# Chapter 21 Homework 1: Properties of Stochastic Process: Problems and Tentative Solutions

Exercise 21.1 (Weakly stationary process) Consider real valued random variable $$A_i,B_i$$, $$i=1,\cdots,k$$ such that $$E(A_i)=E(B_i)=0$$ and $$Var(A_i)=Var(B_i)=\sigma^2>0$$ for $$i=1,\cdots,k$$. Moreover, assume they are mutually uncorrelated, that is, $$E(A_iA_l)=E(B_iB_l)=0$$, for $$i\neq l$$, and $$E(A_iB_l)=0$$, for all $$i,l$$. Define the stochastic process $$X=\{X_t,t\in\mathcal{R}\}$$ by $$X_t=\sum_{i=1}^k(A_i\cos(\omega_i t)+B\sin(\omega_i t))$$, where $$\omega_i$$ are real constants $$i=1,\cdots,k$$. Show that $$X$$ is weakly stationary.

Proof. Firstly, consider the expectation, for any $$t\in\mathcal{R}$$, we have $$$\begin{split} E(X_t)&=E(\sum_{i=1}^k(A_i\cos(\omega_it)+B_i\sin(\omega_it)))\\ &=\sum_{i=1}^k\cos(\omega_it)E(A_i)+\sin(\omega_it)E(B_i)=0 \end{split} \tag{21.1}$$$ As a function of $$t$$, $$E(X_t)$$ is a constant.

Then, for the covariance between $$X_t$$ and $$X_{t+r}$$ of any $$t,t+r\in\mathcal{R}$$, since $$E(X_t)=E(X_{t+r})=0$$, we have $$$\begin{split} Cov&(X_t,X_{t+r})=E(X_tX_{t+r})\\ &=E[\{\sum_{i=1}^k(A_i\cos(\omega_it)+B_i\sin(\omega_it))\}\times\{\sum_{i=1}^k(A_i\cos(\omega_i(t+r))+B_i\sin(\omega_i(t+r))\}]\\ &=\sum_{i=1}^kE(A_i^2)\cos(\omega_it)\cos(\omega_i(t+r))+E(B_i^2)\sin(\omega_it)\sin(\omega_i(t+r))\\ &=\sum_{i=1}^k\sigma_i^2\cos(\omega_ir) \end{split} \tag{21.2}$$$ which is a function of $$r$$ only. Thus, the stochastic process $$X_t$$ is weakly stationary.

Exercise 21.2 (Weakly but not strongly stationary process) Consider a discrete-time, real-valued stochastic process $$X=\{X_n: n\geq 1\}$$ defined by $$X_n=\cos(nU)$$, where $$U$$ is uniformly distributed on $$(-\pi,\pi)$$. Show that $$X$$ is weakly stationary but not strongly stationary.

Proof. Firstly, consider the expectation of $$X_n$$ for any $$n\in\mathbb{N}_+$$, we have $$$\begin{split} E(X_n)&=E(\cos(nU))\\ &=\int_{-\pi}^{\pi}\cos(nu)\cdot\frac{1}{2\pi}du=0 \end{split} \tag{21.3}$$$ which is a constant as a function of $$n$$.

Then, for the covariance between $$X_n$$ and $$X_{n+m}$$ with any $$n\in\mathbb{N}_+$$ and $$m\in\mathbb{N}$$, $$E(X_n)=E(X_{n+m})=0$$, so $$$\begin{split} Cov&(X_n,X_{n+m})=E(X_tX_{t_r})\\ &=\int_{-\pi}^{\pi}\cos(nu)\cos((n+m)u)\cdot\frac{1}{2\pi}du\\ &=\frac{1}{4\pi}[\int_{-\pi}^{\pi}(\cos((2n+m)u)+\cos(mu))du]\\ &=\frac{1}{4\pi}[\frac{\sin((2n+m)u)}{2n+m}|_{-\pi}^{\pi}+\frac{\sin(mu)}{m}|_{-\pi}^{\pi}]\\ &=0 \end{split} \tag{21.4}$$$ which is a function of $$m$$. Therefore, the stochastic process is weakly stationary.

