Print this page and work through it on paper. Revealed answers are never printed, so you can check one question and still print a clean sheet.

1. 1.Computation · 6 min

   Each part asks whether a process has stationary increments, independent increments, both, or neither. Argue from the definitions, and when a property fails exhibit two specific increments that violate it.

   1. (a)Let $Z\sim N(0,1)$ and $Y_t=tZ$ for $t\ge 0$. Decide whether $\{Y_t\}$ has stationary increments and whether it has independent increments.
   2. (b)Let $N$ be a Poisson process with rate $\lambda$ and let $X_t=N(t)-\lambda t$. Show that $X$ has independent and stationary increments, and find $E(X_t)$ and $\mathrm{Var}(X_t)$.
   3. (c)For the same Poisson process, use independent increments to show that $\mathrm{Cov}(N(s),N(t))=\lambda \min (s,t)$.
   Show the solution to question 1

   (a) Stationary, but not independent. The increment over $(t,t+h]$ is $Y_{t+h}-Y_t=(t+h)Z-tZ=hZ$, which is $N(0,h^2)$ whatever $t$ is. Its law depends on $h$ alone, so the increments are stationary.

   For independence take $0<t_1<t_2$ and the two consecutive increments $Y_{t_1}-Y_0=t_{1}Z$ and $Y_{t_2}-Y_{t_1}=(t_2-t_1)Z$. Independent random variables have zero covariance, but

   $$\mathrm{Cov}(t_{1}Z,(t_2-t_1)Z)=t_1(t_2-t_1)\mathrm{Var}(Z)=t_1(t_2-t_1)\ne 0.$$

   So the increments are not independent. In fact any one increment determines $Z$ and hence every other: the sample path is a random straight line through the origin, and knowing its slope on one stretch tells you the slope everywhere. A second symptom is the variance $h^2$, which is not linear in $h$ — independent stationary increments force $\mathrm{Var}(X_{t+h}-X_t)=ch$, and $h^2$ does not have that form.

   (b) Both. For $s<t$, $X_t-X_s=(N(t)-N(s))-\lambda (t-s)$: a Poisson increment shifted by a constant. Increments of $N$ over disjoint intervals are independent by the counting definition, and subtracting constants does not disturb independence, so $X$ has independent increments. The law of $X_t-X_s$ is Poisson$(\lambda (t-s))$ shifted by $-\lambda (t-s)$, which depends on $t-s$ only, so the increments are stationary.

   Since $N(t)\sim$ Poisson$(\lambda t)$ has mean and variance $\lambda t$,

   $$E(X_t)=\lambda t-\lambda t=0,\mathrm{Var}(X_t)=\mathrm{Var}(N(t))=\lambda t,$$

   the variance being unchanged by the subtraction of a constant. $X$ is the Poisson process with its mean removed, and it is the jump-driven counterpart of Brownian motion: mean zero, variance growing linearly in $t$, independent stationary increments.

   (c) Split at the earlier time. Take $s<t$ and write $N(t)=N(s)+(N(t)-N(s))$. The increment over $(s,t]$ is independent of $N(s)=N(s)-N(0)$, the increment over $(0,s]$, so its covariance with $N(s)$ is zero and

   $$\mathrm{Cov}(N(s),N(t))=\mathrm{Cov}(N(s),N(s))+\mathrm{Cov}(N(s),N(t)-N(s))=\lambda s+0=\lambda s.$$

   The expression is symmetric in the roles of $s$ and $t$ once one relabels, so in general $\mathrm{Cov}(N(s),N(t))=\lambda \min (s,t)$. This is the same covariance structure as Brownian motion, $\min (s,t)$, up to the factor $\lambda$ — and for the same reason: independent increments whose variance is proportional to the length of the interval.

2. 2.Computation · 7 min

   A detector registers radioactive decays as a Poisson process with rate $\lambda =3$ per minute. In each part, before writing any probability, state the interval involved, its length in minutes, and the Poisson mean you are using. Give every answer as an exact expression.

   1. (a)The probability of exactly two decays in a $40$-second window.
   2. (b)The probability of at least one decay in a $20$-second window.
   3. (c)The probability of exactly five decays in the first three minutes, given that exactly two occurred in the first minute.
   4. (d)The detector has registered nothing during the last full minute. The probability that it registers nothing during the next $20$ seconds either.
   Show the solution to question 2

   (a) Convert the units first. Forty seconds is $\frac{2}{3}$ of a minute, so the Poisson mean is $\lambda t=3\cdot \frac{2}{3}=2$, a pure number. By the counting definition,

   $$P(N(\frac{2}{3})=2)=e^{-2}\frac{2^2}{2!}=2e^{-2}.$$

   (b) Twenty seconds is $\frac{1}{3}$ minute, mean $3\cdot \frac{1}{3}=1$. The complement of “at least one” is “none”:

   $$P(N(\frac{1}{3})\ge 1)=1-P(N(\frac{1}{3})=0)=1-e^{-1}.$$

   (c) Rewrite the event in terms of an increment. Given $N(1)=2$, the event $\{N(3)=5\}$ is the event $\{N(3)-N(1)=3\}$, a statement about the interval $(1,3]$ of length $2$. By independent increments this increment is independent of $N(1)$, so the conditioning drops away, and the mean over $(1,3]$ is $3\cdot 2=6$:

   $$P(N(3)=5\mid N(1)=2)=P(N(3)-N(1)=3)=e^{-6}\frac{6^3}{3!}=36e^{-6}.$$

   It is not $P(N(3)=5)/P(N(1)=2)$: that ratio would be a conditional probability only if $\{N(3)=5\}$ were contained in $\{N(1)=2\}$, which it is not.

   (d) The past is irrelevant. “Nothing in the last minute” concerns the interval $(-1,0]$ and “nothing in the next twenty seconds” concerns $(0,\frac{1}{3}]$. These are disjoint, so by independent increments the second event is independent of the first, and the answer is the unconditional probability of no decay in an interval of length $\frac{1}{3}$, with mean $1$:

   $$P(\text{none in the next }20\text{ s}\mid \text{none in the last minute})=e^{-1}.$$

   Seen from the inter-arrival definition this is the memorylessness of the exponential gap: the minute already waited is discarded. Note the answer is exactly the complement of (b), as it must be.

3. 3.Computation · 8 min

   Requests reach a server as a Poisson process with rate $\lambda =2$ per hour, and $T_n$ denotes the time of the $n$-th request.

