{
  "deckName": "Exam MAS-I — Poisson Processes",
  "examCode": "Exam MAS-I",
  "cards": [
    {
      "front": "State the three defining properties (axioms) of a **Poisson process** $\\{N(t),t\\ge0\\}$ with rate $\\lambda$.",
      "back": "1. **$N(0)=0$** — the count starts at zero.\n2. **Independent increments** — counts over disjoint time intervals are independent.\n3. **Stationary increments** — the distribution of $N(t+s)-N(s)$ depends only on the length $t$, not on $s$, and equals $\\text{Poisson}(\\lambda t)$.\nEquivalently, in a small interval of length $h$ the probability of one event is $\\lambda h + o(h)$, of two or more is $o(h)$, and these are independent across disjoint intervals.",
      "tag": "Process axioms"
    },
    {
      "front": "What do **stationary increments** and **independent increments** mean for a Poisson process, and why does \"memorylessness\" follow?",
      "back": "**Stationary:** the count in any interval depends only on its length — $N(t+s)-N(s)\\sim\\text{Poisson}(\\lambda t)$ for every $s$.\n**Independent:** counts in non-overlapping intervals are independent random variables.\nTogether they imply the process \"forgets\" its past: the number of future arrivals after time $s$ is independent of the history up to $s$ and behaves like a fresh Poisson process started at $0$. This is the process-level analogue of the exponential interarrival's **memoryless** property.",
      "tag": "Process axioms"
    },
    {
      "front": "Give the small-interval (infinitesimal) characterization of a Poisson process with rate $\\lambda$.",
      "back": "For a small width $h$:\n$\\Pr(N(t+h)-N(t)=1)=\\lambda h + o(h)$,\n$\\Pr(N(t+h)-N(t)\\ge 2)=o(h)$,\n$\\Pr(N(t+h)-N(t)=0)=1-\\lambda h + o(h)$,\nwith increments over disjoint intervals independent. Here $o(h)$ denotes a term with $\\lim_{h\\to0}o(h)/h=0$. This local description is equivalent to the global axioms and is the basis of the Poisson differential equations.",
      "tag": "Process axioms"
    },
    {
      "front": "State the distribution of $N(t)$ for a Poisson process with rate $\\lambda$, with its mean and variance.",
      "back": "$N(t)\\sim\\text{Poisson}(\\lambda t)$, so $\\Pr(N(t)=k)=e^{-\\lambda t}\\frac{(\\lambda t)^{k}}{k!}$ for $k=0,1,2,\\dots$.\nMean $E[N(t)]=\\lambda t$ and variance $\\text{Var}(N(t))=\\lambda t$ — for any Poisson variable the mean equals the variance. The rate $\\lambda$ is the expected number of events per unit time.",
      "tag": "Counting & increments"
    },
    {
      "front": "For a Poisson process, what is the distribution of an increment $N(t+s)-N(s)$, and why does the starting point $s$ not matter?",
      "back": "$N(t+s)-N(s)\\sim\\text{Poisson}(\\lambda t)$ — it depends only on the interval length $t$.\nThe starting point $s$ drops out by **stationary increments**, and by **independent increments** this increment is independent of $N(s)$ (the count before time $s$). So you may always shift any interval to start at $0$ when computing increment probabilities.",
      "tag": "Counting & increments"
    },
    {
      "front": "Claims arrive as a Poisson process at rate $\\lambda=3$ per day. Find the probability of exactly $2$ claims in a given day.",