The process is not strongly stationary. For example, consider the event $$[X_1\leq -1+\epsilon,X_3\geq 1-\epsilon]$$ for some small $$\epsilon>0$$. The event has zero probability because $$X_1\leq 1-\epsilon$$ happens when $$U$$ is near $$-\pi$$ or $$\pi$$, but $$X_3\geq 1-\epsilon$$ happens when $$U$$ is near $$-\frac{2\pi}{3},0$$ or $$-\frac{2\pi}{3}$$. Therefore, we can pick small enough $$\epsilon$$ such that those regions are not overlapped. However, the event $$[X_2\leq -1+\epsilon,X_4\geq 1-\epsilon]$$ have positive probability because we can pick $$U$$ around $$\frac{\pi}{2}$$ to make both conditions hold.

Exercise 21.3 (Find correlation function from spectral density) Consider a weakly stationary process $$X=\{X_t:t\in\mathcal{R}\}$$ with zero mean and unit variance. Find the correlation function of $$X$$ if the spectral density function $$f$$ of $$X$$ is given by:

1. $$f(u)=0.5\exp(-|u|)$$, $$u\in\mathcal{R}$$
1. $$f(u)=\phi(\alpha^2+u^2)^{-1}$$, $$u\in\mathcal{R}$$
1. $$f(u)=\frac{1}{2}\sigma(\pi\alpha)^{-1}\exp(-u^2/(4\alpha))$$, $$u\in\mathcal{R}$$

Proof. Since we have variance equal to 1, the correlation function is just covariance function. Now using the one-to-one correspondence between the spectral density and covariance function $$$c(t)=\int_{-\infty}^{\infty}\exp(itu)f(u)du \tag{21.5}$$$ we can get the correlation function for each case.

For part (a), we have $$$\begin{split} c(t)&=\int_{-\infty}^{\infty}0.5\exp(-|u|+itu)du\\ &=E_{U_1}(\exp(itu))=\frac{1}{1+t^2} \end{split} \tag{21.6}$$$ where since $$U_1\sim Laplace(0,1)$$ we have the last equation. Therefore, in this case, $$c(t)=\frac{1}{1+t^2}$$.

For part(b), we have $$$\begin{split} c(t)&=\int_{-\infty}^{\infty}\frac{\phi\exp(itu)}{\alpha^2+u^2}du\\ &=\frac{\phi\pi}{\alpha}E_{U_2}(\exp(itu))=\frac{\phi\pi}{\alpha\exp(\alpha|t|)} \end{split} \tag{21.7}$$$ where since $$U_2\sim Cauchy(0,\alpha)$$ we have the last equation. Therefore, $$c(t)=\frac{\phi\pi}{\alpha\exp(\alpha|t|)}$$.

Finally, for part (c) $$$\begin{split} c(t)&=\int_{-\infty}^{\infty}\frac{\sigma\exp(itu-u^2/(4\alpha))}{2\pi\alpha}du\\ &=\frac{\sigma}{\sqrt{\pi\alpha}}E_{U_3}(\exp(itu))=\frac{\sigma\exp(-\alpha t^2)}{\sqrt{\pi\alpha}} \end{split} \tag{21.8}$$$ where since $$U_3\sim N(0,2\alpha)$$ we have the last equation. Therefore, $$c(t)=\frac{\sigma\exp(-\alpha t^2)}{\sqrt{\pi\alpha}}$$.

Exercise 21.4 (Strong and weak stationarity for Gaussian process) A stochastic process $$X=\{X_t:t\in\mathcal{R}\}$$ is called the Gaussian process with mean $$\mu$$ and covariance kernel $$\kappa(\cdot,\cdot)$$ if for any $$t_1,\cdots,t_n$$ and any $$n\geq 1$$, $$(X_{t_1},\cdots,X_{t_n})^T\sim N(\mu\mathbf{1},\mathbf{K})$$, where $$\mathbf{K}=(\kappa(t_i,t_j))_{i,j=1}^n$$. Show that the strong and weak stationarity are equivalent for a Gaussian process.

