Modeling "Time Until Event Occurs"

Question

I'm having trouble understanding why "the time until an event occurs" is modeled as:

$S(t) = P[T > t]$ where $T$ is time until event and $t$ is some time.

I don't see how $P[T<t]$ or $P[T=t]$ would be phrased, and I'm confused on how to know if the time till an event occurs is greater/less/equal to some time $t$.

For example, consider the following problem:

In analyzing the risk of a catastrophic event, an insurer uses the exponential distribution with mean $\alpha$ as the distribution of the time until the event occurs. The insured has $n$ independent catastrophe policies of this type. Find the expected time until the insurer will have the first catastrophe claim.

How do I know to use $P[T>t]$ and not $P[T<t]$ or $P[T=t]$?

Answer 1

The reason for this particular formulation lies in the wording of the problem.

We are told that there are $n$ independent and identically distributed (i.i.d.) random variables $T_1, T_2, \ldots, T_n$. We want to find the expected value of the minimum $T_\text{min}$ of these random variables (this might be interpreted as, e.g., the time until the first catastrophe occurs).

We now observe that $T_\text{min} > t$ if and only if $T_1 > t, T_2 > t, \ldots, T_n > t$. This is the critical observation. It permits us to conclude

$$ P(T_\text{min} > t) = P(T_1 > t, T_2 > t, \ldots, T_n > t) $$

From independence, we can then write

$$ P(T_\text{min} > t) = P(T_1 > t) \times P(T_2 > t) \times \cdots \times P(T_n > t) $$

and from the fact that they are identically distributed, we can write

$$ P(T_\text{min} > t) = [P(T_1 > t)]^n $$

(Note that we can not assert $T_\text{min} < t$ if and only if $T_1 < t, T_2 < t, \ldots, T_n < t$, so this useful telescoping does not occur if we approach the problem that way.) Now, for the exponential distribution, fortunately, we have $P(T_1 > t) = 1 - P(T_1 < t) = 1-[1-e^{-\lambda t}] = e^{-\lambda t}$ (where $\lambda = 1/\alpha$), so

$$ P(T_\text{min} > t) = e^{-n\lambda t} $$

That is, the minimum of $n$ i.i.d. exponentially distributed random variables is distributed identically to a single exponentially distributed random variable with $n$ times the rate (or, equivalently, $1/n$ of the mean).

Long story short: The analysis chooses to focus on $P(T > t)$ because the formulation of the problem makes that approach simple. If we were interested in the maximum of those random variables, we would choose $P(T_\text{max} < t)$. Unfortunately, in that case, the exponential distribution does not help us, particularly; the maximum of $n$ i.i.d. exponentially distributed random variables is not itself exponentially distributed.

Answer 2

Are you referring to?: $\mathsf E(T) = \int_0^\infty \mathsf P(T>t) \operatorname d t$

It comes from the definition of expectation and the fact that for strictly positive† continuous random variables.

$$\begin{align}\mathsf E(T) = & ~ \int_0^\infty s ~f_T(s)\operatorname d s \\[1ex] = & ~ \int_0^\infty f_T(s)\int_0^s\operatorname d t\operatorname d s \\[1ex] = & ~ \int_0^\infty \int_t^\infty f_T(s)\operatorname d s\operatorname d t \\[1ex] = & ~ \int_0^\infty \mathsf P(T>t)\operatorname d t \end{align}$$

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Note: † - it only works for strictly positive random variables (those with a support interval not less than zero).