### 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Random Walks

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## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Random Walks in 1D

Consider a “random walker” on a one-dimensional lattice. Time will be discrete, and at every time step the walker will take a step in one of the possible two directions, with equal probability. What is the mean value of, say, coordinate $x$, after $N$ steps? From symmetry, it is clearly zero. But we certainly don’t expect the walker to be precisely at the origin after a large number of steps. Its typical magnitude away from the origin may be found by calculating the RMS (root-mean-square) of the trajectories, i.e., the second moment of the distance from the origin.

Let’s say the walker was at position $x$ after $N$ steps. After an additional step, it could be in one of two possible positions. Averaging the squared distance between these two possibilities, conditioned on being at position $x$ at time $N$, gives us
$$\frac{1}{2}\left[(x+1)^{2}+(x-1)^{2}\right]=x^{2}+1 .$$
Upon averaging over the value of $x^{2}$ at time $N$, we find that
$$\left\langle x_{N+1}^{2}\right\rangle=\left\langle x_{N}^{2}\right\rangle+1$$
therefore, repeating the argument $N-1$ times, we find that
$$\left\langle x_{N}^{2}\right\rangle=N$$

This implies that the typical distance from the origin scales like $\sqrt{N}$. What about the position distribution? If $N$ is even, then it is clear that the probability to be a distance $M$ from the origin after $N$ steps is zero for odd $M$, and for even $M$ it is
$$p_{M}=\frac{1}{2^{N}}\left(\begin{array}{l} N \ R \end{array}\right)=\frac{1}{2^{N}} \frac{N !}{R !(N-R) !}$$
where $R$ is the number of steps to the right, thus $R-(N-R)=2 R-N=M$. We can now evaluate this probability for $N \gg M$ using Stirling’s formula, which provides an (excellent) approximation for $N$ !, namely $N ! \approx \sqrt{2 \pi N}\left(\frac{N}{e}\right)^{N}$. This leads to
\begin{aligned} p_{M} & \approx \frac{1}{\sqrt{2 \pi}} \frac{1}{2^{N}} \frac{N^{N+1 / 2}}{R^{R+1 / 2}(N-R)^{N-R+1 / 2}} \ &=\frac{1}{\sqrt{2 \pi}} e^{-N \log (2)+(N+1 / 2) \log (N)-(R+1 / 2) \log (R)-(N-R+1 / 2) \log (N-R)} \end{aligned}
We can proceed by using our assumption that $N \gg M$, implying that $R$ is approximately equal to $N / 2$.

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Derivation of the Diffusion Equation for Random Walks

We shall now approach the problem with more generality, following nearly precisely the derivation by Einstein. We will work in $3 \mathrm{D}$ but the approach would be the same

in any dimension. The approach will have discrete time steps, but the step direction will be a continuous random variable, described by a probability distribution $g(\vec{\Delta})$ (here $\vec{\Delta}$ is a vector describing the step in the 3D space – not to be confused with the Laplacian operator!). We will not limit the random walker to a lattice, though it is possible to implement such a scenario by taking $g(\vec{\Delta})$ to be a sum of $\delta$-functions (can you see how?).

We will seek to obtain the probability distribution $p(\vec{r})$, i.e., $p(\vec{r}) d V$ will be the probability to find the particle in a volume $d V$ around the point $\vec{r}$ (at some point in time corresponding to a a given number of steps). If the original problem is cast on a lattice, this distribution will be relevant to describe the coarse grained problem, when we shall zoom-out far enough such that we will not care about the details at the level of the lattice constant.

If at time $t$ the probability distribution is described by $p(\vec{r}, t)$, let us consider what it will be a time $\tau$ later, where $\tau$ denotes the duration of each step. As you can guess, in a realistic scenario the time of a step is non-constant, and $\tau$ would be the mean step time. Thus, we haven’t lost too much in making time discrete – but we did make an assumption that a mean time exists. In Chapter 7 we will revisit this point, and show that in certain situations when the mean time diverges, we can get subdiffusion (slower spread of the probability distribution over time compared with diffusion).