Proof. For any stochastic process, the strong stationary implies weak stationary. Therefore, for Gaussian process, we only need to show the weak stationary implies strong stationary.

Consider a weakly stationary Gaussian process $$X_t$$ with mean function $$m(\cdot)$$ and covariance function $$C(\cdot,\cdot)$$. By definition, any finite dimensional distributions of it is $$$F(X_1,\cdots,X_k)=\Phi(m(\mathbf{X}),C(\mathbf{X})) \tag{21.9}$$$ where $$\Phi(m(\mathbf{X}),C(\mathbf{X}))$$ represent the distribution function of a multivariate normal distribution with mean vector $$m(\mathbf{X})=(m(X_1),\cdots,m(X_k))^T$$ and covariance matrix $$C(\mathbf{X})$$, whose $$i,j$$th term is $$C(X_i,X_j)$$ for $$1\leq i,j\leq k$$. Similarly, the distribution for $$X_{1+t_0},\cdots,X_{k+t_0}$$ is also multivariate normal with mean vector $$m_{t_0}(\mathbf{X})=(m(X_{1+t_0}),\cdots,m(X_{k+t_0}))^T$$ and covariance matrix $$C_{t_0}(\mathbf{X})$$, whose $$i,j$$th term is $$C(X_{i+t_0},X_{j+t_0})$$. From the weakly stationary assumption, we have $$m(\mathbf{X})=m_{t_0}(\mathbf{X})$$ and $$C(\mathbf{X})=C_{t_0}(\mathbf{X})$$, and since they are all multivariate normal distribution, we have $$$F(X_1,\cdots,X_k)=F(X_{1+t_0},\cdots,X_{k+t_0}) \tag{21.10}$$$ for any $$k\in\mathbb{N}_+$$ and $$t_0\in\mathcal{T}$$. Thus, the Gaussian process is also strongly stationary.

Exercise 21.5 (Necessary and sufficient condition for a Gaussian process to be Markovian) By definition, a continuous-time real-valued stochastic process $$X=\{X_t:t\in\mathcal{R}\}$$ is called a Markov process if for all $$n$$, for all $$x_1,\cdots,x_n$$, and all increasing sequences $$t_1<\cdots<t_n$$ of index points, $$$Pr(X_{t_n}\leq x|X_{t_1}=x_1,\cdots,X_{t_{n-1}}=x_{n-1})=Pr(X_{t_{n}}\leq x|X_{t_{n-1}}=x_{n-1}) \tag{21.11}$$$ Let $$Z$$ be a real valued Gaussian process. Show that $$Z$$ is a Markov process if and only if $$$E(Z_{t_n}|Z_{t_1}=x_1,\cdots,Z_{t_{n-1}}=x_{n-1})=E(Z_{t_{n}}|Z_{t_{n-1}}=x_{n-1}) \tag{21.12}$$$

Proof. ($$\Longleftarrow$$) This is trivial since if $$Z$$ is a Markov process, then by definition for all $$n$$ and for all $$x,x_1,\cdots,x_{n-1}$$ and any sequence of index points $$t_1<\cdots<t_n$$, we have $$Pr(Z_{t_n}\leq x|Z_{t_1}=x_1,\cdots,Z_{t_{n-1}}=x_{n-1})=Pr(Z_{t_n}\leq x|Z_{t_{n-1}}=x_{n-1})$$. Then of course $$E(Z_{t_n}|Z_{t_1}=x_1,\cdots,Z_{t_{n-1}}=x_{n-1})=E(Z_{t_n}|Z_{t_{n-1}}=x_{n-1})$$. We have the necessarity.