   1. (a)Compute $P(T_2>1)$ by translating it into a statement about $N(1)$.
   2. (b)Write down the density of $T_2$ and compute $P(T_2>1)$ a second time by integrating it. The two answers must agree.
   3. (c)Find $E(T_3)$ and $\mathrm{Var}(T_3)$, and explain in one sentence why $T_3$ is not exponential with mean $E(T_3)$.
   4. (d)Given that the first request arrived at exactly $t=\frac{1}{2}$, find $P(T_2>1)$, naming the property you use.
   Show the solution to question 3

   (a) Counting–waiting duality. The second request arrives after time $1$ exactly when at most one request has arrived by time $1$: $\{T_2>1\}=\{N(1)\le 1\}$, the complement of $\{N(1)\ge 2\}=\{T_2\le 1\}$. With $N(1)\sim$ Poisson$(2)$,

   $$P(T_2>1)=P(N(1)=0)+P(N(1)=1)=e^{-2}(1+2)=3e^{-2}.$$

   (b) The Erlang density. $T_2={\tau }_1+{\tau }_2$ is a sum of two independent Exp$(2)$ gaps, so by the law of the $n$-th arrival time with $n=2$,

   $$f_{T_2}(t)=\frac{{\lambda }^{2}t{e}^{-\lambda t}}{1!}=4t{e}^{-2t},t>0.$$

   Integrate by parts with $u=4t$ and $dv=e^{-2t}dt$, so $v=-\frac{1}{2}{e}^{-2t}$:

   $${\int }_1^{\infty }4t{e}^{-2t}dt=[-2t{e}^{-2t}{]}_1^{\infty }+{\int }_1^{\infty }2e^{-2t}dt=2e^{-2}+e^{-2}=3e^{-2}.✓$$

   The agreement is the equivalence theorem in one line: the count route and the waiting-time route describe the same process.

   (c) $T_3$ is a sum of three independent exponentials, each with mean $\frac{1}{\lambda }=\frac{1}{2}$ and variance $\frac{1}{{\lambda }^2}=\frac{1}{4}$, so by linearity and independence

   $$E(T_3)=\frac{3}{2}\text{hours},\mathrm{Var}(T_3)=\frac{3}{4}{\text{hours}}^2.$$

   An exponential with mean $\frac{3}{2}$ would have variance $\frac{9}{4}$, three times as large, and its density would be largest at $t=0$, whereas $T_3$ cannot be small unless all three gaps are — its Erlang density $\frac{{\lambda }^3{t}^2}{2}{e}^{-\lambda t}$ vanishes at $0$. A sum of exponentials is not a rescaled exponential.

   (d) Memorylessness. By the inter-arrival definition $T_2=T_1+{\tau }_2$ with ${\tau }_2\sim$ Exp$(2)$ independent of $T_1$. Given $T_1=\frac{1}{2}$, the event $\{T_2>1\}$ is $\{{\tau }_2>\frac{1}{2}\}$, so

   $$P(T_2>1\mid T_1=\frac{1}{2})=P({\tau }_2>\frac{1}{2})=e^{-2\cdot \frac{1}{2}}=e^{-1}.$$

   Equivalently, by independent increments it is the probability of no request in $(\frac{1}{2},1]$, an interval of length $\frac{1}{2}$ with mean $1$, which is $e^{-1}$ again. Compare with (a): the conditional value $e^{-1}\approx 0.37$ is below the unconditional $3e^{-2}\approx 0.41$. That is the right direction, because the unconditional probability also includes the paths on which even the first request is still outstanding at $t=1$ (probability $e^{-2}$), and on those $T_2>1$ is certain; fixing $T_1=\frac{1}{2}$ removes them.

4. 4.Computation · 8 min

   Let $N$ be a Poisson process with rate $\lambda =3$ per hour, and suppose it is known that exactly three events occurred in the first two hours: $N(2)=3$. Every answer should be a plain number in which $\lambda$ does not appear — say at the outset why it cannot.

   1. (a)$P(N(1)=1\mid N(2)=3)$.
   2. (b)$P(T_1>\frac{3}{2}\mid N(2)=3)$.
   3. (c)$E(T_1+T_2+T_3\mid N(2)=3)$.
   4. (d)$E(T_3\mid N(2)=3)$. Find the conditional distribution function of $T_3$ first.
   Show the solution to question 4

   Why $\lambda$ cannot appear. By conditional uniformity, given $N(2)=3$ the three arrival times are, as an unordered set, three independent Uniform$(0,2)$ points $U_1,U_2,U_3$; their conditional density $n!/t^n=6/8$ on the ordered region contains no $\lambda$. The rate decides how many events occur, and that has been fixed. Every part below is a question about three uniform points.

   (a) Binomial thinning of the uniforms. $N(1)$ is the number of the $U_i$ that fall in $(0,1]$. Each does so independently with probability $\frac{1}{2}$, so $N(1)\mid N(2)=3\sim$ Binomial$(3,\frac{1}{2})$ and

   $$P(N(1)=1\mid N(2)=3)=\left(\genfrac{}{}{0ex}{}{3}{1}\right)(\frac{1}{2}{)}^3=\frac{3}{8}.$$

   Cross-check with increments, where $\lambda$ enters and then cancels: $P(N(1)=1)P(N(2)-N(1)=2)/P(N(2)=3)=\frac{3e^{-3}\cdot \frac{9}{2}{e}^{-3}}{36e^{-6}}=\frac{27/2}{36}=\frac{3}{8}$. $✓$

   (b) $T_1$ is the smallest of the three uniforms, and $T_1>\frac{3}{2}$ means all three exceed $\frac{3}{2}$. Each does so with probability $\frac{2-3/2}{2}=\frac{1}{4}$, independently:

   $$P(T_1>\frac{3}{2}\mid N(2)=3)=(\frac{1}{4}{)}^3=\frac{1}{64}.$$

   (c) The sum of the arrival times is unchanged by ordering, so it is $U_1+U_2+U_3$, and by linearity

   $$E(T_1+T_2+T_3\mid N(2)=3)=3\cdot \frac{2}{2}=3,$$

   which is the expected sum of arrival times $\frac{nt}{2}$ with $n=3$, $t=2$.

