5.3.1 problem 37

a. We need to find the value of $t$ such that $F\left(t\right)=1∕2$ - this will indicate that there is a 1/2 chance that the particle has decayed before time t.

$1-e^{-\mathrm{\lambda t}}=1/2$ implies $\ln \left(2\right)=\mathrm{\lambda t}$, so $t=\ln \left(2\right)∕\lambda$

b. We need to compute $P(t<T<t+ϵ|T>t)=P(t<T<t+ϵ)∕P(T>t)$. This is $\frac{(1-e^{-\lambda (t+ϵ)})-(1-e^{-\mathrm{\lambda t}})}{e^{-\mathrm{\lambda t}}}=1-e^{-\mathrm{\lambda ϵ}}$. Using the approximation given in the hint and the assumption that $ϵ$ is small enough that $\mathrm{ϵ\lambda }\approx 0$, this is about $1-(1-\mathrm{ϵ\lambda })=\mathrm{ϵ\lambda }$.

c. $P(L>t)=P(T_1>t)P(T_2>t)...P(T_n>t)=e^{-\mathrm{n\lambda t}}$, so $L\sim \mathrm{Expo}\left(\mathrm{n\lambda }\right)$. Therefore, if $X\sim \mathrm{Expo}\left(1\right)$, we have $L=X∕\mathrm{n\lambda }$. Then since $E\left(X\right)=1$ and $V\mathrm{ar}\left(X\right)=1$, we can get $E\left(L\right)=1∕\left(\mathrm{n\lambda }\right)$ and $V\mathrm{ar}\left(L\right)=1/\left(n^2{\lambda }^2\right)$

d. M must be equal to the sum of $D_1+D_2+D_3+...+D_n$, where $D_i$ is the amount of time between the $i-1$th and $i$th decay event. We observe that $D_i$ must then be the minimum of $n-i+1$ $\mathrm{Expo}\left(\lambda \right)$ variables - for example, $D_1$ is the first particle to decay out of n particles, $D_2$ is the first particle to decay out of the remaining n-1 particles, etc. Since $\mathrm{Expo}$ is memoryless, $D_{i+1}$ is independent of $D_i$ as the amount of time it takes for the next particle to decay is not affected by the amount of time it took the previous particle to decay. Therefore, $D_i\sim \mathrm{Expo}\left(\right(n-i+1\left)\lambda \right)$.

Then $E\left(M\right)=E\left(D_1\right)+E\left(D_2\right)+...+E\left(D_n\right)=\frac{1}{\lambda }\left({\sum }_{i=1}^n\right(1/n\left)\right)=\frac{1}{\lambda }\left(\ln \right(n)+0.577)$