \frac{2\beta^{\alpha}}{\Gamma(\alpha)}\int_0^ux^{(2\alpha-1)}e^{-\beta x^2}\text{dx}$$------(3)

I am not getting how to further solve Eq. (3).

Any help in this regard is highly appreciated.

• It doesn't have a closed form solution. see here. What you have can be re-written using the incomplete gamma function. Is your question about how to do that? Commented Feb 20, 2021 at 19:06
• Yes ...my question is about the same i.e., how that CDF expression in terms of lower incomplete Gamma function is derived. Commented Feb 21, 2021 at 2:15

$$F_X(u)=\int_{0}^{u} f_X(x)dx=\int_{0}^{u} \frac{2}{\Gamma(m)}\left(\frac{m}{\Omega}\right)^mx^{(2m-1)}e^{-\left(\frac{m}{\Omega}x^2\right)}dx$$
Now, substitute $$t=\frac{m x^2}{\Omega}$$, $$dt=\frac{2 m x}{\Omega}$$ to derive
$$F_X(u) =\int_{0}^{u} \frac{2}{\Gamma(m)}\left(\frac{m}{\Omega}\right)x \left(\frac{m x^2}{\Omega}\right)^{(m-1)}e^{-\left(\frac{m x^2}{\Omega}\right)}dx =\frac{1}{\Gamma(m)}\int_{0}^{\frac{2u^2}{\Omega}} t^{(m-1)}e^{-t}dt=\frac{\gamma(m,\frac{2u^2}{\Omega})}{\Gamma(m)}=P(m,\frac{2u^2}{\Omega})$$
$$\gamma$$ is the lower incomplete gamma function and $$P$$ is the regularized Gamma function