   (d) The maximum of three uniforms. $T_3=\max (U_1,U_2,U_3)$, and $T_3\le x$ exactly when all three are at most $x$:

   $$P(T_3\le x\mid N(2)=3)=(\frac{x}{2}{)}^3,0\le x\le 2.$$

   Either differentiate to get the density $\frac{3x^2}{8}$ and integrate $x$ against it, or use the tail formula for a non-negative variable:

   $$E(T_3\mid N(2)=3)={\int }_0^2(1-\frac{x^3}{8})dx=2-\frac{1}{8}\cdot \frac{2^4}{4}=2-\frac{1}{2}=\frac{3}{2}.$$

   Sanity check against (c): by the symmetry $x\mapsto 2-x$ of the uniform, $E(T_1\mid \cdot )=2-\frac{3}{2}=\frac{1}{2}$, and the middle point has mean $1$; the three means sum to $\frac{1}{2}+1+\frac{3}{2}=3$, agreeing with (c). $✓$

5. 5.Critique · 8 min

   Jobs arrive at a printer as a Poisson process with rate $\lambda =1$ per hour. A student is asked for (i) the probability that exactly one job arrives in the first hour and exactly three arrive in the first two hours, and (ii) the probability that the second job arrives after hour $3$. The submitted work reads:

   Step 1. $N(1)\sim$ Poisson$(1)$ and $N(2)\sim$ Poisson$(2)$, so $P(N(1)=1)=e^{-1}$ and $P(N(2)=3)=e^{-2}\cdot \frac{2^3}{3!}=\frac{4}{3}{e}^{-2}$.

   Step 2. The process has independent increments, so $N(1)$ and $N(2)$ are independent, and $P(N(1)=1,N(2)=3)=e^{-1}\cdot \frac{4}{3}{e}^{-2}=\frac{4}{3}{e}^{-3}$.

   Step 3. $T_2$ is the time of the second job, so $T_2$ is exponential with mean $2/\lambda =2$ hours, and $P(T_2>3)=e^{-3/2}$.

   One step is correct and two are not.

   1. (a)Identify the correct step and the two flawed ones. For each flaw, name the definition or theorem being misquoted, and give a one-line reason the conclusion cannot be right that does not require computing the true value.
   2. (b)Compute both probabilities correctly, naming the results you use.
   Show the solution to question 5

   (a) Step 1 is correct. With $\lambda =1$ the counting definition gives $N(t)\sim$ Poisson$(t)$, and $e^{-2}\cdot \frac{8}{6}=\frac{4}{3}{e}^{-2}$ is right.

   Step 2 misquotes independent increments. The definition says that increments over disjoint intervals are independent — here $N(1)-N(0)$ over $(0,1]$ and $N(2)-N(1)$ over $(1,2]$. It says nothing about $N(1)$ and $N(2)$, which are counts over the nested intervals $[0,1]\subset [0,2]$. They cannot be independent: $N(2)\ge N(1)$ on every path, so for instance $P(N(2)=0\mid N(1)=1)=0$ while $P(N(2)=0)=e^{-2}>0$. The one-line refutation: the student's product formula would assign the impossible event $\{N(1)=1,N(2)=0\}$ the positive probability $e^{-1}\cdot e^{-2}$.

   Step 3 misquotes the inter-arrival definition. That definition makes the gaps ${\tau }_1,{\tau }_2$ exponential; $T_2={\tau }_1+{\tau }_2$ is a sum of two independent exponentials, whose law is Erlang with density $t{e}^{-t}$, not exponential. The mean $2$ is correct but the shape is not. One-line refutation: $T_2$ is small only if both gaps are small, so its density must vanish at $0$, whereas every exponential density is largest at $0$. (Equivalently, $\mathrm{Var}(T_2)=2$, not the $4$ an exponential of mean $2$ would have.)

   (b) (i) Rewrite the event with a disjoint increment: $\{N(1)=1,N(2)=3\}=\{N(1)=1,N(2)-N(1)=2\}$. Now independent increments applies, and the increment over $(1,2]$ has mean $1$:

   $$P(N(1)=1,N(2)=3)=P(N(1)=1)P(N(2)-N(1)=2)=e^{-1}\cdot e^{-1}\frac{1^2}{2!}=\frac{e^{-2}}{2}.$$

   Numerically this is about $0.068$ against the student's $\frac{4}{3}{e}^{-3}\approx 0.066$ — uncomfortably close, which is exactly why the structural argument in (a) matters more than a numerical one.

   (ii) By the counting–waiting duality, $\{T_2>3\}=\{N(3)\le 1\}$ with $N(3)\sim$ Poisson$(3)$:

   $$P(T_2>3)=e^{-3}(1+3)=4e^{-3}.$$

   Check by integrating the Erlang density: ${\int }_3^{\infty }t{e}^{-t}dt=[-t{e}^{-t}-e^{-t}{]}_3^{\infty }=3e^{-3}+e^{-3}=4e^{-3}$. $✓$ The student's $e^{-3/2}\approx 0.22$ overstates the true $4e^{-3}\approx 0.20$, because an exponential of mean $2$ has a heavier tail than the Erlang of the same mean.

6. 6.Proof · 9 min

   Let $N_1$ and $N_2$ be independent Poisson processes with rates ${\lambda }_1$ and ${\lambda }_2$, let $N=N_1+N_2$, and write $\lambda ={\lambda }_1+{\lambda }_2$. You may quote the superposition theorem.

   1. (a)Prove that, conditional on $N(t)=n$, the count $N_1(t)$ has the Binomial$(n,{\lambda }_1/\lambda )$ distribution. Start from the joint probability $P(N_1(t)=j,N_2(t)=n-j)$.
   2. (b)Deduce from (a), without any integral, that if exactly one event of the merged process has occurred by time $t$, it came from stream $1$ with probability ${\lambda }_1/\lambda$ — for every $t$. Explain how this relates to the proposition on which stream produced the next event, and what that proposition asserts beyond what (a) gives.
   3. (c)Sensors in two rooms report faults as independent Poisson processes with rates $1$ and $3$ per day. Given that four faults were reported in total during one day, find the probability that none came from the first room, and the expected number that did.
   Show the solution to question 6

   (a) Factorise, divide, cancel. The event $\{N_1(t)=j,N(t)=n\}$ is the event $\{N_1(t)=j,N_2(t)=n-j\}$. The two processes are independent, so

   $$P(N_1(t)=j,N_2(t)=n-j)=e^{-{\lambda }_{1}t}\frac{({\lambda }_{1}t{)}^j}{j!}\cdot e^{-{\lambda }_{2}t}\frac{({\lambda }_{2}t{)}^{n-j}}{(n-j)!}.$$

   By the superposition theorem $N(t)\sim$ Poisson$(\lambda t)$, so $P(N(t)=n)=e^{-\lambda t}(\lambda t{)}^n/n!$. Divide:

   $$P(N_1(t)=j\mid N(t)=n)=\frac{e^{-{\lambda }_{1}t}{e}^{-{\lambda }_{2}t}}{e^{-\lambda t}}\cdot \frac{t^j{t}^{n-j}}{t^n}\cdot \frac{n!}{j!(n-j)!}\cdot \frac{{\lambda }_1^j{\lambda }_2^{n-j}}{{\lambda }^n}.$$

   The exponentials cancel because ${\lambda }_1+{\lambda }_2=\lambda$, the powers of $t$ cancel exactly, and what remains is

   $$\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\frac{{\lambda }_1}{\lambda }{)}^j\right(\frac{{\lambda }_2}{\lambda }{)}^{n-j}=\left(\genfrac{}{}{0ex}{}{n}{j}\right)p^j(1-p{)}^{n-j},p=\frac{{\lambda }_1}{\lambda },$$

   since ${\lambda }_2/\lambda =1-{\lambda }_1/\lambda$. This is the Binomial$(n,p)$ mass function. $■$ Note that $t$ has disappeared, as $\lambda$ did in conditional uniformity: given the total, the split depends only on the ratio of the rates.

