Let and be Poisson processes with parameters and , respectively, and suppose that and are mutually independent. Define . Compute , , and , . (Optional: Define . Compute .)

Probability Theory Solution – Problem 1: Compute , , and ,

Question

Let $\{X(t),t\geqslant0\}$ and $\{Y(t),t\geqslant0\}$ be Poisson processes with parameters $\lambda$ and $\mu$ , respectively, and suppose that $\{X(t),t\geqslant0\}$ and $\{Y(t),t\geqslant0\}$ are mutually independent. Define $T_{1}\,=\,\operatorname*{min}\{t\geqslant0:Y(t)=1\}$ . Compute $P\left(X(T_{1})=k\right)$ , $k\,=$ $0,1,2\ldots$ , and $P\left(X(\frac{T_{1}}{2})=k\right)$ , $k=0,1,2\ldots$ .

(Optional: Define $T_{a}=\operatorname*{min}\{t\geqslant0:Y(t)=a\},a\in\mathbb{Z}^{*}$ . Compute $P\left(X(T_{a})=k\right)$ .)

Step-by-step solution

(1) Compute $P(X(T_{1})=k)$ ( $k=0,1,2\ldots$ ): $T_1$ is the time of the first event of the Poisson process $\{Y(t)\}$ . The first arrival time of a Poisson process follows an exponential distribution with parameter $\mu$ , i.e., the probability density function is: $f_{T_1}(t) = \mu e^{-\mu t},\quad t\geqslant0$ The cumulative distribution function is: $F_{T_1}(t) = 1 - e^{-\mu t},\quad t\geqslant0$ Since $\{X(t)\}$ and $\{Y(t)\}$ are mutually independent, given $T_1=t$ , $X(T_1)$ follows a Poisson distribution with parameter $\lambda t$ , namely: $P(X(T_1)=k \mid T_1=t) = \frac{(\lambda t)^k e^{-\lambda t}}{k!},\quad k=0,1,2\ldots$ By the law of total probability, integrating over the continuous random variable $T_1$ : $P(X(T_1)=k) = \int_{0}^{+\infty} P(X(T_1)=k \mid T_1=t) f_{T_1}(t) dt$ Substituting the distributions into the integral: $P(X(T_1)=k) = \int_{0}^{+\infty} \frac{(\lambda t)^k e^{-\lambda t}}{k!} \cdot \mu e^{-\mu t} dt = \frac{\lambda^k \mu}{k!} \int_{0}^{+\infty} t^k e^{-(\lambda + \mu) t} dt$ Using the Gamma function integral formula $\int_{0}^{+\infty} t^k e^{-s t} dt = \frac{k!}{s^{k+1}}$ ( $s>0$ ), with $s=\lambda + \mu$ , we obtain: $P(X(T_1)=k) = \frac{\lambda^k \mu}{k!} \cdot \frac{k!}{(\lambda + \mu)^{k+1}} = \left( \frac{\lambda}{\lambda + \mu} \right)^k \cdot \frac{\mu}{\lambda + \mu}$ (2) Compute $P(X(\frac{T_1}{2})=k)$ ( $k=0,1,2\ldots$ ): Since $T_1 \sim \text{Exp}(\mu)$ , the probability density function of $Z$ is: $f_Z(z) = f_{T_1}(2z) \cdot 2 = 2\mu e^{-2\mu z},\quad z\geqslant0$ Given $Z=z$ , $X(Z)=X(2z)$ follows a Poisson distribution with parameter $\lambda \cdot 2z$ : $P(X(Z)=k \mid Z=z) = \frac{(2\lambda z)^k e^{-2\lambda z}}{k!}$ Applying the law of total probability and integrating: $P(X(\frac{T_1}{2})=k) = \int_{0}^{+\infty} \frac{(2\lambda z)^k e^{-2\lambda z}}{k!} \cdot 2\mu e^{-2\mu z} dz = \frac{(2\lambda)^k \cdot 2\mu}{k!} \int_{0}^{+\infty} z^k e^{-2(\lambda + \mu) z} dz$ Substituting the Gamma function formula ( $s=2(\lambda + \mu)$ ): $\int_{0}^{+\infty} z^k e^{-2(\lambda + \mu) z} dz = \frac{k!}{[2(\lambda + \mu)]^{k+1}}$ Simplifying: $P(X(\frac{T_1}{2})=k) = \frac{(2\lambda)^k \cdot 2\mu}{k!} \cdot \frac{k!}{[2(\lambda + \mu)]^{k+1}} = \left( \frac{\lambda}{\lambda + \mu} \right)^k \cdot \frac{\mu}{\lambda + \mu}$

Final answer

$P(X(T_{1})=k) ={\left( \frac{\lambda}{\lambda + \mu} \right)^k \cdot \frac{\mu}{\lambda + \mu}}$ for $k=0,1,2\ldots$ $P(X(\frac{T_{1}}{2})=k) = {\left( \frac{\lambda}{\lambda + \mu} \right)^k \cdot \frac{\mu}{\lambda + \mu}}$ for $k=0,1,2\ldots$

Marking scheme

The following rubric is tailored to the official solution provided.

