The z-transform of the PMF of a discrete random variable $K$ is given by
$\,\displaystyle G_K\left (z\right )=A\left[\frac{10+8 z^2}{2-z}\right]$
a. What is the expected value of $K$ ?

b. Find $p_K(1)$ , the probability that $K$ has the value 1.

Question

The z-transform of the PMF of a discrete random variable [latex]K[/latex] is given by

[latex]\,\displaystyle G_K\left (z\right )=A\left[\frac{10+8 z^2}{2-z}\right][/latex]

a. What is the expected value of [latex]K[/latex] ?

b. Find [latex]p_K(1)[/latex], the probability that [latex]K[/latex] has the value 1.

Accepted Answer

Before answering both questions we need to obtain the numerical value of [latex]A[/latex]. For [latex]G_K(z)[/latex] to be a valid z-transform, it must satisfy the condition [latex]G_K(1)=1[/latex]. Thus, we have that

[latex]\,\displaystyle G_K\left (1 ight )=A\left[\frac{10+8}{2-1} ight]=18 A=1 \Rightarrow A=\frac{1}{18}[/latex]

a. The expected value of [latex]K[/latex] is

[latex]\,\displaystyle E[K]=\left.\frac{d}{d z} G_K\left (z ight ) ight\vert _{z=1}=A\left[\frac{\left (2-z ight ) 16 z-\left (10+8 z^2 ight )\left (-1 ight )}{\left (2-z ight )^2} ight]_{z=1}=\frac{34}{18}=1.9[/latex]

b. To obtain the PMF of [latex]K[/latex], we can use two methods:

Method 1:
We observe that

[latex]\,\displaystyle \begin{aligned} G_K\left (z ight ) & =A\left[\frac{10+8 z^2}{2-z} ight]=\frac{A}{2}\left[\frac{10+8 z^2}{1-\frac{z}{2}} ight]=\frac{A\left (10+8 z^2 ight )}{2} \sum_{k=0}^{\infty}\left (\frac{z}{2} ight )^k \ & =\frac{A\left (10+8 z^2 ight )}{2}\left\{1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\frac{z^4}{16}+\cdots ight\} \ & =\frac{A}{2}\left\{10+z\left[\frac{10}{2} ight]+z^2\left[\frac{10}{4}+8 ight]+z^3\left[\frac{10}{8}+4 ight]+z^4\left[\frac{10}{16}+2 ight]+\cdots ight\} \end{aligned}[/latex]

Thus, the probability that [latex]K[/latex] has a value 1 is the coefficient of [latex]z[/latex] in [latex]G_K(z)[/latex], which is

[latex]\,\displaystyle p_K(1)=\frac{A}{2} imes \frac{10}{2}=\frac{5}{36}[/latex]

Method 2:
We can also obtain the value of [latex]p_K(1)[/latex] as follows:

[latex]\,\displaystyle \begin{aligned} p_K\left (1 ight ) & =\left.\frac{1}{1 !} \frac{d}{d z} G_K\left (z ight ) ight\vert _{z=0}=\left.A\left\{\frac{\left (2-z ight )\left (16 z ight )-\left (10+8 z^2 ight )\left (-1 ight )}{\left (2-z ight )^2} ight\} ight\vert _{z=0} \ & =A\left\{\frac{10}{4} ight\}=\frac{1}{18} imes \frac{10}{4}=\frac{5}{36} \end{aligned}[/latex]

Question 7.7: The z-transform of the PMF of a discrete random variable K i......

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