   (b) Put $n=1$: $P(N_1(t)=1\mid N(t)=1)=\left(\genfrac{}{}{0ex}{}{1}{1}\right)p={\lambda }_1/\lambda$. On the event $\{N(t)=1\}$ the single event by time $t$ is the first event of $N$, and it belongs to stream $1$ exactly when $N_1(t)=1$. So, given that exactly one event has occurred by time $t$, it came from stream $1$ with probability ${\lambda }_1/\lambda$, and this holds for every $t>0$ because $t$ cancelled.

   The proposition on which stream produced the next event says the same number, ${\lambda }_1/\lambda$, but asserts more: that the identity of the first event is independent of its time $T_1$, not merely that the conditional probability is the same for every window $\{N(t)=1\}$. Those windows overlap and are not a partition of the sample space, so (a) alone does not deliver the unconditional statement; the proof of the proposition supplies it by showing the joint law of $(\min ({\tau }^{(1)},{\tau }^{(2)}),\text{identity})$ factorises. What (a) does show, cleanly and with no integral, is where the number ${\lambda }_1/\lambda$ comes from: it is the binomial success probability of a Poisson split.

   (c) Here ${\lambda }_1=1$, ${\lambda }_2=3$, $\lambda =4$, $p=\frac{1}{4}$ and $n=4$. By (a), $N_1\mid N=4\sim$ Binomial$(4,\frac{1}{4})$, so

   $$P(N_1=0\mid N=4)=(\frac{3}{4}{)}^4=\frac{81}{256},E(N_1\mid N=4)=4\cdot \frac{1}{4}=1.$$

   Sanity check: unconditionally $E(N_1(1))=1$ as well, and $E(N(1))=4$, so the conditioning on the typical total leaves the first room's expectation where it was. $✓$

7. 7.Computation · 10 min

   Two relaxations of the Poisson process. In (a)–(b), claims reach an insurer as a Poisson process with rate $\lambda =4$ per month, and each claim independently has size $Y$ (in thousands) with $P(Y=1)=P(Y=2)=\frac{1}{2}$; $X(t)$ is the total amount claimed by time $t$. In (c)–(d), a website receives visits as a non-homogeneous Poisson process with intensity $\lambda (u)=2u$ visits per hour, $0\le u\le 3$.

   1. (a)Find $E(X(3))$ and $\mathrm{Var}(X(3))$, and say what fraction of the variance is due to randomness in the number of claims.
   2. (b)Find $P(X(1)=2)$ exactly.
   3. (c)Find the mean-value function $m(t)$, the expected number of visits in the second hour, and the probability of no visit at all in the second hour.
   4. (d)Given that exactly one visit occurred in the three hours, find the probability that it occurred during the first hour — by computing with increments, not by quoting a formula — and explain why the answer is not $\frac{1}{3}$.
   Show the solution to question 7

   (a) Moments of a compound Poisson process. The claim size has $\mu =E(Y)=\frac{1+2}{2}=\frac{3}{2}$ and $E(Y^2)=\frac{1+4}{2}=\frac{5}{2}$, so ${\sigma }^2=\frac{5}{2}-\frac{9}{4}=\frac{1}{4}$. Over three months the count has mean $\lambda t=12$. Hence

   $$E(X(3))=\lambda t\mu =12\cdot \frac{3}{2}=18,\mathrm{Var}(X(3))=\lambda tE(Y^2)=12\cdot \frac{5}{2}=30$$

   (in thousands, and thousands squared). The law of total variance splits the $30$ as $\lambda t{\sigma }^2+\lambda t{\mu }^2=12\cdot \frac{1}{4}+12\cdot \frac{9}{4}=3+27$: only $3$ comes from uncertainty about how big the claims are, and $27$ — a fraction $\frac{27}{30}=\frac{9}{10}$ — from uncertainty about how many there are. Even with every claim fixed at $1.5$ the variance would still be $27$.

   (b) Enumerate the ways to total $2$. $X(1)=2$ happens in exactly two ways: one claim of size $2$, or two claims each of size $1$. (No claims gives $0$; three or more claims give at least $3$.) With $N(1)\sim$ Poisson$(4)$ and the sizes independent of the count,

   $$P(X(1)=2)=P(N(1)=1)\cdot \frac{1}{2}+P(N(1)=2)\cdot (\frac{1}{2}{)}^2=4e^{-4}\cdot \frac{1}{2}+\frac{16}{2}{e}^{-4}\cdot \frac{1}{4}=2e^{-4}+2e^{-4}=4e^{-4}.$$

   (c) Integrate the intensity. $m(t)={\int }_0^{t}2udu=t^2$. The second hour is the interval $(1,2]$, and the count there is Poisson with mean

   $$m(2)-m(1)=4-1=3,$$

   so the expected number of visits is $3$ and $P(\text{no visit in }(1,2])=e^{-3}$. Note the first hour has mean $1$ and the third has mean $9-4=5$: the increments are Poisson and independent but not stationary.

   (d) Condition through the increments. The event $\{N(3)=1,\text{the visit is in }(0,1]\}$ is $\{N(1)=1,N(3)-N(1)=0\}$. The two increments are independent with means $m(1)=1$ and $m(3)-m(1)=8$, and $N(3)\sim$ Poisson$(9)$:

   $$P(N(1)=1\mid N(3)=1)=\frac{P(N(1)=1)P(N(3)-N(1)=0)}{P(N(3)=1)}=\frac{e^{-1}\cdot e^{-8}}{9e^{-9}}=\frac{1}{9}.$$

   Conditional uniformity would give $\frac{1}{3}$, but it requires a constant rate: the intensity $2u$ is lowest in the first hour, so a lone visit is less likely to have landed there. The answer $\frac{1}{9}=m(1)/m(3)$ is the conditional-uniformity statement on the $m$-scale — given one event, its time has density $\lambda (u)/m(3)=\frac{2u}{9}$, and ${\int }_0^1\frac{2u}{9}du=\frac{1}{9}$. $✓$

8. 8.Computation · 9 min

   Let $W$ be a standard Brownian motion. In every part, split at the earlier time so that a known quantity and an independent increment appear; do not work with correlated pairs directly.

