Let , and denote , , so that is a renewal process. (1) Prove that . (2) If the distribution function of is , find the exact analytical expression for . (3) Using the Key Renewal Theorem, find .

Stochastic Processes Solution – Problem 8: Prove that

Question

Let $X_{1},\cdots X_{n}\stackrel{iid}{\sim}X$ , and denote $S_{n}=\sum_{i=1}^{n}X_{i}$ , $N(t)=\inf\{n:S_{n}>t\}$ , so that $N(t)$ is a renewal process.

(1) Prove that $\mathbb{P}(X_{N(t)}>x)\geq\mathbb{P}(X>x),\forall x\geq0$ .

(2) If the distribution function of $X$ is $F(x)=1-e^{-x}$ , find the exact analytical expression for $\mathbb{P}(X_{N(t)}>x)$ .

(3) Using the Key Renewal Theorem, find $\lim_{t\to\infty}\mathbb{E}X_{N(t)}$ .

Step-by-step solution

(1) Step 1. Denote $H(t, x) = \mathbb{P}(X_{N(t)} > x)$ . By conditioning on the first renewal epoch $X_1$ , we can establish a renewal equation for $H(t, x)$ . When $X_1 > t$ , $N(t) = 1$ , so $X_{N(t)} = X_1$ , giving $\mathbb{P}(X_{N(t)} > x | X_1 > t) = \mathbb{P}(X_1 > x | X_1 > t)$ . When $X_1 \leq t$ , the process restarts from $t - X_1$ , so $X_{N(t)}$ has the same distribution as $X_{N(t-X_1)}$ . Thus: $H(t, x) = \mathbb{P}(X_1 > \max(t, x)) + \int_0^t H(t-y, x) dF(y).$

Step 2. Denote $\bar{F}(x) = 1 - F(x) = \mathbb{P}(X > x)$ . Let $\Delta(t) = H(t, x) - \bar{F}(x)$ . Substituting $H(t, x) = \Delta(t) + \bar{F}(x)$ into the renewal equation and simplifying: $\Delta(t) = z(t) + \int_0^t \Delta(t-y) dF(y),$ where the source term is $z(t) = \bar{F}(\max(t, x)) - \bar{F}(x) + \bar{F}(x)F(t)$ .

Step 3. Analyze the sign of $z(t)$ . When $t < x$ : $z(t) = \bar{F}(x)F(t) \geq 0$ . When $t \geq x$ : $z(t) = \bar{F}(t)F(x) \geq 0$ . Thus $z(t) \geq 0$ for all $t \geq 0$ .

Step 4. By the renewal equation solution formula, $\Delta(t) = \int_0^t z(t-y) dU(y)$ , where $U(t) = \sum_{n=0}^\infty F^{*n}(t)$ is the renewal function. Since $z(t) \geq 0$ and $dU(y)$ is a nonnegative measure, $\Delta(t) \geq 0$ . Thus $H(t, x) \geq \bar{F}(x)$ , i.e., $\mathbb{P}(X_{N(t)} > x) \geq \mathbb{P}(X > x)$ . QED.

(2) Since $F(x) = 1 - e^{-x}$ , $X$ follows an exponential distribution with parameter $1$ . The renewal process corresponding to the exponential distribution is a Poisson process, with renewal function $U(t) = 1 + t$ (including the unit step at $n=0$ ). From the derivation in (1), $\mathbb{P}(X_{N(t)} > x)$ satisfies a renewal equation whose solution is: $H(t, x) = \bar{F}(\max(t, x)) + \int_0^t \bar{F}(\max(t-y, x)) dy.$ When $x > t$ : $H(t, x) = e^{-x}(1 + t).$ When $x \leq t$ : $H(t, x) = e^{-x}(1 + x).$ In summary, $\mathbb{P}(X_{N(t)} > x) = e^{-x}(1 + \min(x, t))$ .

(3) Denote $g(t) = \mathbb{E} X_{N(t)}$ . Conditioning on $X_1$ , we establish a renewal equation for $g(t)$ : $g(t) = \int_t^\infty y dF(y) + \int_0^t g(t-y) dF(y).$ Let $h(t) = \int_t^\infty y dF(y)$ . This is a standard renewal equation $g(t) = h(t) + \int_0^t g(t-y) dF(y)$ . Verify integrability: $\int_0^\infty h(t) dt = \mathbb{E}[X^2].$ Assuming $\mathbb{E}[X^2] < \infty$ , $h(t)$ is directly Riemann integrable. By the Key Renewal Theorem, $\lim_{t \to \infty} g(t) = \frac{1}{\mu} \int_0^\infty h(t) dt = \frac{\mathbb{E}[X^2]}{\mathbb{E}[X]},$ where $\mu = \mathbb{E}[X]$ .

