Find cumulative distribution function of random variable given probability generating function

Question

Let $f_c(x) := c \cdot \ln\left(1 - \frac{x}{2}\right)$. Find for what $c \in \mathbb{R}$ $f_c$ is a probability generating function.

I wrote $$ f_c(x) = \sum_{k \in \mathbb{N}_0} -c \cdot \frac{x^k}{2^k \cdot k!}. $$ For $f_c$ to be probability generating function, $p(k) := \frac{-c}{2^k \cdot k!}$ has to be a PMF, i.e \begin{equation*} 1 \overset{!}{=} \sum_{k \in \mathbb{N}_0} p(k) = \sum_{k \in \mathbb{N}_0} \frac{-c}{2^k \cdot k!} = -c \cdot \sqrt{e}, \end{equation*} yielding $c = - e^{-\frac{1}{2}}$.

Let $X_c$ be a random variable with the distribution obtained from $f_c$. Calculate $E[X_c]$ and $P(X_c = 1)$.

Similarly to the above I wrote \begin{align*} E[X_c] & = \sum_{k \in \mathbb{N}_0} k p(k) = e^{-\frac{1}{2}} \sum_{k \in \mathbb{N}_0} \frac{k}{2^k \cdot k!} \\ & = \frac{e^{-\frac{1}{2}}}{2} \sum_{k \in \mathbb{N}_0} \frac{1}{2^{k - 1} \cdot (k - 1)!} = \frac{e^{-\frac{1}{2}}}{2} e^{\frac{1}{2}} = \frac{1}{2}. \end{align*} and \begin{equation*} P(X_c = 1) = p(1) = \frac{e^{-\frac{1}{2}}}{2^1 \cdot 1!} = \frac{1}{2 \sqrt{e}}. \end{equation*}

Find the cumulative distribution function of $X_c$.

I wrote \begin{align*} P(X_c \le x) & = \sum_{k = 0}^{\lfloor x \rfloor} p(k) = e^{-\frac{1}{2}} \cdot \sum_{k = 0}^{\lfloor x \rfloor} \frac{1}{2^k \cdot k!} \end{align*}

Questions

1. Is there a closed form for the above result (not utilising the incomplete gamma function, which we did not cover in class?
2. Are my results correct? Is there an easier way to arrive at them?

Answer 1

Using standard probability notation you have probability generating function $G_X(z) = c \cdot \ln(1-\tfrac{z}{2})$ and the corresponding random variable is $X$. From the sums in your question, I will assume that $X$ is a non-negative integer random variable. With this condition, one of the properties of the PGF is that:

$$\mathbb{P}(X=k) = \frac{G_X^{(k)}(0)}{k!}.$$

These probability mass values must sum to one, which imposes a constraint on the PGF, which we can use to find the constant $c$. Now, substituting your specified PGF you get:

$$\begin{equation} \begin{aligned} G_X^{(k)}(z) = c \cdot \Big( \frac{d}{dz} \Big)^k \ln(1 - z/2) &= \begin{cases} c \cdot \ln(1 - z/2) & & & \text{for } k=0, \\[6pt] - c \cdot (k-1)! \cdot ( 2-z )^{-k} & & & \text{for } k>0, \\[6pt] \end{cases} \\[6pt] \end{aligned} \end{equation}$$

which gives the mass function:

$$\begin{equation} \begin{aligned} \mathbb{P}(X=k) = \frac{G_X^{(k)}(0)}{k!} &= \begin{cases} 0 & & & \text{for } k=0, \\[6pt] - (c/k) \cdot 2^{-k} & & & \text{for } k>0. \\[6pt] \end{cases} \\[6pt] \end{aligned} \end{equation}$$

The constraint equation therefore reduces to:

$$\begin{equation} \begin{aligned} 1 = \sum_{k=0}^\infty \mathbb{P}(X=k) = -c \cdot \sum_{k=1}^\infty \frac{1}{k \cdot 2^k} = -c \cdot \ln (2). \\[6pt] \end{aligned} \end{equation}$$

From this constraint we have $c=-1 / \ln (2)$ so your PGF is:

$$G_X(z) = - \frac{\ln(1-\tfrac{z}{2})}{\ln(2)} = 1 - \frac{\ln(2-z)}{\ln(2)},$$

and the corresponding probability mass function is:

$$p_X(k) \equiv \mathbb{P}(X=k) = \frac{1}{\ln (2)} \cdot \frac{1}{k \cdot 2^k} \quad \quad \quad \text{for all } k \in \mathbb{N}.$$

This is the logarithmic series distribution with probability parameter $p=\tfrac{1}{2}$. The CDF, moments, and other properties, can be derived from the mass function.