finding probability generating function of odd values of $X$ $X$ is defined as a discrete random variable. How can we show that its probability generating function of $X$ taking an odd value is 
$$G(t) =\frac12\big(1 - G(-1)\big)\;?$$
 A: Your question as asked does not make much sense, since you are saying that $G(t)$ is given by a constant expression.  I assume that you mean the probability of $X$ being odd is given by the expression.  Here is a proof.
We have
$$\eqalign{
  G(t)&=\sum_{k=0}^\infty p_kt^k\cr
  G(1)&=\sum_{k=0}^\infty p_k=1\cr
  G(-1)&=\sum_{k=0}^\infty p_k(-1)^k\ .\cr}$$
Now $1-(-1)^k$ is equal to $0$ if $k$ is even, $2$ if $k$ is odd.  Therefore
$$1-G(-1)=\sum_{k=0}^\infty p_k(1-(-1)^k)=\sum_{k\ \rm odd}2p_k=2P(\hbox{$X$ is odd})\ .$$
A: We have that $$G(t)=P(X=x_1)t^{x_1}+P(X=x_2)t^{x_2}...$$
$$G(1)=P(X=x_1)(1)^{x_1}+P(X=x_2)(1)t^{x_2}...= P(X=x_1)+P(X=x_2)+...=1$$
(The last piece comes from knowing that the sum of all probabilites equals 1.)
From that we get that  $1=P(X=e)+P(X=o)$, where e is even and o is odd. 
We do the same for G(-1)
$$G(-1)= P(X=x_1)(-1)^{x_1}+P(X=x_2)(-1)^{x_2}+...=-P(X=x_1)+P(X=x_2)+...$$
So $G(-1)=P(X=e)-P(X=o)$, since $x_n$ when $n$ is even results in $P(X=x_n)$ being positive, but when $n$ is odd, $P(X=x_n)$ is negative.
Now you have a system of equations with these two: 
$$1=P(X=e)+P(X=o)$$ $$G(-1)=P(X=e)-P(X=o)$$
Solve for $P(X=o)$ and you're done. here it is.
$$P(X=e)=1-P(X=o)$$ $$G(-1)=(1-P(X=o))-P(X=o)$$ $$G(-1)=1-2P(X=o)$$ $$2P(X=o)=1-G(-1)$$ $$P(X=o)= \frac{1}{2}(1-G(-1))$$
A: 
Assume that $X$ is a discrete random variable taking non-negative integer values. If $G$ is the probability generating function of $X$ defined by $G(t)=E(t^X)$ for every $|t|\leqslant1$, show that the probability of $X$ taking an odd value is $\frac12(1−G(−1))$.

Integrate the pointwise identity, valid for every integer valued random variable $X$, $$2\cdot\mathbf 1_{X\ \text{odd}}=1-(-1)^X.$$
