EX 3.14 (EX 3.12 continued)

(EX 3.12 gender of newborn child continued) The pmf of $X=$ the number of births up to and including that of the first boy had $p\left(x\right)=\{\begin{array}{cc}{\left(1-p\right)}^{x-1}p& x=1,2,3,\dots \\ 0& \text{ otherwise }\end{array}$

For any positive integer $x$ ,

$$\begin{array}{rl}F(x)& =\sum _{y\le x}p(y)\\ & =\sum _{y=1}^x(1-p{)}^{y-1}p\\ & =p\sum _{y=0}^{x-1}(1-p{)}^y\end{array}\text{\hspace{1em}(3.4)}$$

To evaluate this sum, recall that the partial sum of a geometric series is $\sum _{y=0}^k{a}^y=\frac{1-a^{k+1}}{1-a}$

Using this in Equation (3.4), with $a=1-p$ and $k=x-1$ , gives $F\left(x\right)=p\cdot \frac{1-{\left(1-p\right)}^x}{1-\left(1-p\right)}=1-{\left(1-p\right)}^x$ where $x$ a positive integer.

Since $F$ is constant in between positive integers, F\left( x\right) = \left\{ \begin{matrix} 0 & x < 1 \\ 1 - {\left( 1 - p\right) }^{\left\lbrack x\right\rbrack } & x \geq 1 \end{matrix}\right. \tag{3.5}

where $\left[x\right]$ is the largest integer $\le x$

• e.g., $\left[2.7\right]=2$

Thus if $p=.51$ as in the birth example, then the probability of having to examine at most five births to see the first boy is $F\left(5\right)=1-{\left(.49\right)}^5=1-.0282=.9718$ whereas $F\left(10\right)\approx 1.0000$. This cdf is graphed in Figure 3.6.

Figure 3.6 A graph of $F\left(x\right)$ for Example 3.14