1. Checkpoints (max 7 pts total)

Score exactly one chain for Part 1 (A or B) | Part 2 is additive to Part 1.

Part 1: Compute $P(X(T_{1})=k)$ (4 pts)

Chain A: Integration / Law of Total Probability (Official Approach)
Distribution of $T_1$ (1 pt): Correctly identify that $T_1$ follows an exponential distribution with parameter $\mu$ and write down the density $f_{T_1}(t) = \mu e^{-\mu t}$ or the distribution function.
Setting up the total probability integral (1 pt): Establish the correct integral expression $P(X(T_1)=k) = \int_{0}^{+\infty} \frac{(\lambda t)^k e^{-\lambda t}}{k!} \cdot \mu e^{-\mu t} dt$ .
Evaluating the integral (1 pt): Use the Gamma function $\Gamma(k+1)=k!$ or integration by parts to correctly evaluate the core integral $\int t^k e^{-(\lambda+\mu)t} dt$ .
Final result (1 pt): Obtain $\left( \frac{\lambda}{\lambda + \mu} \right)^k \frac{\mu}{\lambda + \mu}$ .
Chain B: Competing Events / Probabilistic Argument (Alternative)
Competing event probability (2 pts): Identify that in the merged Poisson process, the probability that the next event belongs to $Y$ (rather than $X$ ) is $\frac{\mu}{\lambda+\mu}$ , or equivalently that the probability for $X$ is $\frac{\lambda}{\lambda+\mu}$ .
Sequential analysis (1 pt): Argue that the event sequence must consist of $k$ events from $X$ followed immediately by one event from $Y$ (using independence and the memoryless property).
Final result (1 pt): Obtain $\left( \frac{\lambda}{\lambda + \mu} \right)^k \frac{\mu}{\lambda + \mu}$ .

Part 2: Compute $P(X(\frac{T_{1}}{2})=k)$ (3 pts)

Change of variable and distribution (1 pt): Let $Z = T_1/2$ (or derive directly) and obtain the correct density $f_Z(z) = 2\mu e^{-2\mu z}$ .
Setting up the integral / probability expression (1 pt): Following the logic of the official solution, write down the integral involving $Z$ .
Note: Credit is awarded as long as the total probability integral is set up correctly (combining the Poisson kernel with the exponential density), regardless of whether the Poisson parameter is taken as $\lambda z$ or $2\lambda z$ as in the official solution.
Final result (1 pt): Compute and obtain the result consistent with the official solution: $\left( \frac{\lambda}{\lambda + \mu} \right)^k \frac{\mu}{\lambda + \mu}$ .
Note: Per the official solution, this result should be identical to that of Part 1.

Total (max 7)

2. Zero-credit items

Merely copying the definitions of $X(t)$ and $Y(t)$ from the problem statement or writing down the Poisson distribution formula without performing any substitution or computation involving $T_1$ .
Writing down the formula for $T_a$ without completing the computation for $T_1$ or $T_1/2$ (the optional part does not earn credit unless it substitutes for the main parts with consistent logic).
Stating the answer by intuition alone without any derivation (e.g., writing down the geometric distribution formula without justification).

3. Deductions

Integral parameter error (-1): Incorrect manipulation of the Gamma function coefficients when evaluating $\int t^k e^{-s t} dt$ (e.g., missing the factorial or an error in the exponent).
Logical gap (-1): In Part 2, directly asserting that "the probability is unchanged" without showing the change of variable or the integration (unless a rigorous scaling argument is provided).
Notational confusion (-1): Confusing random variables with deterministic values (e.g., writing the constant $T$ inside the integral instead of the integration variable $t$ ).
Missing range (no deduction): Omitting conditions such as $k=0,1,2\ldots$ or $t \ge 0$ does not incur a penalty as long as the core computation is correct.

Probability Theory – Problem 1: Compute , , and ,