Poisson Processes and Brownian Motion

(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.

(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.

(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.

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). $✓$

(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.

(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. $✓$

(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 ${\int }_0^1\frac{2u}{9}du=\frac{1}{9}$. $✓$

(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 $T_a$.

(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.

(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.

(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.)