($$\Longrightarrow$$) Since the f.d.d.s. of Guassian process is multivariate normal distribution, we can actually write down the explict form of the distribution of $$Z_{t_n}|Z_{t_1},\cdots,Z_{t_{n-1}}$$ and $$Z_{t_n}|Z_{t_{n-1}}$$ using the properties of multivariate normal distribution. Actually, let us denote the joint distribution of $$(Z_{t_n},Z_{t_{n-1}})$$ as $$$\begin{pmatrix} Z_{t_n}\\Z_{t_{n-1}}\end{pmatrix}\sim N(\begin{pmatrix} \mu_{t_n}\\ \mu_{t_{n-1}}\end{pmatrix},\begin{pmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22}\end{pmatrix}) \tag{21.13}$$$ Then, $$Z_{t_n}|Z_{t_{n-1}}=x_{n-1}\sim N(\mu_{t_n}+\Sigma_{12}\Sigma_{22}^{-1}(x_{n-1}-\mu_{n-1}),\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21})$$ for any $$x_{n-1}$$.

Similarly, denote the joint distribution of $$(Z_{t_n},\mathbf{Z}_{t_{n-1}})$$ where $$\mathbf{Z}_{t_{n-1}}=(Z_{t_{n-1}},\cdots,Z_{t_1})$$ as $$$\begin{pmatrix} Z_{t_n}\\\mathbf{Z}_{t_{n-1}}\end{pmatrix}\sim N(\begin{pmatrix} \mu_{t_n}\\ \boldsymbol{\mu}_{t_{n-1}}\end{pmatrix},\begin{pmatrix} \Lambda_{11} & \Lambda_{12}\\ \Lambda_{21} & \Lambda_{22}\end{pmatrix}) \tag{21.14}$$$ Then, $$Z_{t_n}|\mathbf{Z}_{t_{n-1}}=\mathbf{x}_{n-1}\sim N(\mu_{t_n}+\Lambda_{12}\Lambda_{22}^{-1}(\mathbf{x}_{n-1}-\boldsymbol{\mu}_{n-1}),\Lambda_{11}-\Lambda_{12}\Lambda_{22}^{-1}\Lambda_{21})$$ for any vector $$\mathbf{x}_{n-1}=(x_1,\cdots,x_{n-1})$$.

Now, from the assumption, we know $$$\mu_{t_n}+\Sigma_{12}\Sigma_{22}^{-1}(x_{n-1}-\mu_{n-1})=\mu_{t_n}+\Lambda_{12}\Lambda_{22}^{-1}(\mathbf{x}_{n-1}-\boldsymbol{\mu}_{n-1}) \tag{21.15}$$$ for any $$x_{n-1}$$ and $$\mathbf{x}_{n-1}$$. Then specifically we can choose $$x_{n-1}=\mu_{n-1}+\Sigma_{21}$$ and $$\mathbf{x}_{n-1}=\boldsymbol{\mu}_{n-1}+\Lambda_{21}$$ and subsitituted in () we have $$$\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}=\Lambda_{12}\Lambda_{22}^{-1}\Lambda_{21} \tag{21.16}$$$ from which we can see the two distributions in (21.13) and (21.14) are identical. Therefore, the condition for Markov process is satisfied and $$Z$$ is indeed a Markov process. We have the sufficiency.

Exercise 21.6 (Condition for Markovian) Show that any stochastic process $$X=\{X_n,n=0,\cdots\}$$ with independent increments is a Markov process.