      "back": "With $t=1$ day, $N(1)\\sim\\text{Poisson}(\\lambda t)=\\text{Poisson}(3)$.\n$\\Pr(N(1)=2)=e^{-3}\\frac{3^{2}}{2!}=e^{-3}\\frac{9}{2}=4.5\\,e^{-3}$.\nSince $e^{-3}\\approx 0.049787$, $\\Pr(N(1)=2)\\approx 4.5(0.049787)\\approx 0.2240$.",
      "tag": "Counting & increments"
    },
    {
      "front": "Calls arrive as a Poisson process at rate $\\lambda=4$ per hour. Find the probability of **at least one** call in the next $30$ minutes.",
      "back": "Take $t=0.5$ hour, so $\\lambda t = 4(0.5)=2$ and $N(0.5)\\sim\\text{Poisson}(2)$.\n$\\Pr(N(0.5)\\ge 1)=1-\\Pr(N(0.5)=0)=1-e^{-2}\\frac{2^{0}}{0!}=1-e^{-2}$.\nSince $e^{-2}\\approx 0.135335$, the probability $\\approx 1-0.135335 = 0.8647$.",
      "tag": "Counting & increments"
    },
    {
      "front": "Accidents follow a Poisson process at rate $\\lambda=2$ per week. Find the probability of exactly $5$ accidents over a $3$-week span.",
      "back": "Over $t=3$ weeks the count is $N(3)\\sim\\text{Poisson}(\\lambda t)=\\text{Poisson}(6)$.\n$\\Pr(N(3)=5)=e^{-6}\\frac{6^{5}}{5!}=e^{-6}\\frac{7776}{120}=64.8\\,e^{-6}$.\nSince $e^{-6}\\approx 0.00247875$, $\\Pr(N(3)=5)\\approx 64.8(0.00247875)\\approx 0.1606$.",
      "tag": "Counting & increments"
    },
    {
      "front": "Events occur as a Poisson process at rate $\\lambda=1.5$ per hour. Find the probability of exactly $3$ events in the first hour **and** exactly $1$ in the second hour.",
      "back": "The two hours are disjoint intervals, so by **independent increments** the counts are independent, each $\\sim\\text{Poisson}(1.5)$.\n$\\Pr(N_1=3)=e^{-1.5}\\frac{1.5^{3}}{3!}=e^{-1.5}\\frac{3.375}{6}=0.5625\\,e^{-1.5}\\approx 0.5625(0.223130)\\approx 0.12551$.\n$\\Pr(N_2=1)=e^{-1.5}(1.5)\\approx 1.5(0.223130)\\approx 0.33470$.\nProduct $\\approx 0.12551 \\times 0.33470 \\approx 0.0420$.",
      "tag": "Counting & increments"
    },
    {
      "front": "Why is $\\Pr(N(5)=8 \\mid N(3)=3)$ computed using only the increment over $(3,5]$?",
      "back": "By independent increments, $N(5)-N(3)$ is independent of $N(3)$, and we need $N(5)-N(3)=8-3=5$ events in the interval $(3,5]$ of length $2$. By stationary increments that increment is $\\text{Poisson}(\\lambda\\cdot 2)$.\nFor $\\lambda=2$: $N(5)-N(3)\\sim\\text{Poisson}(4)$, so $\\Pr=e^{-4}\\frac{4^{5}}{5!}=e^{-4}\\frac{1024}{120}\\approx 8.5333(0.018316)\\approx 0.1563$.",
      "tag": "Counting & increments"
    },
    {
      "front": "State the distribution of the **interarrival times** in a Poisson process with rate $\\lambda$.",
      "back": "The interarrival times $T_1,T_2,\\dots$ (gaps between consecutive events) are i.i.d. $\\text{Exponential}(\\lambda)$ with density $f(t)=\\lambda e^{-\\lambda t}$ for $t\\ge 0$.\nEach has mean $E[T_i]=\\frac{1}{\\lambda}$ and variance $\\frac{1}{\\lambda^{2}}$. The exponential interarrival is equivalent to the Poisson-counting description — either fully characterizes the process.",
      "tag": "Interarrival & waiting times"
    },
    {
      "front": "State the **memoryless** property of the exponential interarrival time and what it means for waiting.",
