0
$\begingroup$

Am trying to show that the probability a count r.v $X$ takes even values is given by

$\frac 1 2( 1 + G(-1)),$ where $G(t)$ is its probability generating function.

I know that due to possible symmetry then the $P(\text{odd}) + P(\text{even}) =1.$

Kindly help any initial stages.

$\endgroup$

1 Answer 1

0
$\begingroup$

$$ G(t) = \sum_{n=0}^\infty t^n \Pr(X=n). $$ $$ G(-1) = \sum_{n=0}^\infty (-1)^n \Pr(X=n). $$ \begin{align} & 1 + G(-1) \\[10pt] = {} & \left(\sum_{n=0}^\infty 1\Pr(X=n)\right) + \left( \sum_{n=0}^\infty (-1)^n \Pr(X=n) \right) & & \text{since the first sum is 1} \\[10pt] = {} & \sum_{n=0}^\infty \Big( 1 + (-1)^n\Big) \Pr(X=n). \end{align} Then observe that $\displaystyle 1+(-1)^n = \begin{cases} 2 & \text{if $n$ is even,} \\ 0 & \text{if $n$ is odd.} \end{cases}$

$\endgroup$
1
  • $\begingroup$ Thank you, now I see the use of 1/2. $\endgroup$ Jan 25, 2018 at 0:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .