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 $$\begin{equation} \begin{split} E(X_t)&=E(\sum_{i=1}^k(A_i\cos(\omega_it)+B_i\sin(\omega_it)))\\ &=\sum_{i=1}^k\cos(\omega_it)E(A_i)+\sin(\omega_it)E(B_i)=0 \end{split} \tag{21.1} \end{equation}$$ 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 $$\begin{equation} \begin{split} Cov&(X_t,X_{t+r})=E(X_tX_{t+r})\\ &=E[\{\sum_{i=1}^k(A_i\cos(\omega_it)+B_i\sin(\omega_it))\}\times\{\sum_{i=1}^k(A_i\cos(\omega_i(t+r))+B_i\sin(\omega_i(t+r))\}]\\ &=\sum_{i=1}^kE(A_i^2)\cos(\omega_it)\cos(\omega_i(t+r))+E(B_i^2)\sin(\omega_it)\sin(\omega_i(t+r))\\ &=\sum_{i=1}^k\sigma_i^2\cos(\omega_ir) \end{split} \tag{21.2} \end{equation}$$ 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 $$\begin{equation} \begin{split} E(X_n)&=E(\cos(nU))\\ &=\int_{-\pi}^{\pi}\cos(nu)\cdot\frac{1}{2\pi}du=0 \end{split} \tag{21.3} \end{equation}$$ 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 $$\begin{equation} \begin{split} Cov&(X_n,X_{n+m})=E(X_tX_{t_r})\\ &=\int_{-\pi}^{\pi}\cos(nu)\cos((n+m)u)\cdot\frac{1}{2\pi}du\\ &=\frac{1}{4\pi}[\int_{-\pi}^{\pi}(\cos((2n+m)u)+\cos(mu))du]\\ &=\frac{1}{4\pi}[\frac{\sin((2n+m)u)}{2n+m}|_{-\pi}^{\pi}+\frac{\sin(mu)}{m}|_{-\pi}^{\pi}]\\ &=0 \end{split} \tag{21.4} \end{equation}$$ 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. $f(u)=0.5\exp(-|u|)$, $u\in\mathcal{R}$
• 2. $f(u)=\phi(\alpha^2+u^2)^{-1}$, $u\in\mathcal{R}$
• 3. $f(u)=\frac{1}{2}\sigma(\pi\alpha)^{-1}\exp(-u^2/(4\alpha))$, $u\in\mathcal{R}$
     

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 $$\begin{equation} c(t)=\int_{-\infty}^{\infty}\exp(itu)f(u)du \tag{21.5} \end{equation}$$ we can get the correlation function for each case.

For part (a), we have $$\begin{equation} \begin{split} c(t)&=\int_{-\infty}^{\infty}0.5\exp(-|u|+itu)du\\ &=E_{U_1}(\exp(itu))=\frac{1}{1+t^2} \end{split} \tag{21.6} \end{equation}$$ 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 $$\begin{equation} \begin{split} c(t)&=\int_{-\infty}^{\infty}\frac{\phi\exp(itu)}{\alpha^2+u^2}du\\ &=\frac{\phi\pi}{\alpha}E_{U_2}(\exp(itu))=\frac{\phi\pi}{\alpha\exp(\alpha|t|)} \end{split} \tag{21.7} \end{equation}$$ 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) $$\begin{equation} \begin{split} c(t)&=\int_{-\infty}^{\infty}\frac{\sigma\exp(itu-u^2/(4\alpha))}{2\pi\alpha}du\\ &=\frac{\sigma}{\sqrt{\pi\alpha}}E_{U_3}(\exp(itu))=\frac{\sigma\exp(-\alpha t^2)}{\sqrt{\pi\alpha}} \end{split} \tag{21.8} \end{equation}$$ 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 $$\begin{equation} F(X_1,\cdots,X_k)=\Phi(m(\mathbf{X}),C(\mathbf{X})) \tag{21.9} \end{equation}$$ 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 $$\begin{equation} F(X_1,\cdots,X_k)=F(X_{1+t_0},\cdots,X_{k+t_0}) \tag{21.10} \end{equation}$$ 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, $$\begin{equation} Pr(X_{t_n}\leq x|X_{t_1}=x_1,\cdots,X_{t_{n-1}}=x_{n-1})=Pr(X_{t_{n}}\leq x|X_{t_{n-1}}=x_{n-1}) \tag{21.11} \end{equation}$$ Let $Z$ be a real valued Gaussian process. Show that $Z$ is a Markov process if and only if $$\begin{equation} E(Z_{t_n}|Z_{t_1}=x_1,\cdots,Z_{t_{n-1}}=x_{n-1})=E(Z_{t_{n}}|Z_{t_{n-1}}=x_{n-1}) \tag{21.12} \end{equation}$$