      "back": "If $T\\sim\\text{Exponential}(\\lambda)$, then $\\Pr(T>s+t \\mid T>s)=\\Pr(T>t)=e^{-\\lambda t}$ for all $s,t\\ge0$.\nHaving already waited $s$ with no event gives no information about the remaining wait — the residual time is again $\\text{Exponential}(\\lambda)$. The exponential is the only continuous distribution with this property; it is why the Poisson process \"restarts\" at every instant.",
      "tag": "Interarrival & waiting times"
    },
    {
      "front": "State the distribution of the **waiting time** $S_n$ to the $n$-th arrival in a Poisson process with rate $\\lambda$.",
      "back": "$S_n=T_1+\\cdots+T_n$ is the sum of $n$ i.i.d. $\\text{Exponential}(\\lambda)$ gaps, so $S_n\\sim\\text{Gamma}(n,\\lambda)$ (an Erlang) with density $f_{S_n}(t)=\\frac{\\lambda^{n}t^{n-1}e^{-\\lambda t}}{(n-1)!}$ for $t\\ge0$.\nMean $E[S_n]=\\frac{n}{\\lambda}$ and variance $\\text{Var}(S_n)=\\frac{n}{\\lambda^{2}}$.",
      "tag": "Interarrival & waiting times"
    },
    {
      "front": "Explain the duality $\\Pr(S_n>t)=\\Pr(N(t)<n)$ linking waiting times and counts.",
      "back": "The $n$-th arrival occurs after time $t$ **iff** fewer than $n$ events have happened by time $t$: $\\{S_n>t\\}=\\{N(t)<n\\}=\\{N(t)\\le n-1\\}$.\nHence $\\Pr(S_n>t)=\\sum_{k=0}^{n-1}e^{-\\lambda t}\\frac{(\\lambda t)^{k}}{k!}$, which is exactly the survival function of a $\\text{Gamma}(n,\\lambda)$. This identity converts gamma-tail questions into Poisson sums.",
      "tag": "Interarrival & waiting times"
    },
    {
      "front": "Buses arrive as a Poisson process at rate $\\lambda=6$ per hour. Find the expected wait until the **next** bus and the probability you wait more than $20$ minutes.",
      "back": "The wait to the next event is $\\text{Exponential}(\\lambda)$ with $\\lambda=6$ per hour.\nExpected wait $=\\frac{1}{\\lambda}=\\frac{1}{6}$ hour $=10$ minutes.\nFor more than $20$ minutes $=\\frac{1}{3}$ hour: $\\Pr(T>\\tfrac{1}{3})=e^{-\\lambda t}=e^{-6/3}=e^{-2}\\approx 0.1353$.",
      "tag": "Interarrival & waiting times"
    },
    {
      "front": "Claims arrive as a Poisson process at rate $\\lambda=2$ per day. Find the expected time until the **4th** claim and its standard deviation.",
      "back": "The $4$th-arrival time is $S_4\\sim\\text{Gamma}(4,2)$.\n$E[S_4]=\\frac{n}{\\lambda}=\\frac{4}{2}=2$ days.\n$\\text{Var}(S_4)=\\frac{n}{\\lambda^{2}}=\\frac{4}{4}=1$, so the standard deviation is $\\sqrt{1}=1$ day.",
      "tag": "Interarrival & waiting times"
    },
    {
      "front": "Events occur as a Poisson process at rate $\\lambda=3$ per hour. Find the probability that the **3rd** event occurs after time $t=1$ hour.",
      "back": "Use $\\Pr(S_3>1)=\\Pr(N(1)<3)=\\Pr(N(1)\\le 2)$ with $N(1)\\sim\\text{Poisson}(3)$.\n$\\Pr(N(1)=0)=e^{-3}=0.049787$,\n$\\Pr(N(1)=1)=3e^{-3}=0.149361$,\n$\\Pr(N(1)=2)=4.5e^{-3}=0.224042$.\nSum $\\approx 0.049787+0.149361+0.224042=0.4232$.",
      "tag": "Interarrival & waiting times"
    },
    {
      "front": "Events occur as a Poisson process at rate $\\lambda=1$ per minute. Given that no event occurred in the first $2$ minutes, find the expected additional wait until the first event.",
