Exponential decay, show that the probability that a particle decays in an interval is proportional to the length of the interval

Question

Let $T$ be the time until a radioactive particle decays, and suppose that $T$~$Expo(\lambda)$.

(a) Find the half-life of the particle.

Median (half-life) occurs when CDF = $\frac{1}{2}, F_T(m) = \frac{1}{2}$.

For any positive number $t$, we have

$F_T(t) = P(T\leq t) = \int_{0}^{t} \lambda e^{-\lambda x} dx = 1-e^{-\lambda x}$

To find the median, we set the above equation equal to $\frac{1}{2}$ and solve for m.

$\frac{1}{2} = e^{-\lambda m}$ $\rightarrow$ $ln(\frac{1}{2}) = -\lambda m$ $\rightarrow$ $m = \frac{ln(2)}{\lambda}$

(b) Show that for $\epsilon$, a small positive constant, the probability that the particle decays in the time interval $[t, t + \epsilon]$, given that it has survived until time t, does not depend on $t$ and is approx. proportional to $\epsilon$.

This is where I am stuck. My first inclination is to find $P(t<T<t+\epsilon)$ but I am not even sure how to compute this. Any and all help is appreciated.

Answer 1

Let $A$ be the event it dies in the interval $[t,t+\epsilon]$ and let $B$ be the event it has survived until time $t$. We want $\Pr(A\mid B)$, which is $\frac{\Pr(A\cap B)}{\Pr(B)}$.

Now we have two probabilities to compute. Recall that for positive $t$ we have $\Pr(T\le t)=1-e^{-\lambda t}$. Thus $\Pr(B)=e^{-\lambda t}$.

To find $\Pr(A\cap B)$, note that this is the probability the lifetime is $\le t+\epsilon$ minus the probability it is less than $t$. Compute. We get $$(1-e^{-\lambda(t+\epsilon)})-(1-e^{-\lambda t}).$$ This simplifies to $e^{-\lambda t}-e^{-\lambda(t+\epsilon)}$.

For the conditional probability, divide by $e^{-\lambda t}$. We get $$\Pr(A\mid B)=1-e^{-\lambda \epsilon}.\tag{1}$$

For the memorylessness, note that the right side of (1) does not involve $t$.

For the other part, note that by the Maclaurin expansion of $e^x$, we have that $e^{-\lambda \epsilon }\approx 1-\lambda \epsilon$ when $\epsilon$ is close to $0$. Thus from (1) we have $\Pr(A\mid B)\approx \lambda \epsilon$.