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Consider one individual present at time $t = 0$. It is given to us that the waiting time for any one individual to give birth is exponentially distributed with parameter $m$. Initially only $1$ individual is present, hence the waiting time till the first birth is exponentially distributed with parameter $m$.

After the first birth, there are now $2$ individuals present. We need to find the waiting time till the birth of the next (third) individual.

My textbook's solution says that the time distribution will be exponential yet again, but this time with a parameter $2m$ and my professor agrees with that.

However, when I tried to prove it myself, I initially thought it (waiting time for third individual to be born) would be the sum of two exponential variables (which is gamma/Erling distributed), but then I realized it would be the minimum of the two individual waiting times. How do I prove that this is an exponential distribution itself with parameter $2m$ mathematically?

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  • Minimum of exponential variables with parameters $\lambda_1, ..., \lambda_n$ is exponential distribution with parameter $\sum_i \lambda_i$. In this problem, distribution for $n+1$th birth is minimum for previous $n$, therefore it is exponential distribution with $nm$. – Snowball Jun 16 '21 at 14:39
  • An easy way is to find the probability that neither of the two give birth by time $t$ - the square of the probability one does not - subtract that from $1$ and then differentiate – Henry Jun 16 '21 at 14:39

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