      "back": "By the **memoryless** property, conditioning on no event in the first $2$ minutes leaves the residual waiting time again $\\text{Exponential}(\\lambda=1)$.\nExpected additional wait $=\\frac{1}{\\lambda}=1$ minute — the same as the unconditional mean. The two minutes already elapsed are irrelevant.",
      "tag": "Interarrival & waiting times"
    },
    {
      "front": "State the **conditional distribution of arrival times** given $N(t)=n$ in a Poisson process.",
      "back": "Given $N(t)=n$, the $n$ unordered arrival times are distributed as $n$ independent $\\text{Uniform}(0,t)$ random variables; the ordered arrival times $S_1<S_2<\\cdots<S_n$ are distributed as the **order statistics** of those $n$ uniforms.\nThe joint density of the ordered times is $\\frac{n!}{t^{n}}$ on $0<s_1<\\cdots<s_n<t$. Crucially this is **free of $\\lambda$** — the rate affects how many events occur, not where they land given the count.",
      "tag": "Conditioning on arrivals"
    },
    {
      "front": "Given $N(t)=n$, what is the distribution of the number of those $n$ arrivals that fall in a subinterval $(0,s]$ with $s<t$?",
      "back": "Since each of the $n$ arrivals is independently $\\text{Uniform}(0,t)$, each lands in $(0,s]$ with probability $p=\\frac{s}{t}$ independently. So the count in $(0,s]$ given $N(t)=n$ is $\\text{Binomial}\\!\\left(n,\\frac{s}{t}\\right)$.\nIts conditional mean is $n\\cdot\\frac{s}{t}$. This is the standard way conditioning-on-the-total problems are solved.",
      "tag": "Conditioning on arrivals"
    },
    {
      "front": "Given exactly $1$ event of a Poisson process occurred in $(0,t]$, what is the distribution of its arrival time, and what is its expected value?",
      "back": "With $n=1$, the single arrival time is $\\text{Uniform}(0,t)$ — equally likely anywhere in the interval.\nIts expected value is $\\frac{t}{2}$ (the midpoint), independent of $\\lambda$. For example, if exactly one claim occurred in a $10$-hour shift, its expected time of occurrence is $5$ hours in.",
      "tag": "Conditioning on arrivals"
    },
    {
      "front": "A Poisson process has rate $\\lambda=4$ per hour. Given that exactly $3$ events occurred in $(0,2]$ hours, find the probability that exactly $2$ of them occurred in the first hour.",
      "back": "Given $N(2)=3$, each event is $\\text{Uniform}(0,2)$, landing in $(0,1]$ with $p=\\frac{1}{2}$. The count in the first hour is $\\text{Binomial}(3,\\tfrac{1}{2})$.\n$\\Pr(2\\text{ in first hour})=\\binom{3}{2}\\left(\\tfrac{1}{2}\\right)^{2}\\left(\\tfrac{1}{2}\\right)^{1}=3\\cdot\\frac{1}{8}=\\frac{3}{8}=0.375$.\nNote the rate $\\lambda=4$ never enters — only the count and the interval split matter.",
      "tag": "Conditioning on arrivals"
    },
    {
      "front": "Given $N(10)=5$ for a Poisson process, find the **expected sum** of the $5$ arrival times.",
      "back": "Conditioned on $N(10)=5$, the $5$ arrival times are i.i.d. $\\text{Uniform}(0,10)$, each with mean $\\frac{10}{2}=5$.\nExpected sum $=5\\times 5 = 25$.\nMore generally, given $N(t)=n$ the expected sum of arrival times is $n\\cdot\\frac{t}{2}$.",
      "tag": "Conditioning on arrivals"
    },
    {