Proof. Consider a stochastic process $$X=\{X(n,\omega):n\in\mathbb{N}_+,\omega\in\Omega\}$$ that has independent increments. Then for any $$n\geq 1$$ and any $$s$$, we have $$$\begin{split} &P(X_n=s|X_0=x_0,X_1=x_1,\cdots,X_{n-1}=x_{n-1})\\ &=P(X_n-X_{n-1}=s-x_{n-1}|X_0=x_0,X_1-X_0=x_1-x_0,\\ &X_2-X_1=x_2-x_1,\cdots,X_{n-1}-X_{n-2}=x_{n-1}-x_{n-2})\\ &=P(X_n-X_{n-1}=s-x_{n-1})\quad (by\,independence)\\ &=P(X_n-X_{n-1}=s-x_{n-1}|X_{n-1}=x_{n-1})\\ &=P(X_n=s|X_{n-1}=x_{n-1}) \end{split} \tag{21.17}$$$ Therefore, by defintion of Markov process, we know that $$X$$ is actually a Markov process.

Exercise 21.7 (View Brownian motion as a special Gaussian process) Let $$W=\{W_t:t\geq 0\}$$ be a Brownian motion (see Definition 2.5). Show that a Brownian motion can be viewed as a Gaussian process with mean 0 and $$Cov(W_s,W_t)=\min\{s,t\}$$.

Proof. Consider any finite dimensional distribution of $$W$$, denoted as $$F_{W_{t_1},\cdots,W_{t_k}}(w_{t_1},\cdots,w_{t_k})$$, w.l.o.g. we can assume $$t_1<\cdots<t_k$$ and for simplicity of notation, we can surpass $$t$$ and just denote the f.d.d.s. as $$F_{W_1,\cdots,W_k}(w_1,\cdots,w_k)$$. Firstly, from the independent increment property and $$W_t-W_s\sim N(0,t-s)$$ for $$0\leq s\leq t$$, since $$W_1,\cdots,W_k$$ can be expressed as a linear combination of independent normally distributed random variable $$W_1,W_2-W_1,\cdots,W_k-W_{k-1}$$, we have the f.d.d.s. of $$W$$ is normally distributed. The mean is $$E(W_t)=0$$ and variance $$Var(W_t)=t$$, while the covariance is, assume w.l.o.g. $$s<t$$ $$$\begin{split} Cov(W_sW_t)&=E(W_sW_t)=E(W_s(W_t-W_s+W_s))\\ &=E(W_s^2)+E(W_s)E(W_t-W_s)=s \end{split} \tag{21.18}$$$ Thus, the Brownian motion is a Gaussian process with mean function 0 and covariance function $$Cov(W_s,W_t)=\min\{s,t\}$$.

Exercise 21.8 (Moments of Brownian motion) Show that for a Brownian motion $$E(|W_s-W_t|^{2n})=C_n|s-t|^n$$, where $$C_n=\frac{2n!}{2^nn!}$$.

Proof. Suppose $$Y\sim N(0,\nu)$$, we compute $$E(|Y|^{2n})$$ as follows: $$$\begin{split} E(|y|^{2n})&=\frac{1}{\sqrt{2\pi\nu}}\int_{-\infty}^{\infty}|y|^{2n}\exp(-\frac{y^2}{2\nu})dy\\ &=\frac{2}{\sqrt{2\pi\nu}}\int_{0}^{\infty}y^{2n}\exp(-\frac{y^2}{2\nu})dy\quad (y=\sqrt{2\nu z})\\ &=\frac{2}{\sqrt{2\pi\nu}}\int_{0}^{\infty}(2\nu z)^{n}\exp(-\frac{y^2}{2\nu})\frac{\nu dz}{\sqrt{2\nu z}}\\ &=\frac{\nu^n2^n}{\pi}\int_{0}^{\infty}z^{n-\frac{1}{2}}e^{-z}dz\\ &=\frac{\nu^n2^n}{\pi}\Gamma(n+\frac{1}{2})\\ &=\frac{(2n)!}{2^nn!}\nu^n=C_n\nu^n \end{split} \tag{21.19}$$$

Now since $$W_s-W_t$$ follows $$N(0,|s-t|)$$, using the above result we immediately have $$E(|W_s-W_t|^{2n})=C_n|s-t|^n$$ where $$C_n=\frac{(2n)!}{2^nn!}$$.