To find $p(\vec{r}, t+\tau)$, we need to integrate over all space, and consider the probability to have the “right” jump size to bring us to $\vec{r}$. This leads to
$$p(\vec{r}, t+\tau)=\int p(\vec{r}-\vec{\Delta}, t) d^{3} \vec{\Delta} g(\vec{\Delta}) .$$
To proceed, we will Taylor expand $p$, assuming that the probability distribution is smooth on the scale of the typical jump. If we expand it to first order, we will find that
$$p(\vec{r}-\vec{\Delta}) \approx p(\vec{r})-(\nabla p) \cdot \vec{\Delta} .$$
If the diffusion process is isotropic in space, there is no difference between a jump in the $\pm \vec{\Delta}$ direction, so the integral associated with the second term trivially vanishes:
$$\int((\nabla p) \cdot \vec{\Delta}) g(\vec{\Delta}) d^{3} \vec{\Delta}=0 .$$
This means we have to expand to second order:
$$p(\vec{r}-\vec{\Delta}) \approx p(\vec{r})-(\nabla p) \cdot \vec{\Delta}+\left.\frac{1}{2} \sum_{i, j} \frac{\partial^{2} p}{\partial x_{i} \partial x_{j}}\right|{\vec{r}} \Delta{i} \Delta_{j}$$

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Markov Processes and Markov Chains

Let us approach our analysis of the random walker from a new perspective, which will be easier to generalize to other networks. Within our model for diffusion on a 1D lattice, the probability to go to site $j$ does not depend on the history of the random walker, but only on its current state – this is an example of a Markov process (named after Andrey Markov). A familiar childhood example of a Markov process is the game Chutes and Ladders (as well as many other board games – see also Problem 2.15). But we emphasize that whether a process is Markovian depends on the space: Consider, for example, a game where we throw a die at each round, and keep track of the running sum. This is clearly a Markov process – knowing the current sum determines the probabilities to go to the next states. But what about a game in which we reset the sum to zero each time we get two identical numbers in a row? In this case, the process is memory dependent, so it is non-Markovian. But, if we work in the space where a state is defined by a pair, the running sum, and the result of the last throw – then the process becomes Markovian again. It is also worth noting that in cases where time steps are discrete, the process is referred to as a Markov chain. In what follows we will deal with Markov chains, though in the next chapter we will study Markov processes with continuous time. For an extended discussion of Markov chains, see Feller (2008).

Let us denote by $\boldsymbol{P}$ the matrix describing the transition probabilities, i.e., $\boldsymbol{P}{i j}$ will be the probability to go from $i$ to $j$. The key insight to note is that for a Markov chain if we know the current probabilities to be at every site, which we will denote by the vector $\vec{p}$, and we know the matrix $P$, we can easily find the vector of probabilities to be at every site after an additional move. By the definition of the matrix $P$, the probability to be in the $i$ th site after an additional move is given by $$p{i}^{n+1}=\sum_{j} p_{j}^{n} \boldsymbol{P}_{j i}$$

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Random Walks in 1D

12[(X+1)2+(X−1)2]=X2+1.

⟨Xñ+12⟩=⟨Xñ2⟩+1

⟨Xñ2⟩=ñ

p米=12ñ(ñ R)=12ññ!R!(ñ−R)!

p米≈12圆周率12ñññ+1/2RR+1/2(ñ−R)ñ−R+1/2 =12圆周率和−ñ日志⁡(2)+(ñ+1/2)日志⁡(ñ)−(R+1/2)日志⁡(R)−(ñ−R+1/2)日志⁡(ñ−R)

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Derivation of the Diffusion Equation for Random Walks

p(r→,吨+τ)=∫p(r→−Δ→,吨)d3Δ→G(Δ→).

p(r→−Δ→)≈p(r→)−(∇p)⋅Δ→.

∫((∇p)⋅Δ→)G(Δ→)d3Δ→=0.

p(r→−Δ→)≈p(r→)−(∇p)⋅Δ→+12∑一世,j∂2p∂X一世∂Xj|r→Δ一世Δj

p一世n+1=∑jpjn磷j一世

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