Convolution Integrals — Question 9

Question 9 – Solution

Strategy. Work with reciprocal rates. Convolution moment identities reduce identification to a quadratic equation.

Step 1: Prove the moment identities. In the convolution integral use $t=u+w$, $u,w\ge 0$. Nonnegativity permits interchange, and finite second moments make the following values finite: $$\begin{aligned} \int(p*q)&=(\int p)(\int q)=1,\\ \mu_{p*q}&=\iint(u+w)p(u)q(w)\,du\,dw=\mu_p+\mu_q,\\ \int t^2(p*q)(t)\,dt&=\int u^2p(u)\,du+ 2\mu_p\mu_q+\int w^2q(w)\,dw. \end{aligned}$$ Subtracting $(\mu_p+\mu_q)^2$ from the last line proves $\boxed{v_{p*q}=v_p+v_q}$.

Step 2: Calculate the stage statistics. Integration by parts gives mass $1$, first moment $1/\alpha$ and second moment $2/\alpha^2$ for $k_\alpha$. Its variance is therefore $1/\alpha^2$. A delay replaces time by $t=d+u$: mass stays $1$, the mean increases by $d$, and the centered second moment is unchanged. Hence $$\mu=d+\frac 1\alpha+\frac 1\beta,\qquad v=\frac 1{\alpha^2}+\frac 1{\beta^2}.$$

Step 3: Recover the parameters. Both stage kernels are positive for positive times, so their convolution is positive at every positive time and zero at zero. The observed onset thus forces $d=1$. Put $x=1/\alpha$, $z=1/\beta$. Then $$x+z=\tfrac 32,\quad x^2+z^2=\tfrac 54,\quad xz=\tfrac 12,\quad (X-x)(X-z)=X^2-\tfrac 32X+\tfrac 12.$$ Its roots are $1$ and $1/2$, giving $\boxed{\{\alpha,\beta\}=\{1,2\}}$. The delay and unordered rate pair are unique. Stage order cannot be identified even from the entire output because convolution is commutative.

Step 4: Give the full feasibility condition. Write $m=\mu-d$. Positive $x,z$ require and are possible exactly when $$\boxed{m>0,\qquad \frac{m^2}{2}\le v<m^2.}$$ Indeed $xz=(m^2-v)/2>0$ and $(x-z)^2=2v-m^2\ge 0$; conversely these conditions make $(m\pm\sqrt{2v-m^2})/2$ real and strictly positive. At the lower equality the rates coincide. At the upper boundary one reciprocal rate would be zero, requiring an infinite rate, excluded here.