      "front": "Customers arrive Poisson at rate $\\lambda=5$ per hour. Given exactly $4$ arrivals in the first $2$ hours, find the probability all $4$ came in the **second** hour.",
      "back": "Given $N(2)=4$, each arrival is $\\text{Uniform}(0,2)$; the chance one falls in the second hour $(1,2]$ is $p=\\frac{1}{2}$.\nThe count in the second hour is $\\text{Binomial}(4,\\tfrac{1}{2})$, so $\\Pr(\\text{all }4)=\\left(\\tfrac{1}{2}\\right)^{4}=\\frac{1}{16}=0.0625$.",
      "tag": "Conditioning on arrivals"
    },
    {
      "front": "State the **thinning** (decomposition) theorem for a Poisson process.",
      "back": "If each event of a rate-$\\lambda$ Poisson process is independently classified as type-1 with probability $p$ (and type-2 with probability $1-p$), then the type-1 events form a Poisson process with rate $\\lambda p$ and the type-2 events form a Poisson process with rate $\\lambda(1-p)$, and these two processes are **independent** of each other.\nWith $k$ categories of probabilities $p_1,\\dots,p_k$ (summing to $1$), you get $k$ independent Poisson processes with rates $\\lambda p_1,\\dots,\\lambda p_k$.",
      "tag": "Thinning & superposition"
    },
    {
      "front": "State the **superposition** theorem for independent Poisson processes.",
      "back": "If $N_1(t),\\dots,N_m(t)$ are **independent** Poisson processes with rates $\\lambda_1,\\dots,\\lambda_m$, then their sum $N(t)=N_1(t)+\\cdots+N_m(t)$ is a Poisson process with rate $\\lambda=\\lambda_1+\\cdots+\\lambda_m$.\nGiven an event of the merged process, it came from stream $j$ with probability $\\frac{\\lambda_j}{\\lambda}$, independently of everything else — the dual of thinning.",
      "tag": "Thinning & superposition"
    },
    {
      "front": "Claims arrive Poisson at rate $\\lambda=10$ per day. Independently, $30\\%$ are \"large.\" Find the distribution and expected number of large claims per day, and per $5$ days.",
      "back": "By thinning, large claims form a Poisson process with rate $\\lambda p = 10(0.30)=3$ per day, independent of the small-claim process.\nLarge claims in a day: $\\text{Poisson}(3)$, expected $3$.\nOver $5$ days: $\\text{Poisson}(3\\times 5)=\\text{Poisson}(15)$, expected $15$.",
      "tag": "Thinning & superposition"
    },
    {
      "front": "Cars pass a point Poisson at rate $\\lambda=20$ per minute; each is a truck with probability $0.15$ independently. Find the probability of exactly $2$ trucks in a given minute.",
      "back": "By thinning, trucks form a Poisson process with rate $\\lambda p = 20(0.15)=3$ per minute, so the truck count in a minute is $\\text{Poisson}(3)$.\n$\\Pr(2\\text{ trucks})=e^{-3}\\frac{3^{2}}{2!}=4.5\\,e^{-3}\\approx 4.5(0.049787)\\approx 0.2240$.",
      "tag": "Thinning & superposition"
    },
    {
      "front": "Two independent claim streams arrive Poisson at rates $\\lambda_1=4$ and $\\lambda_2=6$ per hour. Find the probability of exactly $3$ total claims (either stream) in a given hour.",
      "back": "By superposition the combined process is Poisson with rate $\\lambda=\\lambda_1+\\lambda_2=4+6=10$ per hour, so the total in one hour is $\\text{Poisson}(10)$.\n$\\Pr(3\\text{ total})=e^{-10}\\frac{10^{3}}{3!}=e^{-10}\\frac{1000}{6}\\approx 166.667(0.0000453999)\\approx 0.00757$.",
