Exam MAS-I — Poisson Processes Practice Flashcards

**Answer: (E).** Properties (A)-(D) are exactly the standard axioms and the small-interval characterization of a Poisson process. The interarrival times are i.i.d. $\text{Exponential}(\lambda)$, **not** uniform: $f(t)=\lambda e^{-\lambda t}$ with mean $\frac{1}{\lambda}$. A uniform interarrival would violate the memoryless property that underlies the whole construction, so (E) is the false statement. Distractor note: (E) is tempting because the mean interarrival time is indeed $\frac{1}{\lambda}$, but matching a mean does not make the distribution uniform.

**Answer: (B).** By **stationary increments**, $N(7)-N(4)$ depends only on the interval length $7-4=3$, so it is $\text{Poisson}(\lambda\cdot 3)=\text{Poisson}(3\lambda)$. By **independent increments**, the count over the disjoint interval $(4,7]$ is independent of $N(4)$, the count over $(0,4]$. Distractors: (A) misuses the right endpoint $7$ for the mean; (C) adds the two interval lengths; (D) drops independence; (E) replaces the exact Poisson with a normal — the normal is only an approximation for large means.

**Answer: (A).** The interarrival time is $\text{Exponential}(\lambda)$, which has the **memoryless** property: $\Pr(T>s+t\mid T>s)=\Pr(T>t)=e^{-\lambda t}$. Conditioning on no event by time $s$ leaves the residual wait again $\text{Exponential}(\lambda)$ with mean $\tfrac{1}{\lambda}$, free of $s$. Distractor (B) wrongly adds the elapsed time to the mean; (E) reflects the gambler's-fallacy idea that an event becomes "due," which memorylessness explicitly contradicts.

**Answer: (B).** With $t=1$ day, $N(1)\sim\text{Poisson}(\lambda t)=\text{Poisson}(3)$. $\Pr(N(1)=2)=e^{-3}\dfrac{3^{2}}{2!}=e^{-3}\dfrac{9}{2}=4.5\,e^{-3}$. Since $e^{-3}\approx 0.049787$, $\Pr(N(1)=2)\approx 4.5(0.049787)\approx 0.2240$. Distractor (A) is $\Pr(N=1)=3e^{-3}$; (E) drops the exponential and reports the coefficient $4.5\,e^{-3}\to 4.5$ mis-scaled, or $9/2$ read as a probability.

**Answer: (D).** Take $t=0.5$ hour, so $\lambda t = 4(0.5)=2$ and $N(0.5)\sim\text{Poisson}(2)$. $\Pr(N(0.5)\ge 1)=1-\Pr(N(0.5)=0)=1-e^{-2}\dfrac{2^{0}}{0!}=1-e^{-2}$. Since $e^{-2}\approx 0.135335$, the probability $\approx 1-0.135335=0.8647$. Distractor (B) is $\Pr(N=0)=e^{-2}$ (the complement); (E) uses $\lambda t=4$ (forgetting to halve the hour) in $1-e^{-4}$.

**Answer: (C).** Over $t=3$ weeks the count is $N(3)\sim\text{Poisson}(\lambda t)=\text{Poisson}(6)$. $\Pr(N(3)=5)=e^{-6}\dfrac{6^{5}}{5!}=e^{-6}\dfrac{7776}{120}=64.8\,e^{-6}$. Since $e^{-6}\approx 0.00247875$, $\Pr(N(3)=5)\approx 64.8(0.00247875)\approx 0.1606$. Distractor (A) uses $\text{Poisson}(2)$ for one week instead of $\text{Poisson}(6)$ for three weeks — a common rate-scaling error.

**Answer: (B).** The two hours are disjoint, so by **independent increments** the counts are independent, each $\sim\text{Poisson}(1.5)$. $\Pr(N_1=3)=e^{-1.5}\dfrac{1.5^{3}}{3!}=e^{-1.5}\dfrac{3.375}{6}=0.5625\,e^{-1.5}\approx 0.12551$. $\Pr(N_2=1)=e^{-1.5}(1.5)\approx 0.33470$. Product $\approx 0.12551 \times 0.33470 \approx 0.0420$. Distractor (E) adds the two probabilities instead of multiplying; (C) and (D) are the individual factors.

**Answer: (B).** By stationary and independent increments, the count over $(2,6]$ depends only on the length $6-2=4$, so $N(6)-N(2)\sim\text{Poisson}(\lambda\cdot 4)=\text{Poisson}(0.5\times 4)=\text{Poisson}(2)$. $\Pr=e^{-2}\dfrac{2^{3}}{3!}=e^{-2}\dfrac{8}{6}\approx 1.3333(0.135335)\approx 0.1804$. Distractor (D) uses $\text{Poisson}(0.5\times 6=3)$ by taking the right endpoint as the length; the correct length is the $4$-hour gap, not $6$.