Mutually exclusive exponential distributions

Question

Suppose that, in a memoryless way, an object A can suddenly transform into object B or object C. Once it transforms, it can no longer transform again (so if it becomes B, it cannot become C, and visa versa)

Suppose that the pdf of an object A becoming object B is

$$\lambda e^{-\lambda t}$$

Where $t$ is the time of the transition

And that same object A becoming object C instead has a pdf of

$$\mu e^{-\mu t}$$

We can integrate over the timespan $\tau$ of the experiment to derive

P(A -> B over timespan $\tau$) = $1-e^{-\lambda \tau}$

P(A -> C over timespan $\tau$) = $1-e^{-\mu \tau}$

But the events A -> B over timespan $\tau$ and A -> C over timespan $\tau$ are mutually exclusive, so in theory

P(A transitions to B or C over timespan $\tau$) = P(A -> B over timespan $\tau$) + P(A -> C over timespan $\tau$) = $1-e^{-\lambda \tau} + 1-e^{-\mu \tau} = 2-(e^{-\lambda \tau} + e^{-\mu \tau})$

But this is clearly incorrect, because as the timespan $\tau$ increases without bound, the probability of transition increases to be greater than 1, which is impossible.

Where did I go wrong? At first pass, everything I did seems correct, but it obviously isn't.

Answer 1

The way you have set up the question does not work, and you have demonstrated that it does not work.

So let's create a system that does work involving memorylessness and your two rates of $\lambda$ and $\mu$:

• Suppose you have two exponentially distributed random variables: $X$ with rate $\lambda$ and $Y$ with rate $\mu$. Let $Z=\min(X,Y)$

  • $Z$ is exponentially distributed with rate $\lambda+\mu$, so with pdf $(\lambda+\mu)e^{-(\lambda+\mu) t}$ and cdf $1-e^{-(\lambda+\mu) t}$ for $z \gt 0$

  • $\mathbb P(Z=X) = \mathbb P(X\lt Y ) = \frac{\lambda}{\lambda+\mu}$ and $\mathbb P(Z=Y) = \mathbb P(X\gt Y ) = \frac{\mu}{\lambda+\mu}$

• Now let's say $A$ transforms at time $Z$, and transforms into $B$ if $X \lt Y$ but into $C$ if $X \gt Y$

  • in a sense this has a memoryless property since $Z$ is exponentially distributed, and if $Z \gt s$ then the conditional distribution of $Z-s$ is the same as the original distribution of $Z$, while the conditional probabilities of $A$ transforming into $B$ or into $C$ have not changed from the original probabilities

• In this form

  • The probability $A$ changes into $B$ by time $\tau$ is $\frac{\lambda}{\lambda+\mu}\left(1-e^{-(\lambda+\mu) \tau}\right)$, which has a derivative of $\lambda e^{-(\lambda+\mu) \tau}$
  • The probability $A$ changes into $C$ by time $\tau$ is $\frac{\mu}{\lambda+\mu}\left(1-e^{-(\lambda+\mu) \tau}\right)$, which has a derivative of $\mu e^{-(\lambda+\mu) \tau}$
  • These are not in the usual sense cdfs (individually they do not approach $1$ as $\tau$ increases, though their sum does) and pdfs (individually they do not integrate to $1$ over positive $\tau$, though their sum does)

Answer 2

Altough the underlying processes are indepenedent the events $A \rightarrow B$ and $A \rightarrow C$ are not independent. $A \rightarrow B$ can occur only if $A \rightarrow C$ has not already occurred and vice versa. Thus you are looking for the probabilities of the events: $A \rightarrow B$ occurs before $A \rightarrow C$ and before $\tau:$ $P((A \rightarrow B) \cap (A \rightarrow C)^c) = \int_{0}^{\tau}\lambda e^{-\lambda t} e^{-\mu t} dt$ and

$A \rightarrow C$ occurs before $A \rightarrow B$ and before $\tau:$ $P((A \rightarrow C) \cap (A \rightarrow B)^c) = \int_{0}^{\tau}\mu e^{-\mu t} e^{-\lambda t} dt$.

As $\tau \rightarrow \infty$ one of the events will occur.