   1. (a)Find the distribution of $W_3-2W_1$.
   2. (b)Find $P(W_4>0\mid W_1=-1)$, leaving the answer in terms of $\Phi$.
   3. (c)Find $\mathrm{Corr}(W_2,W_8)$, and state what $\mathrm{Corr}(W_{2c},W_{8c})$ is for an arbitrary $c>0$.
   4. (d)Fix $s>0$ and let $B_t=W_{s+t}-W_s$ for $t\ge 0$. Prove that $B$ is a standard Brownian motion, checking all four clauses of the definition.
   Show the solution to question 8

   (a) Split at time $1$. Write $W_3-2W_1=(W_3-W_1)-W_1$. The increment $W_3-W_1\sim N(0,2)$ is independent of $W_1\sim N(0,1)$ by independent increments, and a difference of independent normals is normal with the variances added:

   $$W_3-2W_1\sim N(0,2+1)=N(0,3).$$

   Check through the covariance $\mathrm{Cov}(W_1,W_3)=\min (1,3)=1$: $\mathrm{Var}(W_3-2W_1)=3+4\cdot 1-2\cdot 2\cdot 1=3$. $✓$

   (b) $W_4=W_1+(W_4-W_1)$ with $W_4-W_1\sim N(0,3)$ independent of $W_1$. Given $W_1=-1$, therefore, $W_4\sim N(-1,3)$, and

   $$P(W_4>0\mid W_1=-1)=P(Z>\frac{0-(-1)}{\sqrt{3}})=1-\Phi \left(\frac{1}{\sqrt{3}}\right).$$

   (c) $\mathrm{Cov}(W_2,W_8)=\min (2,8)=2$, and the standard deviations are $\sqrt{2}$ and $\sqrt{8}=2\sqrt{2}$, with product $4$. So $\mathrm{Corr}(W_2,W_8)=\frac{2}{4}=\frac{1}{2}$, which is $\sqrt{2/8}$. For $(W_{2c},W_{8c})$ the covariance is $2c$ and the product of standard deviations is $\sqrt{2c}\sqrt{8c}=4c$, so the correlation is again $\frac{1}{2}$: it depends only on the ratio of the two times, which is the scaling property showing itself.

   (d) Verify the four clauses. Start: $B_0=W_s-W_s=0$. Independent increments: for $0\le u<t$, $B_t-B_u=W_{s+t}-W_{s+u}$ is the increment of $W$ over $(s+u,s+t]$. Disjoint intervals in $t$ shift to disjoint intervals in $s+t$, so increments of $B$ over disjoint intervals are increments of $W$ over disjoint intervals, hence independent. Gaussian increments: $B_t-B_u=W_{s+t}-W_{s+u}\sim N(0,(s+t)-(s+u))=N(0,t-u)$. Continuity: $t\mapsto W_{s+t}$ is a continuous path shifted in time, and subtracting the constant $W_s$ preserves continuity, so the paths of $B$ are continuous almost surely. All four hold, so $B$ is a standard Brownian motion. $■$

   Moreover $B$ is independent of $\{W_u:u\le s\}$, since its increments live on intervals beyond $s$. That is the Markov property of $W$ at the fixed time $s$; the strong Markov property invoked by the reflection principle is the same restart at a random time $T_a$.

9. 9.Proof · 10 min

   Fix $t>0$ and let ${\Pi }_n$ be partitions $0=t_0<t_1<\cdots <t_{k_n}=t$ of $[0,t]$ with mesh $\Vert {\Pi }_n\Vert ={\max }_i(t_i-t_{i-1})\to 0$. Write ${\delta }_i=t_i-t_{i-1}$. You may quote the theorem that $Q_n={\sum }_i(W_{t_i}-W_{t_{i-1}}{)}^2\to t$ in $L^2$.

   1. (a)Let $f$ be continuously differentiable on $[0,t]$. Prove that ${\sum }_i(f(t_i)-f(t_{i-1}){)}^2\to 0$.
   2. (b)Let $X_t=\mu t+\sigma W_t$ be Brownian motion with drift. Prove that ${\sum }_i(X_{t_i}-X_{t_{i-1}}{)}^2\to {\sigma }^{2}t$ in $L^2$. (Expand the square and treat the three resulting sums separately.)
   3. (c)Explain how (a) and (b) together prove that, with probability $1$, the path of $X$ is not continuously differentiable on $[0,t]$, and say what (b) shows about whether quadratic variation detects the drift.
   Show the solution to question 9

   (a) A smooth function has zero quadratic variation. Since $f^{\prime }$ is continuous on the compact interval $[0,t]$ it is bounded: $|f^{\prime }(u)|\le K$ for all $u$. By the mean value theorem $f(t_i)-f(t_{i-1})=f^{\prime }({\xi }_i){\delta }_i$ for some ${\xi }_i\in (t_{i-1},t_i)$, so $|f(t_i)-f(t_{i-1})|\le K{\delta }_i$ and

   $$\sum _i(f(t_i)-f(t_{i-1}){)}^2\le K^2\sum _i{\delta }_i^2\le K^2(\underset{i}{\max }{\delta }_i)\sum _i{\delta }_i=K^2\Vert {\Pi }_n\Vert t⟶0.■$$

   The mechanism is $(\Delta f{)}^2\sim (\Delta t{)}^2$, one order smaller than $\Delta t$; summing $k_n$ such terms gives something of order $\Vert {\Pi }_n\Vert$.

   (b) Expand and treat three sums. Write ${\Delta }_i=W_{t_i}-W_{t_{i-1}}$, so the ${\Delta }_i$ are independent with ${\Delta }_i\sim N(0,{\delta }_i)$, and $X_{t_i}-X_{t_{i-1}}=\mu {\delta }_i+\sigma {\Delta }_i$. Squaring and summing,

   $$\sum _i(X_{t_i}-X_{t_{i-1}}{)}^2=\underset{A_n}{\underbrace{{\mu }^2\sum _i{\delta }_i^2}}+\underset{B_n}{\underbrace{2\mu \sigma \sum _i{\delta }_i{\Delta }_i}}+{\sigma }^2{Q}_n.$$

   The drift-only sum $A_n$ is deterministic, and by the estimate of (a) applied to $f(u)=\mu u$, $0\le A_n\le {\mu }^2\Vert {\Pi }_n\Vert t\to 0$.

