I'm interested in finding the CDF and PDF of $U_i$ defined as follows,
$$U_i=\frac g{d^{\alpha}}$$
where $g$ is a gamma distributed random variable with shape $k$ and scale $\theta$, and $d$ is a random variable with distribution $f_d(x)=\frac {2x}{R^2}, 0\le x \le R$.
To find the CDF of $U_i$
$$F_{U_i}(y)=P(U_i<y)=\int_0^R F_g(yx^\alpha)dx$$
where $F_g(y)$ is the CDF of gamma distributed random variable.
After performing the integration, which requires the use of the expansion of incomplete gamma function, I came to the following CDF:
$$F_{U_i}(y)=\sum_{n=0}^\infty \frac {2(-1)^nR^{k\alpha+n\alpha}y^{k+n}}{\Gamma(k)n!\theta^k(k+n)(k\alpha+n\alpha+2)}$$
I want the CDF in this format because my ultimate goal is to find the CDF of $U=\sum_{i=1}^NU_i$ which requires me to find the Laplace transform of $F_{U_i}(y)$ and raise it to power $N$.
Now it's easy to find the PDF of $U_i$, i.e. $f_{U_i}(y)=\frac {d}{dy}F_{U_i}(y)$.
I'm looking for someone to clarify these points to me:
Probability distribution function expressed in terms of a divergent series
William
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