1. **State the problem:** We need to find the value of $$R + \mathrm{Var}(Y) + P(Y < 2)$$ where: - $X$ is a discrete random variable with geometric distribution and mean 0.8. - $R$ is the radius of convergence of the probability generating function (PGF) of $X$. - $Y$ is a random variable with characteristic function $$\varphi_Y(t) = \frac{1}{1 - it}, \quad t \in \mathbb{R}.$$ 2. **Find $R$, the radius of convergence of the PGF of $X$:** - For a geometric random variable with success probability $p$, the mean is $$E[X] = \frac{1-p}{p}.$$ - Given mean $0.8$, solve for $p$: $$0.8 = \frac{1-p}{p} \implies 0.8p = 1 - p \implies 1.8p = 1 \implies p = \frac{1}{1.8} = \frac{5}{9}.$$ - The PGF of $X$ is $$G_X(s) = \frac{p}{1 - (1-p)s} = \frac{5/9}{1 - (4/9)s}.$$ - The radius of convergence $R$ is the distance from 0 to the singularity at $$s = \frac{1}{1-p} = \frac{1}{4/9} = \frac{9}{4} = 2.25.$$ - So, $$R = 2.25.$$ 3. **Find $\mathrm{Var}(Y)$:** - The characteristic function of $Y$ is $$\varphi_Y(t) = \frac{1}{1 - it}.$$ - This is the characteristic function of an exponential distribution with rate parameter $\lambda = 1$ (shifted to the right). - For an exponential distribution with rate $\lambda$, $$\mathrm{Var}(Y) = \frac{1}{\lambda^2} = 1.$$ 4. **Find $P(Y < 2)$:** - Since $Y \sim \mathrm{Exponential}(1)$, the CDF is $$F_Y(y) = 1 - e^{-y}, \quad y \geq 0.$$ - Thus, $$P(Y < 2) = F_Y(2) = 1 - e^{-2}.$$ 5. **Sum all parts:** $$R + \mathrm{Var}(Y) + P(Y < 2) = 2.25 + 1 + (1 - e^{-2}) = 4.25 - e^{-2}.$$ **Final answer:** $$\boxed{4.25 - e^{-2}}.$$