Final answer

(1) QED.; (2) $\mathbb{P}(X_{N(t)} > x) = e^{-x}(1 + \min(x, t))$ ; (3) $\dfrac{\mathbb{E}[X^2]}{\mathbb{E}[X]}$

Marking scheme

The following is the rubric designed for this problem and its official solution.

1. Checkpoints (Total 7 pts)

Part (1) (3 pts)

Establish the renewal equation for $H(t,x)$ [1 pt]: Correctly write the renewal equation satisfied by $\mathbb{P}(X_{N(t)}>x)$ , namely $H(t, x) = \mathbb{P}(X_1 > \max(t, x)) + \int_0^t H(t-y, x) dF(y)$ . Note: The source term $\mathbb{P}(X_1 > \max(t, x))$ must correctly reflect the relationship between the first renewal epoch $X_1$ and both $t$ and $x$ .
Analyze the sign of the difference function or source term [1 pt]: Construct the difference $\Delta(t) = H(t,x) - \mathbb{P}(X>x)$ and derive its renewal equation, OR directly analyze the iteration properties of the original equation. Must show the key algebraic steps proving the effective source term (such as $z(t)$ in the solution) is nonnegative in both cases $t<x$ and $t \ge x$ .
Derive the inequality conclusion [1 pt]: Based on the nonnegativity of the source term and the sign-preserving property of the renewal function (or renewal equation solution), rigorously conclude $\Delta(t) \ge 0$ or $H(t,x) \ge \mathbb{P}(X>x)$ .

Part (2) (2 pts)

Exponential distribution setup and integral formulation [1 pt]: Identify the renewal process properties corresponding to the exponential distribution (such as $U(t)=1+t$ or $dU(y) = \delta(y)dy + dy$ ), and correctly establish the specific integral expression. If the specific form of $U(t)$ is not substituted and only the general formula is given, no credit.
Compute the piecewise analytical expression [1 pt]: Correctly compute the integrals for both cases $x > t$ and $x \le t$ , and write the final piecewise result (or unified form $e^{-x}(1+\min(x,t))$ ). Both parts must be correct for credit; only one case correct receives no credit.

Part (3) (2 pts)

Mean renewal equation and source term integral [1 pt]: Write the renewal equation for $g(t) = \mathbb{E}X_{N(t)}$ , and verify that the integral of its source term $h(t)=\int_t^\infty y dF(y)$ over $(0,\infty)$ equals $\mathbb{E}[X^2]$ . Key point: must show the computation of $\int_0^\infty h(t)dt$ or explain why it equals the second moment.
Apply the Key Renewal Theorem (KRT) [1 pt]: Cite the Key Renewal Theorem, derive the limit as $\frac{\int h(t)dt}{\mathbb{E}[X]}$ , and give the final result $\frac{\mathbb{E}[X^2]}{\mathbb{E}[X]}$ .

Total (max 7)

2. Zero-credit items

Part (1): Only intuitively describing that "the interval covering $t$ tends to be longer than a typical interval" (length-biased sampling intuition) without providing a rigorous mathematical proof (such as renewal equation or coupling argument), receives 0 pts.
Part (1): Only verifying the cases $t=0$ or $t \to \infty$ without proving for all $t$ , receives 0 pts.
Part (3): Without deriving via the renewal theorem, directly writing the expectation formula for the "length-biased distribution" without derivation, receives 0 pts (the problem explicitly requires "using the Key Renewal Theorem").

3. Deductions

Concept confusion: Mistakenly computing $X_{N(t)}$ (the total lifetime covering $t$ ) as the residual life ( $S_{N(t)} - t$ ) or the current age, leading to incorrect results in Part (2) or (3) (e.g., obtaining the residual life limit $\frac{\mathbb{E}[X^2]}{2\mathbb{E}[X]}$ ): that part receives 0 pts.
Lack of existence verification: In Part (3), if $\mathbb{E}[X^2] < \infty$ or Riemann integrability conditions are not mentioned, no deduction.
Sign error: If the logic is reversed when proving the inequality (e.g., assuming the conclusion first then deriving), deduct 1 pt.
Integration computation error: Pure arithmetic/integration errors in Part (2) or (3) with completely correct approach, deduct 1 pt.

Stochastic Processes – Problem 8: Prove that