      "tag": "Thinning & superposition"
    },
    {
      "front": "Two independent Poisson streams have rates $\\lambda_1=3$ and $\\lambda_2=2$ per hour. Given that an event occurred, find the probability it came from stream 1.",
      "back": "By superposition the merged rate is $\\lambda=3+2=5$. Each event is independently attributed to stream $j$ with probability $\\frac{\\lambda_j}{\\lambda}$.\n$\\Pr(\\text{from stream 1})=\\frac{\\lambda_1}{\\lambda_1+\\lambda_2}=\\frac{3}{5}=0.6$.",
      "tag": "Thinning & superposition"
    },
    {
      "front": "Customers arrive Poisson at rate $\\lambda=12$ per hour and each independently buys with probability $0.25$. Find the probability that exactly $4$ buyers arrive in the next half hour.",
      "back": "Buyers are a thinned Poisson process with rate $\\lambda p = 12(0.25)=3$ per hour. Over $t=0.5$ hour the buyer count is $\\text{Poisson}(3\\times 0.5)=\\text{Poisson}(1.5)$.\n$\\Pr(4\\text{ buyers})=e^{-1.5}\\frac{1.5^{4}}{4!}=e^{-1.5}\\frac{5.0625}{24}=0.210938\\,e^{-1.5}\\approx 0.210938(0.223130)\\approx 0.04707$.",
      "tag": "Thinning & superposition"
    },
    {
      "front": "Define a **nonhomogeneous Poisson process** with intensity $\\lambda(t)$ and its mean (cumulative-intensity) function.",
      "back": "A nonhomogeneous Poisson process has a time-varying rate $\\lambda(t)\\ge0$. It keeps independent increments but **drops** stationary increments.\nThe **mean function** is $m(t)=\\int_0^{t}\\lambda(u)\\,du$, and the count over $(a,b]$ is $\\text{Poisson}\\!\\left(\\int_a^{b}\\lambda(u)\\,du\\right)=\\text{Poisson}(m(b)-m(a))$.\nIn particular $N(t)\\sim\\text{Poisson}(m(t))$ with $E[N(t)]=m(t)$.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "For a nonhomogeneous Poisson process, how do you compute the probability of $k$ events in $(a,b]$?",
      "back": "Compute the expected count over the interval, $\\mu=\\int_a^{b}\\lambda(u)\\,du = m(b)-m(a)$, then use the Poisson law:\n$\\Pr(N(b)-N(a)=k)=e^{-\\mu}\\frac{\\mu^{k}}{k!}$.\nUnlike the homogeneous case, the interval's location matters because $\\lambda(t)$ varies — only the integrated intensity $\\mu$ enters the formula.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "A nonhomogeneous Poisson process has intensity $\\lambda(t)=2t$ for $t\\ge0$ (time in hours). Find the expected number of events in the first $3$ hours.",
      "back": "$E[N(3)]=m(3)=\\int_0^{3}2t\\,dt=\\left[t^{2}\\right]_0^{3}=9$.\nSo $N(3)\\sim\\text{Poisson}(9)$ with expected count $9$ events.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "For the nonhomogeneous process with $\\lambda(t)=2t$, find the probability of exactly $2$ events between $t=1$ and $t=2$.",
      "back": "The expected count over $(1,2]$ is $\\mu=\\int_1^{2}2t\\,dt=\\left[t^{2}\\right]_1^{2}=4-1=3$, so the increment is $\\text{Poisson}(3)$.\n$\\Pr(2\\text{ events})=e^{-3}\\frac{3^{2}}{2!}=4.5\\,e^{-3}\\approx 4.5(0.049787)\\approx 0.2240$.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "A nonhomogeneous Poisson process has $\\lambda(t)=3+t$ per hour. Find the probability of **no** events in the first $2$ hours.",
