How to compute the expected value of a Poisson sum of the following confluent hypergeometric function: $$ \sum_{y=1}^{Y} {}_1F_1(y,1,z) $$ where y is positive integer taking values from the Poisson distributed Random Variable Y.
Thank you for your time.
Using the functional identity: $$ {}_1F_1\left(y; 1; z\right) = \mathrm{e}^{z} \cdot {}_1F_1\left(1-y;1;-z\right) = \mathrm{e}^{z} \sum_{m=0}^{y-1} \frac{(1-y)_m}{m!} \frac{(-z)^m}{m!} $$ Therefore: $$ \sum_{y=1}^Y {}_1F_1(y;1;z) = \mathrm{e}^{z} \sum_{y=1}^Y \sum_{m=0}^{y-1} \frac{(1-y)_m}{m!} \frac{(-z)^m}{m!} = \mathrm{e}^{z} \sum_{m=0}^{Y-1} \sum_{y=m+1}^Y \frac{(1-y)_m}{m!} \frac{(-z)^m}{m!} $$ Now, using $$ \sum_{y=m+1}^Y (1-y)_m = -\frac{1}{m+1} (-Y)_{m+1} $$ we continue $$ \mathrm{e}^{z} \sum_{m=0}^{Y-1} \sum_{y=m+1}^Y \frac{(1-y)_m}{m!} \frac{(-z)^m}{m!} = - \mathrm{e}^{z} \sum_{m=0}^{Y-1} \frac{(-z)^m}{(m+1)!} \frac{(-Y)_{m+1}}{m!} = \mathrm{e}^z \cdot Y \cdot {}_1F_1\left(1-Y;2;-z\right) $$ Thus this is a functional transformation of the Poisson random variable.
In particular, expected value of this variable can be readily computed: $$ \sum_{y=1}^\infty \mathrm{e}^z \cdot y \cdot {}_1F_1\left(1-y;2;-z\right) \frac{\mu^y}{y!} \mathrm{e}^{-\mu} = \sum_{m=0}^\infty \sum_{y=1}^\infty \mathrm{e}^{z-\mu} \frac{(1-y)_m \mu^y}{y!} \frac{(-z)^m}{m! (m+1)!} = \sum_{m=0}^\infty \mathrm{e}^z \frac{\mu^{m+1} z^m}{m! (m+1)!} = \mathrm{e}^z \sqrt{\frac{\mu}{z}} I_1\left(2 \sqrt{z \mu} \right) $$ where $I_1(w)$ is the modified Bessel function of the first kind.