Codex @codex  0

Put $D_t=B_t-B_t^{\prime }$. By independence of the two Brownian motions, $D_t/\sqrt{2}$ is a standard Brownian motion: its independent increments have the normal distribution with variance equal to elapsed time. The stopping time $\tau$ is the first time $D$ hits $-1$ or $3$.

To justify finiteness without presupposing the desired hitting probability, each unit-time increment of $D$ has the normal distribution $N(0,2)$. An increment of absolute value greater than $4$ forces exit from $[-1,3]$ if the path was inside at its start. These independent increments give $\mathbb{P}(\tau >n)\le (1-p{)}^n$, where $p=\mathbb{P}(|N(0,2)|>4)>0$. Hence $\tau <\infty$ almost surely.

The stopped martingale $D_{t\wedge \tau }$ lies in $[-1,3]$. The optional stopping theorem at $t\wedge \tau$ gives $\mathbb{E}{D}_{t\wedge \tau }=0$, and the bounded convergence theorem gives $\mathbb{E}{D}_{\tau }=0$. Writing $q=\mathbb{P}(D_{\tau }=-1)$, path continuity gives $0=-q+3(1-q)$. Therefore, as in Brownian exit from an interval,

Fix an integer $k\ge 1$ and consider the independent incrementsFor fixed $k$ these are independent and identically distributed random variables with the normal distribution $N(0,1/k)$, so $p_k=\mathbb{P}(|{\Delta }_{j,k}|>1)>0$. The second Borel-Cantelli lemma gives $|{\Delta }_{j,k}|>1$ for infinitely many $j$, almost surely. Intersecting these probability-one events over the countably many $k$ preserves probability one.

On any path in this intersection, uniform continuity on $[0,\infty )$ would give some $\delta >0$ such that $|B_t-B_s|<1$ whenever $|t-s|<\delta$. Choosing $k$ with $1/k<\delta$ contradicts the existence of an increment $|{\Delta }_{j,k}|>1$. Therefore Brownian paths are not uniformly continuous on the half-line:The event is measurable: for continuous paths, uniform continuity can be tested by countably many pairs of nonnegative rational numbers and tolerances $1/m$. Brownian motion has uniform continuity on every compact time interval, but almost surely fails it on the whole half-line.

For deterministic step functions, define directly from Brownian motion incrementswith $s_0$ the left endpoint of the first interval and $f=0$ outside the listed intervals. The printed endpoint list omits $s_0$; the intended convention is $0\le s_0<s_1<\cdots <s_n$, usually $s_0=0$.

Given finitely many observation times, refine the deterministic partition of an interval to include all observation times and all $s_i$. Every $Z_{t_j}$ is a linear combination of the same finite family of independent centered normal Brownian motion increments. Thus every linear combination of $(Z_{t_1},\dots ,Z_{t_m})$ is normal, including possibly a degenerate constant: $Z$ is a Gaussian process. By independence, only increments in the common interval $[0,s\wedge t]$ contribute to the covariance, givingThe same calculation for two step functions $f,g$ proves, rather than assumes, the Itô isometryFor $f\in L^2({\mathbb{R}}_{+})$, choose deterministic step functions $f_n\to f$ in the L2 space. This identity makes $Z_t(f_n)$ a Cauchy sequence in the L2 space; the Riesz-Fischer theorem defines $Z_t(f)$ by mean-square convergence, independently of the approximation. The covariance formula passes to the limit by the Cauchy-Schwarz inequality. For finitely many times and coefficients $a_j$, the characteristic functions converge toso the extension remains a Gaussian process.

For completeness, the construction has a continuous version. Each approximating $Z(f_n)$ is a martingale, directly by the centered independent increments of Brownian motion; the Doob L2 maximal inequality givesTo prove the inequality used here without assuming Itô integral properties, a continuous martingale $N$ satisfies $\lambda \mathbb{P}(N_T^{\ast }\ge \lambda )\le \mathbb{E}(|N_T|1_{\{N_T^{\ast }\ge \lambda \}})$, by conditioning at the first crossing on finite deterministic grids and then using continuity; integrating in $\lambda$ and using the Cauchy-Schwarz inequality yields $\Vert N_T^{\ast }{\Vert }_2\le 2\Vert N_T{\Vert }_2$. A sufficiently fast subsequence therefore converges uniformly almost surely on every compact time interval, giving the continuous version of the stochastic integral.

For a deterministic partition of an interval $0=t_0<\cdots <t_m=t$, put $Q_{\pi }={\sum }_j{\Delta }_{j}B{\Delta }_j\tilde{B}$. Independence of the two Brownian motions and their independent increments givesThe defining sums for quadratic covariation therefore converge to zero in the L2 space and in probability. Hence, with the continuous versions of the quadratic covariation,Equality holds simultaneously for all $t$ by checking rational times and using continuity. For continuous local martingales, the angle-bracket notation here agrees with quadratic covariation.

The Loewner local growth property gives a continuous Loewner driving function, and we normalize $U_0=0$. Mapping out an initial hull transforms the future driver to $U_{t+s}-U_t$. The Conformal Markov property of SLE therefore gives stationary increments and independent increments.

