I have gotten stuck on the following problem. We wish to show the following equality holds:
We want to show that the following equality holds:
$$ \frac{g_N(t) - 1}{t - 1} = \sum_{n \geq 0} P(N > n) t^n $$
where (g_N(t)) is the probability generating function for some unknown random variable (N). N assumes non-negative integer values and $|t|<1$
I have developed (g_N(t)) into
$$ \sum_{n \geq 0} t^n P(N = n) $$
giving us on the left-hand side:
$$ \frac{\sum_{n \geq 0} t^n P(N = n) - 1}{t - 1} $$
and on the right-hand side:
$$ \sum_{n \geq 0} (1 - P(N = n)) t^n $$
equating both sides gives us:
$$ \frac{\sum_{n \geq 0} t^n P(N = n) - 1}{t - 1} = \sum_{n \geq 0} (1 - P(N = n)) t^n $$
From here on I have tried multiplying over (t-1), I have tried to write $t^nP(N=n)$ as a geometric series in closed form $\frac{P(N=n)}{1-t}$ since $n>= 0$ and |t|<1. But I have just gotten more confused in the process. Would appreciate some help in showing this equality. Thanks.
$$\sum_{n}P(N>n)t^{n}=\sum_{n=0}^{\infty}\sum_{m=n+1}^{\infty}P(N=m)t^{n}$$
Change the order of summation(Valid by Tonelli's Theorem as summands are positive):
$$\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}P(N=m)t^{n}$$
Use Geometric sum formula,
$$\sum_{m=1}^{\infty}P(N=m)\bigg(\frac{1-t^{m}}{1-t}\bigg)$$
Rearrange (start the summation at $0$) now that the numerator is :-
$$1-P(N=0)+P(N=0)-\sum_{m=0}^{\infty}P(N=m)t^{m}=1-g_{N}(t)$$
And so you get the result.