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.
Best Answer
$$F_X(u)=\int_{0}^{u} f_X(x)dx=\int_{0}^{u} \frac{2}{\Gamma(m)}\left(\frac{m}{\Omega}\right)^mx^{(2m-1)}e^{-\left(\frac{m}{\Omega}x^2\right)}dx$$
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