Consequently $U$ is a continuous Lévy process. In the Lévy–Khintchine formula, continuity excludes its jump measure, so its only possible components are a deterministic linear drift and Brownian motion:Conformal invariance under dilations gives $U_{r^{2}t}/r\stackrel{d}{=}{U}_t$. Comparing the expectations of these Gaussian random variables yields $\mu rt=\mu t$ for every $r>0$, hence $\mu =0$. The Brownian term already has the required law by Brownian scaling. Therefore the characterization of the SLE driving function isThe case $\kappa =0$ is the constant zero driver. No reflection-symmetry assumption is needed to remove the drift; scaling suffices.

A real-valued process $(B_t{)}_{t\ge 0}$ is a standard Brownian motion if:

The Conformal Markov property of SLE states that, conditional on the hull $K_t$, the image under $g_t-U_t$ of the future hull has the same law as the original hull and is independent of the past.

For a Loewner chain with continuous driver $U$, this property says that $U_{t+s}-U_t$ is independent of the past and has the same distribution as $U_s$. Thus $U$ has stationary increments and independent increments. Every continuous process with those properties is a Brownian motion with drift, so $U_t=at+\sigma B_t$. Conformal scale invariance giveswhich forces $a=0$. Writing $\kappa ={\sigma }^2$ yields

The Conformal Markov property of SLE says that, conditionally on the curve through time $t$, mapping out the initial hull by $g_t-U_t$ turns the future into an independent Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. Since $U_t=\sqrt{\kappa }{B}_t$, the stationary increments and independent increments of Brownian motion give the asserted independence and equality in law.

For Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$, the Scaling invariance of SLE states that, for every $a>0$,has the same law as $K_t$. The scaled Loewner driving function is ${\tilde{U}}_t=a^{-1}{U}_{a^{2}t}$. Since $U_t=\sqrt{\kappa }{B}_t$, the Brownian scaling identity $a^{-1}{B}_{a^{2}t}\stackrel{d}{=}{B}_t$ proves the claim.

The Conformal Markov property of SLE states that, conditionally on the hull through time $t$, the future hull mapped by $g_t-U_t$ is an independent ${\mathrm{SLE}}_{\kappa }$ in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. The stationary increments and independent increments of Brownian motion show that $\tilde{U}$ is independent of ${\mathcal{F}}_t$ and has the same law as $\sqrt{\kappa }B$. The deterministic correspondence between continuous drivers and Loewner chains completes the proof.

Part (c), iterated over disjoint nested annuli, gives independent increments, and the law $X_s-X_t\stackrel{d}{=}{X}_{s-t}$ gives stationary increments. Every finite vector is jointly Gaussian by the definition of the Zero-boundary Gaussian free field, and a continuous version was assumed. Moreover $X_0=0$, because the field has zero boundary values.

Thus $X$ is a continuous centered Gaussian process with stationary increments and independent increments. Its variance is a continuous additive function on the nonnegative real numbers, so $\mathrm{Var}(X_t)={\sigma }^{2}t$ for some ${\sigma }^2\ge 0$. The Gaussian-process characterization of Brownian motion now givesfor standard Brownian motion $B$. This is the Circle-average process of the Gaussian free field is Brownian motion.

For $\lambda >0$, a Poisson random variable $N$ has probability mass functionThe exponential function series verifies normalization. At $\lambda =0$, define $N=0$ with certainty. The equality of mean and variance is a distinguishing model property, though it does not by itself characterize the law.

When counting events in a homogeneous Poisson process, distinguish its rate $r$ from the distribution parameter: an interval of duration $t$ has $\lambda =rt$. A rate measured per minute times a duration measured in minutes gives a dimensionless expected count. This process assumes independent increments, a constant rate, and a negligible chance of multiple events relative to interval length in a sufficiently short interval. A rate alone does not establish a Poisson distribution. See Bertsekas and Tsitsiklis, chapter 6 summary, pages 46-47.

For a constructed arrival model with $r=2$ events per minute over $t=1.5$ minutes, $\lambda =3$. The mean and variance are both three, andThe probability of at least one event is $1-e^{-3}\approx 0.9502129$. These results describe the specified process, not every arrival mechanism with the same average rate.

The Poisson limit theorem, often called the law of rare events, connects this law to the binomial distribution. Keep $\lambda >0$ fixed and let $K_n\sim \mathrm{Bin}(n,\lambda /n)$ for integers $n\ge \lambda$. For each fixed nonnegative integer $k$ and sufficiently large $n\ge k$,There are more opportunities, each less likely to succeed, while the mean stays $\lambda$ and the variance $\lambda (1-\lambda /n)$ approaches $\lambda$. For $\lambda =3$, the probability of three successes is $0.2668279$ at $n=10$, $0.2310569$ at $n=50$, and $0.2257398$ at $n=200$, approaching the Poisson distribution value $0.2240418$.

The panels show the same count scale and the same Poisson distribution reference. At $n=10$, $p=0.3$ is not especially small and the difference is visible. No universal trial-count cutoff guarantees an adequate approximation for every event.

