Let $N_t$ be a renewal process and $B_t=t-S_{N_t}$ with $S_{N_t}=X_1+...+X_{N_t}$. If $F$ is the distribution function of $X_1$, show that $$\int_0^tE(B_t\mid X_1=x)F(dx)=\int_0^tE(B_{t-x})F(dx)$$

My proof:

\begin{align}\int_0^tE(B_t|X_1=x)F(dx)&=\int_0^t E\left(t-(x+\dots+X_{N_t})\right)F(dx)\\[0.2cm]&=\int_0^tE\left(t-x-(X_2+\dots+X_{N_t})\right)F(dx)\\[0.2cm]&=\int_0^tE\left(t-x-S_{N_{t-x}}\right)F(dx)=\int_0^tE\left(B_{t-x}\right)F(dx)\end{align}

Is it correct?

Thank you

  • $\begingroup$ I think you need to justify the transition from the second line to the third line. Why is $X_2+\dots X_{N_t}=S_{N_{t-x}}$? $\endgroup$ – Jimmy R. Feb 16 '16 at 12:37
  • $\begingroup$ Because I don't take the jump $X_1$ $\endgroup$ – user314891 Feb 16 '16 at 13:54

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