# Essentials of Survival Time Analysis

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This post aims to clarify the relationship between rates and probabilities.

## Introduction

It is a common situation that individuals would change their state as time goes, such as from susceptible to infectious, or from juvenile to mature, etc. We are interested in the expected time that an individual stays in a given state.

## Terminology

- $a$: the time that an individual spends in a given state. It can be called soujourn time, the survival time.
- $F(a)$: the probability that an individual has not left the state before or at time $a$. It is non-increasing and $F(0)=1$.
- $T$: the time to exit a given state, then \(F(a)=P(T>a)\)
- $G(a)=1-F(a)=P(T\le a)$: the probability to have left before time $a$.

It is worthy noting that the definition is the same as the definition in this post Cox Regression, just differ in notation.

## Conditional Probabilities and Exit Rates

Let us consider the conditional probabilities that individuals still remain in the state for $h$ time units longer, given that the individual stayed already up to time $a$. The conditional probability is given by

\[F(h\mid a)=\frac{F(a+h)}{F(a)}\]The conditional probability to exit exactly between time $a$ and $a+h$, given that the individual was in the state at time $a$ is then

\[\frac{F(a)-F(a+h)}{F(a)}=1-F(h\mid a)\]If $F$ is differential, then we can define the exit rate as

\[\mu(a) =\lim_{h\rightarrow 0}\frac{F(a)-F(a+h)}{hF(a)}=-\frac{F'(a)}{F(a)}\]### Exponential Distribution

Assuming the exit time is exponentially distributed and the survival function is given by

\[F(a)=e^{-\gamma a}\]In that case, we can obtain

\[\mu(a)=\gamma\]and the conditional probability

\[F(h\mid a)=e^{-\gamma h}=F(h)\]### Recovery Process

Denote $I(t)$ as the random variable for the number of infected individuals at time $t$, then

\[I(t+\Delta t)=I(t)-G(\Delta t)I(t)\]By subtracting $I(t)$ and divide by $\Delta t$, we have

\[\frac{I(t+\Delta t)-I(t)}{\Delta t}=-\frac{G(\Delta t)}{\Delta t}I(t)\]Passing to the limit $\Delta t \rightarrow 0$, we arrive at an ODE

\[\dot I(t)=-\mu I(t)\]