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