Probability and lifetime What is the relation between the lifetime of a particle and the probability of decaying that particle?
Here it says that the probability of survival is exponential if the decay process is a Poisson process:
$P(t) = e^{-\frac{t}{\tau}}$
However, this function is not normalized so it cannot be a probability distribution.
If the decay lifetime of a particle is $\tau$, what is the probability of detection a decay an time t?
 A: Expanding a bit on what David Hammen wrote in comments:
If a particle has a probability of decay per unit time that is
$$P(t)\propto e^{-t/\tau}$$
then we can state "between now (when the particle exists) and a long time from now (when the particle has decayed) there must be exactly a probability of 1 that the particle has decayed - this is the normalization of the expression you are looking for.
This means that we have to find $A$ in
$$\begin{align}
A \int_0^\infty P(t)dt &= 1\\
A\int_0^\infty e^{-t/\tau}dt &= 1\\
\frac{A}{\tau}&=1\\
A&=\tau
\end{align}$$
Now the mean life of the particle is given by
$$\int_0^{\infty}\frac{t}{\tau}e^{-t/\tau}dt\\
=\tau$$
(you evaluate the second integral using integration by parts)
And that is how you find the relationship between the decay probability and the lifetime of the particle.
A: An exponential distribution has a PDF of $$f(t) = \frac 1 \tau \exp\left(-\frac t \tau\right)$$
where $t$ ranges from $[0,\infty)$ and $\tau$ is the expected value of the random variable.
To see that this is a PDF, compute $F_t(t) = \int_0^t f(x)\,dx = 1-\exp\left(-\frac t \tau\right)$. This increases monotonically from 0 to 1 over the interval $[0,\infty)$, which is exactly what's needed. $F_t(t)$ is the CDF for the exponential distribution.
To see that $\tau$ is indeed the mean (expected value), compute $E(t) = \int_0^{\infty} t\,f(t)\,dt$ This integrates to $\left.(t+\tau)\exp\left(- \frac t \tau\right)\right|_{t=0}^{\infty} = \tau$.
Some people prefer to use half-life $\tau_{1/2} = \tau \ln 2$ instead of the mean lifetime $\tau$. With half-life, it makes more sense to use a base of 2 rather than $e$ in the PDF: $$f(t) = \frac {\ln 2}{\tau_{1/2}} 2^{-t/\tau_{1/2}}$$
When you add relativistic time dilation into the mix, simply scale the mean lifetime or the half-life by the relativistic gamma. This time dilation pertains to the time measured by observer watching a moving radioactive particle. This time dilation is why we can see muons created in the upper atmosphere on the surface of the Earth.  There is no time dilation from the particle's perspective. From the muon's perspective, the Earth is rushing straight toward them at relativistic speeds. The muons see a foreshortened distance to the Earth thanks to length contraction.
A: To put these arguments in words rather than math, the claim

[the falling exponential] function is not normalized so it cannot be a probability distribution

depends on a implicit understanding that the range of application is all time. 
However, the physical situation to which the decay rule applies has a definite start time, so the range of integration should not extend back without bound. Once that is the case, the integral is bounded and normalizable.