   The cross sum $B_n$ has $E(B_n)=0$, and by independence of the ${\Delta }_i$,

   $$E(B_n^2)=\mathrm{Var}(B_n)=4{\mu }^2{\sigma }^2\sum _i{\delta }_i^2\mathrm{Var}({\Delta }_i)=4{\mu }^2{\sigma }^2\sum _i{\delta }_i^3\le 4{\mu }^2{\sigma }^2\Vert {\Pi }_n{\Vert }^{2}t⟶0,$$

   so $B_n\to 0$ in $L^2$.

   The Brownian sum ${\sigma }^2{Q}_n\to {\sigma }^{2}t$ in $L^2$ by the quoted theorem, since multiplying by the constant ${\sigma }^2$ multiplies $E[(Q_n-t{)}^2]$ by ${\sigma }^4$.

   Combine with the triangle inequality for the $L^2$ norm $\Vert Y\Vert =(E(Y^2){)}^{1/2}$ (Minkowski's inequality):

   $$\Vert \sum _i(X_{t_i}-X_{t_{i-1}}{)}^2-{\sigma }^{2}t\Vert \le \Vert A_n\Vert +\Vert B_n\Vert +{\sigma }^2\Vert Q_n-t\Vert =A_n+\Vert B_n\Vert +{\sigma }^2\Vert Q_n-t\Vert ⟶0,$$

   using that $A_n$ is a non-negative constant. This is $L^2$ convergence to ${\sigma }^{2}t$; convergence in probability follows by Chebyshev's inequality. $■$

   (c) Roughness, and blindness to drift. Suppose the path of $X$ were continuously differentiable on $[0,t]$ with some probability $p>0$. On that event, (a) says the quadratic sums tend to $0$ along every sequence of partitions. But (b) says they tend to ${\sigma }^{2}t>0$ in probability, so along some subsequence of partitions they tend to ${\sigma }^{2}t$ almost surely — in particular on a subset of the smooth event of probability $p$, where they also tend to $0$. Contradiction, so $p=0$: almost surely the path is not $C^1$ on $[0,t]$, and adding a drift, however large, does not smooth it.

   The limit ${\sigma }^{2}t$ contains no $\mu$. Quadratic variation is blind to the drift: the drift is a smooth function, and (a) shows smooth functions contribute nothing to it, whereas $\sigma$ is recovered exactly from any stretch of path. Two models differing only in $\mu$ produce paths that no quadratic-variation measurement can tell apart; two models differing in $\sigma$ produce paths that are distinguishable with certainty.

10. 10.Sketch · 10 min

    Let $W$ be a standard Brownian motion with running maximum $M_t={\max }_{0\le s\le t}{W}_s$, and write $\phi$ for the standard normal density. You may quote the reflection principle.

    1. (a)Show that for $a\ge 0$, $P(M_t\le a)=2\Phi (a/\sqrt{t})-1$, and deduce that $M_t$ has density $f_{M_t}(a)=\frac{2}{\sqrt{t}}\phi \left(\frac{a}{\sqrt{t}}\right)$ for $a>0$ and $0$ for $a<0$. Explain why there is no atom at $a=0$, and what that says about the path just after time $0$.
    2. (b)Show that $M_t$ and $|W_t|$ have the same distribution, and use this to compute $E(M_t)$ exactly.
    3. (c)On one pair of axes, sketch the densities of $W_1$ and of $M_1$ for $-3\le a\le 3$. The sketch must show and justify: the value of each density at $a=0$; the relationship between the two curves on $a>0$; why both areas can equal $1$; the discontinuity of one of them; and the position of each mean.
    Show the solution to question 10

    (a) Complement the reflection principle. The reflection principle gives $G(a)=P(M_t\ge a)=2(1-\Phi (a/\sqrt{t}))$ for $a>0$. This $G$ is continuous in $a$, and $P(M_t=a)=P(M_t\ge a)-P(M_t>a)=G(a)-{\lim }_{b\downarrow a}G(b)=0$, so $M_t$ has no atoms and

    $$P(M_t\le a)=1-P(M_t>a)=1-G(a)=2\Phi \left(\frac{a}{\sqrt{t}}\right)-1,a>0.$$

    For $a<0$ the probability is $0$, since $M_t\ge W_0=0$ on every path. Differentiating on $a>0$ by the chain rule,

    $$f_{M_t}(a)=2\phi \left(\frac{a}{\sqrt{t}}\right)\cdot \frac{1}{\sqrt{t}}=\frac{2}{\sqrt{t}}\phi \left(\frac{a}{\sqrt{t}}\right).$$

    At $a=0$ the formula gives $P(M_t\le 0)=2\Phi (0)-1=0$, so $P(M_t=0)=0$: with probability $1$ the maximum over $[0,t]$ is strictly positive, however small $t$ is. The path does not linger at or below $0$ after starting — it takes positive values immediately, and by symmetry negative ones too, so it crosses zero infinitely often in every interval $(0,\epsilon )$.

    (b) Same distribution function. For $a\ge 0$,

    $$P(|W_t|\le a)=P(-a\le W_t\le a)=\Phi \left(\frac{a}{\sqrt{t}}\right)-\Phi (-\frac{a}{\sqrt{t}})=2\Phi \left(\frac{a}{\sqrt{t}}\right)-1,$$

    using $\Phi (-x)=1-\Phi (x)$; and both variables have distribution function $0$ for $a<0$. Equal distribution functions mean equal laws, so $M_t\stackrel{d}{=}|W_t|$. (Only in distribution: pathwise $M_t\ge W_t$ but $M_t\ge |W_t|$ fails whenever the path finishes well below where it started.)

    Therefore $E(M_t)=E|W_t|=\sqrt{t}E|Z|$ with $Z$ standard normal, and by symmetry

    $$E|Z|=2{\int }_0^{\infty }z\phi (z)dz=2[-\phi (z){]}_0^{\infty }=2\phi (0)=\frac{2}{\sqrt{2\pi }}=\sqrt{\frac{2}{\pi }},$$

    using ${\phi }^{\prime }(z)=-z\phi (z)$. Hence $E(M_t)=\sqrt{2t/\pi }$, growing like $\sqrt{t}$ as scaling demands.

    (c) The sketch. Draw the $a$-axis from $-3$ to $3$ and a vertical axis up to about $0.9$.

