Consider a random experiment where a single positive integer (0, 1, 2, ...) is drawn at random such that the probability of selecting a number $k \space\space \epsilon \space \space \{1, 2, 3, ...\} $ is $\alpha \space\space \epsilon \space\space(0, 1) $ times the probability of the number $k-1$ preceding it. ($k = 0$ is not included here)
Define the probability space $(\Omega,F, P)$ for this experiment
Attempt: $\Omega = \{0, 1, 2, ...\}$
$F = \{0,1\}^\Omega$ (Power Series)
$P_{k+1} = \alpha * P_{k} = \alpha ^2 * P_{k-1} = ... = \alpha^{k+1}P_0$
and
$\sum_{k=0}^{\infty} P_k = 1$ or $\sum_{k=0}^{\infty} \alpha^k*P_0 = 1$
Now my question is how do you solve the above summation for $P_0$? If I google it, the answer seems to be $P_0 = (1 - \alpha)$ but this is generally stated and not explained
Any help would be appreciated!
As you concluded, $$\sum_{k=0}^{\infty} {\alpha^k \bullet P_0} = 1$$ Thus, $$\sum_{k=0}^{\infty} {\alpha^k} = \frac{1}{P_0}$$ But as we know (geometric series) $$\sum_{k=0}^{\infty} {\alpha^k} = \frac{1}{1-\alpha}$$ and we got that $$\frac{1}{P_0}=\frac{1}{1-\alpha} \rightarrow P_0=1-\alpha$$