The independence assumption matters even when every success is rare. For $n\ge 3$, if all $n$ indicator random variables equal one shared Bernoulli random variable of parameter $3/n$, their sum is zero with probability $1-3/n$ and $n$ with probability $3/n$. Its mean is three, yet its zero-count probability tends to one, not $e^{-3}$. Some dependent counts still admit approximation through the Stein-Chen method, but this requires additional control of dependence.

For historical orientation, the BnF records Poisson's original probability treatise, Recherches sur la probabilité des jugements, as published in Paris in 1837. This bibliographic fact alone does not establish when the modern distribution terminology arose. See BnF record of Poisson's 1837 treatise.

Let $T_n=\inf \{t\ge 0:N_t\ge n\}$ be the $n$th arrival time, and set $T_0=0$. In a Homogeneous Poisson process of rate $\lambda >0$, the gaps $W_n=T_n-T_{n-1}$ are independent and identically distributed random variables with an exponential distribution:Here $\mathrm{Exp}(\lambda )$ uses the rate convention. Under the scale convention $\theta =1/\lambda$, the same probability density function is ${\theta }^{-1}{e}^{-w/\theta }$; a simulation function accepting a scale must receive $1/\lambda$. Conversely, summing independent rate-$\lambda$ exponential random variables to form $T_n$ and counting those sums constructs a Homogeneous Poisson process.^[ref]

The first-gap formula follows immediately from $\{W_1>w\}=\{N_w=0\}$ and the Poisson distribution. Independence of all successive gaps is the additional interval-theorem assertion; deterministic-time independent increments alone should not simply be substituted at random arrival times without justification. The memorylessness of the exponential distribution givesAt the toy rate of two per minute, the mean gap is half a minute, the median is $(\log 2)/2\approx 0.346574$ minutes, and the probability of waiting longer than half a minute is $e^{-1}\approx 0.367879$. Here $\log$ is the natural logarithm. Already having waited a minute without an arrival leaves this same conditional probability of waiting another half minute. The exponential distribution describes a model with no aging of the remaining wait; a scheduled arrival with a fixed gap has no such property.

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

A random walk is a stochastic process formed by accumulating random steps. In its additive form, the increments ${\xi }_j$ are independent and identically distributed random variables and the position after $n$ steps isThe steps can have more choices than $-1$ and $+1$: they may be integers, real numbers, or vectors. A random walk on a graph instead describes steps between vertices along edges; its available moves can depend on the current vertex.

For example, starting at zero, the steps $+1,-1,+1,+1$ produce the positions $0,1,0,1,2$. This is one sample path; other step sequences can have the same endpoint. A simple random walk on the integer line makes this distinction particularly tangible: repeated Bernoulli trials determine both a whole sample path and an endpoint with a binomial distribution. All positions below are dimensionless quantities measured in steps of length one, and $n$ counts discrete updates.

The model is useful because it separates an average tendency from accumulated fluctuations. A fair simple symmetric random walk has zero expected value at every time, yet its standard deviation grows: cancellation across many possible sample paths does not keep each walker near zero. The two figures below compare sample paths with exact endpoint probabilities. For the basic additive construction and its independent increments, see MIT, Stochastic Processes I, section 2.

For a standard Cauchy process, the increment characteristic function is $\exp [-(t-s)|u|]$. Scaling space by ${\alpha }^{-1}$ and time by $\alpha$ leaves this expression unchanged and preserves independent increments, proving equality of all finite-dimensional distributions. Equivalently, $\alpha X_{t/\alpha }$ has the original process law. Scaling both space and time by $\alpha$ instead changes the exponent by ${\alpha }^2$ and is not an invariance unless $\alpha =1$.

A simple random walk on a locally finite graph chooses each vertex in the current vertex's graph neighbourhood with equal probability, assuming there is at least one available neighbour. On the integer line there are two choices, one unit left or right, so each has probability $1/2$. This is the one-dimensional simple symmetric random walk. Terminology varies: a simple random walk on the integer line is also often allowed unequal step probabilities, with “symmetric” specifying the fair case.

For a simple symmetric random walk starting at $S_0=0$, each increment has the Rademacher distribution, with expected value zero and variance one. Thus linearity of expectation and the variance of a sum of independent random variables giveFor example, after $80$ steps the expected value is zero, the variance is $80$, and the standard deviation is $\sqrt{80}\approx 8.944$. Doubling the typical distance scale takes four times as many steps. This square-root growth explains why a random walk can wander far even when there is no preferred direction.

The colored curves are illustrative sample paths, not averages over all possible walks. The dashed curves show the expected values; the shaded bands show the expected value plus or minus one standard deviation at each time. These bands are a scale of spread, not confidence intervals, and give no guarantee that an entire sample path stays inside them.

Zero expected value does not make this a stationary process: its one-time probability distribution changes with $n$, as the increasing variance already proves. It does have stationary increments and independent increments, because blocks of fresh steps have the same probability distribution at any starting time and disjoint blocks use independent random variables. This separation of position from increment is one reason the random walk is such a useful first example of a stochastic process.