    The density of $W_1$ is the standard bell $\phi$: symmetric about $0$, height $\phi (0)=1/\sqrt{2\pi }\approx 0.40$ at the centre, inflection points at $a=\pm 1$, effectively zero by $a=\pm 3$. Its mean is at $a=0$ by symmetry; mark it.

    The density of $M_1$ is $0$ for all $a<0$ — draw it flat along the axis there — and equals $2\phi (a)$ for $a>0$. So on $a>0$ it is the bell stretched vertically by a factor of exactly $2$: at $0^{+}$ it starts at $2\phi (0)=\sqrt{2/\pi }\approx 0.80$, twice the height of the other curve, and it lies above the $W_1$ curve at every positive $a$, decreasing and tending to $0$ by $a=3$. Draw the two curves so that the vertical gap between them at each $a>0$ equals the height of the lower one.

    Areas. The bell has area $1$, of which $\frac{1}{2}$ lies on $a>0$. Doubling that half gives area $2\cdot \frac{1}{2}=1$ for the $M_1$ density, which is why the taller curve on a half-line and the shorter curve on the whole line can both be densities.

    Discontinuity. At $a=0$ the $M_1$ density jumps from $0$ to $\approx 0.80$; draw it as an open circle at $(0,0)$ and a filled one at $(0,0.80)$, or simply as a vertical wall. The distribution function of $M_1$ is continuous (no atom, from (a)); it is the density that jumps.

    Means. Mark the mean of $M_1$ at $a=\sqrt{2/\pi }\approx 0.80$ on the axis, to the right of the curve's peak because the density is decreasing on its whole support and so is right-skewed. It is a coincidence of the standard normal that this number equals the density's height at $0^{+}$; the two marks should sit on different axes.

11. 11.Computation · 11 min

    Let $W$ be a standard Brownian motion, $a>0$, and $T_a=\inf \{s\ge 0:W_s=a\}$ its first hitting time of $a$. Leave every answer exact.

    1. (a)Differentiate $P(T_a\le t)=2(1-\Phi (a/\sqrt{t}))$ to find the density of $T_a$, and confirm it is a genuine density by examining the distribution function as $t\downarrow 0$ and as $t\to \infty$.
    2. (b)Using the density from (a), prove directly that $E(T_a)=\infty$.
    3. (c)Prove that $T_a$ has the same distribution as $a^2{T}_1$, and explain the result through the scaling property of Brownian motion.
    4. (d)Adapt the reflection argument to show that $P(M_t\ge a,W_t\le 0)=P(W_t\ge 2a)$, and then compute $P(W_1\le 0\mid T_1\le 1)$.
    Show the solution to question 11

    (a) Chain rule. Write $F(t)=2-2\Phi (a{t}^{-1/2})$. Then, with ${\Phi }^{\prime }=\phi$ and $\frac{d}{dt}{t}^{-1/2}=-\frac{1}{2}{t}^{-3/2}$,

    $$f_{T_a}(t)=F^{\prime }(t)=-2\phi \left(\frac{a}{\sqrt{t}}\right)\cdot a\cdot (-\frac{1}{2}{t}^{-3/2})=\frac{a}{t^{3/2}}\phi \left(\frac{a}{\sqrt{t}}\right)=\frac{a}{\sqrt{2\pi }{t}^{3/2}}{e}^{-a^2/(2t)},t>0.$$

    It is non-negative. As $t\downarrow 0$, $a/\sqrt{t}\to \infty$ so $F(t)\to 2(1-1)=0$; as $t\to \infty$, $a/\sqrt{t}\to 0$ so $F(t)\to 2(1-\frac{1}{2})=1$. A non-decreasing function running from $0$ to $1$ whose derivative is $f_{T_a}$ means ${\int }_0^{\infty }{f}_{T_a}=1$, so this is a genuine density. (Near $t=0$ the factor $e^{-a^2/(2t)}$ crushes the $t^{-3/2}$: the path almost never reaches $a$ immediately.)

    (b) The tail is too fat for a mean. $E(T_a)={\int }_0^{\infty }t{f}_{T_a}(t)dt=\frac{a}{\sqrt{2\pi }}{\int }_0^{\infty }{t}^{-1/2}{e}^{-a^2/(2t)}dt$. For $t\ge a^2$ the exponent satisfies $\frac{a^2}{2t}\le \frac{1}{2}$, so $e^{-a^2/(2t)}\ge e^{-1/2}$, and dropping the part of the integral below $a^2$,

    $$E(T_a)\ge \frac{a{e}^{-1/2}}{\sqrt{2\pi }}{\int }_{a^2}^{\infty }{t}^{-1/2}dt=\frac{a{e}^{-1/2}}{\sqrt{2\pi }}\underset{R\to \infty }{\lim }2(\sqrt{R}-a)=\infty .■$$

    So the level is reached with probability $1$ but the expected time to reach it is infinite, for every $a>0$: the density decays like $t^{-3/2}$, and $t\cdot t^{-3/2}=t^{-1/2}$ is not integrable at infinity.

    (c) Equal distribution functions. For $t>0$,

    $$P(a^2{T}_1\le t)=P(T_1\le \frac{t}{a^2})=2(1-\Phi (\frac{1}{\sqrt{t/a^2}}\left)\right)=2(1-\Phi (\frac{a}{\sqrt{t}}\left)\right)=P(T_a\le t),$$

    so $T_a\stackrel{d}{=}{a}^2{T}_1$. $■$ The scaling property explains it: ${\tilde{W}}_s=a^{-1}{W}_{a^{2}s}$ is again a standard Brownian motion, and ${\tilde{W}}_s=1$ exactly when $W_{a^{2}s}=a$. So the first time $\tilde{W}$ hits $1$ is $T_a/a^2$, and since $\tilde{W}$ is a standard Brownian motion that time has the law of $T_1$. Doubling the level quadruples the time, because Brownian motion spreads like $\sqrt{t}$.

    (d) Reflect, and integrate over the hitting time. Condition on $T_a=s\le t$. By the strong Markov property the process restarts at level $a$ as a fresh Brownian motion, so $W_t-a$ is symmetric about $0$ given $T_a=s$. Finishing at or below $0$ is $\{W_t-a\le -a\}$ and finishing at or above $2a$ is $\{W_t-a\ge a\}$, mirror images, hence equally likely given $T_a=s$. Integrating over $s\in [0,t]$,

    $$P(T_a\le t,W_t\le 0)=P(T_a\le t,W_t\ge 2a)=P(W_t\ge 2a),$$

    the last step because $W_t\ge 2a>a$ forces $T_a\le t$ by continuity of the path (the intermediate value theorem). Finally $\{T_a\le t\}=\{M_t\ge a\}$, so $P(M_t\ge a,W_t\le 0)=P(W_t\ge 2a)=1-\Phi (2a/\sqrt{t})$. This is the reflection principle with the endpoint tracked: the ordinary principle reflects paths ending anywhere below $a$; here only those ending below $0$, which reflect to paths ending above $2a$.

