Probability of Detecting an Event Given No Previous Detections I'm trying to figure out:
Given a probability distribution of how likely an event is to be detected at time $t=1,t=2,t=3,t=4$ etc, what is the probability that it will be detected at t=n+1 if it is not detected at t=n.
To make sure I'm doing it correctly I've attempted to do this specifically for the example of exponential decay, since the probability of it being detected in any interval is constant, so it's easy to check whether or not I got the correct solution.
To attempt this I've first calculate the cumulative distribution function, CDF. The probability of no detection at t=n is then simply $1-CDF(n)$.
Then using Bayes rule(where Dt=n means detection at time t=n, and Nt=n means no detection at time t=n)
$$P(Dt=n|Nt=n-1) = \frac{P(Nt=n-1|Dt=n)P(Dt=n)}{P(Nt=n-1)} $$
For something to of been detected at time t=n, it must of not been detected previously so $P(Nt=n-1|Dt=n)P(Dt=n)=1$
$$P(Dt=n|Nt=n-1) = \frac{P(Dt=n)}{P(Nt=n-1)} $$
Substiuting in $P(Nt=n-1) = 1-CDF(n-1)$
$$P(Dt=n|Nt=n-1) = \frac{P(Dt=n)}{1-CDF(n-1)} $$
However,this applied to an exponential decay does not result in $P(Dt=n|Nt=n-1)$ being constant. Could anyone point out what mistake I have made? Thank you for your time.
 A: There are a number of confusions in the post. I'm not sure which of them led to the incorrect conclusion at the end, so I'll try to clear them all up.
First, you seem to be using the expression "it is not detected at $t=n$" in a confusing way, namely to mean "it has not been detected by $t=n$", whereas the more natural interpretation would seem to be "it is not detected exactly at $t=n$". So I believe what you're actually trying to calculate is the probability that the decay will be detected at $t=n+1$ if it has not been detected by $t=n$. Similarly, "no detection at $t=n$" should be "no detection by $t=n$".
A possibly related confusion is in your notation for the events. You write "$Dt=n$ means detection at time $t=n$, and $Nt=n$ means no detection at time $t=n$". That makes no sense, since while there is a time at which the decay is detected, there is no such thing as the time at which the decay isn't detected. The notation "$X=x$" is used to denote the event that the random variable $X$ takes the value $x$; but there is no random variable "the time at which the decay isn't detected", and hence no event that such random variable takes the value $n$. The standard way to represent these events would be with a random variable $T$, the time of detection, with $T=t$ denoting the event that the decay is detected at time $t$ and $T\gt t-1$ (or equivalently $T\ge t$) denoting the event that the decay hasn't been detected by time $t-1$.
With this notation, your equation becomes
$$
P(T=t\mid T\gt t-1)=\frac{P(T\gt t-1\mid T=t)P(T=t)}{P(T\gt t-1)}\;.
$$
You're right that $P(T\gt t-1\mid T=t)=1$, so
$$
P(T=t\mid T\gt t-1)=\frac{P(T=t)}{P(T\gt t-1)}\;.
$$
You didn't write which expressions you substituted at this point to derive a contradiction. If you substitute the correct expressions,
$$
P(T=t)=p(1-p)^{t-1}
$$
and
$$
P(T\gt t-1) = \sum_{k=t}^\infty p(1-p)^{k-1}=(1-p)^{t-1}\;,
$$
you recover the constant conditional decay probability as expected:
$$
P(T=t\mid T\gt t-1)=\frac{P(T=t)}{P(T\gt t-1)}=\frac{p(1-p)^{t-1}}{(1-p)^{t-1}}=p\;.
$$
