I know that the formal defintion of the probability of failure for a discrete random variable, $T$, is $$h(t) = \frac{P(T=t)}{1-P(T \leq t-1)}$$
This is all good, but I was told that for geometric distributions specifically, the probability of failure is equal to $p$.
I am just wondering if anyone can formulate the proof for why this is... I can't figure it out myself.
I got to the following but I don't think I can get anywhere with it:
$$h(t) = \frac{(1-p)^{t-1}}{\sum \limits_{w=t}^{\infty} (1-p)^{w-1}}$$
Thanks for the help!
One approach is to use the formula for an infinite geometric series: $a + ar + ar^2 + \dots = \frac{a}{1-r}$. In particular, $\sum_{w=t}^\infty (1-p)^{w-1} = \frac{(1-p)^{t-1}}{1 - (1-p)} = (1-p)^{t-1}/p$ by this formula, so $$ h(t) = \frac{(1-p)^{t-1}}{\sum_{w=t}^\infty (1-p)^{w-1}} = \frac{(1-p)^{t-1}}{(1-p)^{t-1}/p} = p. $$ There is also a quicker way to see this. Recall the definition of a Geometric random variable from its random experiment: we do a sequence of independent random trials with success probability $p$, and count the number of trials it takes to see a success. Therefore the denominator $1 - \Pr[T \le t-1] = \Pr[T \ge t]$ is the probability that the first $t-1$ trials were all failures: which is $(1-p)^{t-1}$. We conclude that $$ h(t) = \frac{\Pr[T=t]}{1 - \Pr[T\le t-1]} = \frac{p(1-p)^{t-1}}{(1-p)^{t-1}} = p. $$