    With $a=1$ and $t=1$, and $P(T_1\le 1)=2(1-\Phi (1))$ from the reflection principle,

    $$P(W_1\le 0\mid T_1\le 1)=\frac{P(M_1\ge 1,W_1\le 0)}{P(M_1\ge 1)}=\frac{1-\Phi (2)}{2(1-\Phi (1))}.$$

    Sanity check on the size: $1-\Phi (2)\approx 0.023$ and $2(1-\Phi (1))\approx 0.317$, so the conditional probability is about $0.07$, far below the unconditional $P(W_1\le 0)=\frac{1}{2}$. A path known to have reached $1$ is very unlikely to have fallen all the way back past $0$ by time $1$, and the formula quantifies exactly how unlikely.

12. 12.Proof · 10 min

    A component's lifetime $\tau$ is uniform on $[0,2]$ years. It is replaced instantly on failure, the replacement's lifetime is an independent copy, and so on. Let $N(t)$ count replacements by time $t$, so $N$ is a renewal process with $T_n={\tau }_1+\cdots +{\tau }_n$ and $N(t)=\max \{n:T_n\le t\}$.

    1. (a)Prove that $N$ does not have independent increments, by computing $P(N(2)-N(1)=0\mid N(1)=0)$ and $P(N(2)-N(1)=0)$ exactly and showing they differ. Show also that the increments are not stationary.
    2. (b)Find the distribution of $N(2)$ exactly, together with $E(N(2))$, and compare both with the Poisson process of the same long-run rate.
    3. (c)Compute $P(\tau >\frac{3}{2}\mid \tau >1)$ and $P(\tau >\frac{1}{2})$. Explain how their disagreement is the root cause of the failure in (a), which defining properties of a Lévy process $N$ therefore lacks, and what the elementary renewal theorem nevertheless guarantees about $N(t)/t$.
    Show the solution to question 12

    (a) One conditional probability is forced to zero. The event $\{N(1)=0\}$ is $\{{\tau }_1>1\}$. But ${\tau }_1\le 2$ always, so on this event $T_1={\tau }_1\in (1,2]$: the first replacement is certain to fall in the second year. Hence

    $$P(N(2)-N(1)=0\mid N(1)=0)=0.$$

    Unconditionally, no replacement in $(1,2]$ can only happen with ${\tau }_1\le 1$ (the case ${\tau }_1>1$ was just excluded) and then $T_2={\tau }_1+{\tau }_2>2$, since $T_1\le 1$ and every later $T_n\ge T_2$. Condition on ${\tau }_1=u\in [0,1]$, which has density $\frac{1}{2}$: we need ${\tau }_2>2-u$, of probability $\frac{2-(2-u)}{2}=\frac{u}{2}$. Therefore

    $$P(N(2)-N(1)=0)={\int }_0^1\frac{u}{2}\cdot \frac{1}{2}du=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}\ne 0.$$

    The increment over $(1,2]$ has a different law given the increment over $(0,1]$ than it has unconditionally, so the two increments are dependent. $■$

    Not stationary either: $P(N(1)-N(0)=0)=P({\tau }_1>1)=\frac{1}{2}$, while $P(N(2)-N(1)=0)=\frac{1}{8}$. Two intervals of the same length carry different increment laws.

    (b) Counting–waiting duality for a renewal process. From $N(t)=\max \{n:T_n\le t\}$ the identity $\{N(2)\ge n\}=\{T_n\le 2\}$ holds exactly as for the Poisson process. Now $({\tau }_1,\dots ,{\tau }_n)$ is uniform on the cube $[0,2{]}^n$ of volume $2^n$, and the region $\{u_1+\cdots +u_n\le 2,u_i\ge 0\}$ is a simplex of volume $2^n/n!$ that lies entirely inside the cube (each coordinate is automatically at most $2$). So

    $$P(N(2)\ge n)=\frac{2^n/n!}{2^n}=\frac{1}{n!},P(N(2)=n)=\frac{1}{n!}-\frac{1}{(n+1)!},$$

    giving $P(N(2)=0)=0$, $P(N(2)=1)=\frac{1}{2}$, $P(N(2)=2)=\frac{1}{3}$, $P(N(2)=3)=\frac{1}{8}$, and so on. The mean, by the tail-sum formula,

    $$E(N(2))=\sum _{n\ge 1}P(N(2)\ge n)=\sum _{n\ge 1}\frac{1}{n!}=e-1.$$

    The long-run rate is $1/\mu =1$ per year (see (c)), so the comparison process is Poisson with $\lambda =1$, for which $N(2)\sim$ Poisson$(2)$: $P(N(2)=0)=e^{-2}\approx 0.135$ against the true $0$ (a component cannot last beyond two years), and $E(N(2))=2$ against the true $e-1\approx 1.72$. Same long-run rate, quite different law at $t=2$.

    (c) Memorylessness fails, and that is the whole story.

    $$P(\tau >\frac{3}{2}\mid \tau >1)=\frac{P(\tau >\frac{3}{2})}{P(\tau >1)}=\frac{1/4}{1/2}=\frac{1}{2},P(\tau >\frac{1}{2})=\frac{3}{4}.$$

    A memoryless lifetime would make these equal; here a component that has survived a year is more likely to fail in the next half-year than a new one is, because it is wearing out. That is exactly the information $N(1)$ carries about the future: $N(1)=0$ says the current component is at least a year old, which changes the law of what happens in $(1,2]$ — the dependence exhibited in (a). Independent increments would require the future after any fixed time to be blind to the age of the current component, which is memorylessness, which only the exponential has.

    So $N$ lacks both independent increments and stationary increments, two of the three clauses of the Lévy definition (it does start at $0$). This is the general fact that a renewal process is Lévy only when its gaps are exponential, seen in one concrete case.

    What survives is the long-run behaviour. The gaps are i.i.d., strictly positive, with finite mean $\mu =\frac{0+2}{2}=1$, so by the elementary renewal theorem $N(t)/t\to 1/\mu =1$ almost surely: one replacement per year in the long run, about $40$ over a $40$-year horizon, even though every exact Poisson formula is wrong for this process at every finite $t$.