= Solution
Convolve three rate-$\lambda$ <exponential distributions>. The first two have convolution density $\lambda^2xe^{-\lambda x}\mathbf1_{x>0}$; convolving once more gives
$$
\int_0^x\lambda^2s e^{-\lambda s}\lambda e^{-\lambda(x-s)}ds=\frac12\lambda^3x^2e^{-\lambda x}.
$$
Thus $X$ is an <Erlang distribution> of shape three, equal in law to the sum of three independent exponentials. Addition of independent means and variances yields
$$
\boxed{\mathbb EX=3/\lambda,\quad\operatorname{Var}X=3/\lambda^2,\quad M_X(t)=\left(\frac\lambda{\lambda-t}\right)^3\ (t<\lambda).}
$$
For $t\ge\lambda$ the <moment-generating function> is infinite.
Realize each <holding time> as a consecutive block of three exponential holding times from one <Poisson process> $N_t$. The renewal count is then $X_t=\lfloor N_t/3\rfloor$ pathwise under this coupling. If $R_t$ is the remainder of $N_t$ modulo three, $N_t=3X_t+R_t$ and $\mathbb ER_t=p_1(t)+2p_2(t)$. Taking expectations gives
$$
\boxed{m(t)=\frac{\lambda t}{3}-\frac13p_1(t)-\frac23p_2(t).}
$$
This proves the formula without assuming that the renewal count itself is Poisson.
Back to article page