      "back": "$\\mu=m(2)=\\int_0^{2}(3+u)\\,du=\\left[3u+\\tfrac{u^{2}}{2}\\right]_0^{2}=6+2=8$, so $N(2)\\sim\\text{Poisson}(8)$.\n$\\Pr(N(2)=0)=e^{-8}\\frac{8^{0}}{0!}=e^{-8}\\approx 0.000335$.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "Define a **compound Poisson process** $X(t)$ and state its mean.",
      "back": "$X(t)=\\sum_{i=1}^{N(t)}Y_i$, where $N(t)$ is a Poisson process with rate $\\lambda$ and the $Y_i$ are i.i.d. (the per-event sizes, e.g. claim amounts), independent of $N(t)$.\nBy the compound-mean formula, $E[X(t)]=E[N(t)]\\,E[Y]=\\lambda t\\,E[Y]$ — the expected number of events times the expected size.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "State the **variance** of a compound Poisson process $X(t)=\\sum_{i=1}^{N(t)}Y_i$.",
      "back": "Using the compound-variance formula $\\text{Var}(X)=E[N]\\text{Var}(Y)+\\text{Var}(N)E[Y]^{2}$ with $E[N(t)]=\\text{Var}(N(t))=\\lambda t$:\n$\\text{Var}(X(t))=\\lambda t\\,\\text{Var}(Y)+\\lambda t\\,E[Y]^{2}=\\lambda t\\,E[Y^{2}]$.\nThe two terms collapse because the Poisson count has equal mean and variance, leaving the clean result $\\text{Var}(X(t))=\\lambda t\\,E[Y^{2}]$.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "Claims arrive Poisson at rate $\\lambda=5$ per day; each claim amount has mean $E[Y]=\\$200$ and standard deviation $\\$150$. Find the expected total claims $X(t)$ over a $1$-day period.",
      "back": "$X(1)=\\sum_{i=1}^{N(1)}Y_i$ is compound Poisson with $\\lambda t = 5(1)=5$.\n$E[X(1)]=\\lambda t\\,E[Y]=5(200)=\\$1{,}000$.\nThe expected daily aggregate claim is $\\$1{,}000$.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "For the same daily claims ($\\lambda=5$, $E[Y]=\\$200$, $\\text{SD}(Y)=\\$150$), find the variance and standard deviation of the aggregate $X(1)$.",
      "back": "First $E[Y^{2}]=\\text{Var}(Y)+E[Y]^{2}=150^{2}+200^{2}=22500+40000=62500$.\n$\\text{Var}(X(1))=\\lambda t\\,E[Y^{2}]=5(62500)=312500$.\nStandard deviation $=\\sqrt{312500}\\approx \\$559.02$.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "A compound Poisson process has rate $\\lambda=3$ per hour and per-event sizes uniform on $[0,10]$. Find $E[X(2)]$ and $\\text{Var}(X(2))$ over $2$ hours.",
      "back": "For $Y\\sim\\text{Uniform}(0,10)$: $E[Y]=5$ and $E[Y^{2}]=\\frac{10^{2}}{3}=\\frac{100}{3}\\approx 33.333$ (since $E[Y^2]=\\text{Var}(Y)+E[Y]^2=\\frac{100}{12}+25$).\nWith $\\lambda t = 3(2)=6$:\n$E[X(2)]=\\lambda t\\,E[Y]=6(5)=30$.\n$\\text{Var}(X(2))=\\lambda t\\,E[Y^{2}]=6\\left(\\frac{100}{3}\\right)=200$.",
      "tag": "Nonhomogeneous & compound"
    },
    {
      "front": "A compound Poisson aggregate over a period has $\\lambda t = 100$ and claim sizes with $E[Y]=500$, $E[Y^{2}]=400000$. Estimate $\\Pr(X>60000)$ via the normal approximation.",
      "back": "$E[X]=\\lambda t\\,E[Y]=100(500)=50000$.\n$\\text{Var}(X)=\\lambda t\\,E[Y^{2}]=100(400000)=40{,}000{,}000$, so $\\text{SD}(X)=\\sqrt{4\\times10^{7}}\\approx 6324.56$.\nStandardize: $z=\\frac{60000-50000}{6324.56}\\approx 1.5811$.\n$\\Pr(X>60000)\\approx 1-\\Phi(1.5811)\\approx 1-0.9431=0.0569$.",
      "tag": "Nonhomogeneous & compound"
    }
  ]
}