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 $$\begin{equation} \begin{pmatrix} Z_{t_n}\\Z_{t_{n-1}}\end{pmatrix}\sim N(\begin{pmatrix} \mu_{t_n}\\ \mu_{t_{n-1}}\end{pmatrix},\begin{pmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22}\end{pmatrix}) \tag{21.13} \end{equation}$$ 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 $$\begin{equation} \begin{pmatrix} Z_{t_n}\\\mathbf{Z}_{t_{n-1}}\end{pmatrix}\sim N(\begin{pmatrix} \mu_{t_n}\\ \boldsymbol{\mu}_{t_{n-1}}\end{pmatrix},\begin{pmatrix} \Lambda_{11} & \Lambda_{12}\\ \Lambda_{21} & \Lambda_{22}\end{pmatrix}) \tag{21.14} \end{equation}$$ 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 $$\begin{equation} \mu_{t_n}+\Sigma_{12}\Sigma_{22}^{-1}(x_{n-1}-\mu_{n-1})=\mu_{t_n}+\Lambda_{12}\Lambda_{22}^{-1}(\mathbf{x}_{n-1}-\boldsymbol{\mu}_{n-1}) \tag{21.15} \end{equation}$$ 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 $$\begin{equation} \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}=\Lambda_{12}\Lambda_{22}^{-1}\Lambda_{21} \tag{21.16} \end{equation}$$ 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 $$\begin{equation} \begin{split} &P(X_n=s|X_0=x_0,X_1=x_1,\cdots,X_{n-1}=x_{n-1})\\ &=P(X_n-X_{n-1}=s-x_{n-1}|X_0=x_0,X_1-X_0=x_1-x_0,\\ &X_2-X_1=x_2-x_1,\cdots,X_{n-1}-X_{n-2}=x_{n-1}-x_{n-2})\\ &=P(X_n-X_{n-1}=s-x_{n-1})\quad (by\,independence)\\ &=P(X_n-X_{n-1}=s-x_{n-1}|X_{n-1}=x_{n-1})\\ &=P(X_n=s|X_{n-1}=x_{n-1}) \end{split} \tag{21.17} \end{equation}$$ 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. $s<t$ $$\begin{equation} \begin{split} Cov(W_sW_t)&=E(W_sW_t)=E(W_s(W_t-W_s+W_s))\\ &=E(W_s^2)+E(W_s)E(W_t-W_s)=s \end{split} \tag{21.18} \end{equation}$$ 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: $$\begin{equation} \begin{split} E(|y|^{2n})&=\frac{1}{\sqrt{2\pi\nu}}\int_{-\infty}^{\infty}|y|^{2n}\exp(-\frac{y^2}{2\nu})dy\\ &=\frac{2}{\sqrt{2\pi\nu}}\int_{0}^{\infty}y^{2n}\exp(-\frac{y^2}{2\nu})dy\quad (y=\sqrt{2\nu z})\\ &=\frac{2}{\sqrt{2\pi\nu}}\int_{0}^{\infty}(2\nu z)^{n}\exp(-\frac{y^2}{2\nu})\frac{\nu dz}{\sqrt{2\nu z}}\\ &=\frac{\nu^n2^n}{\pi}\int_{0}^{\infty}z^{n-\frac{1}{2}}e^{-z}dz\\ &=\frac{\nu^n2^n}{\pi}\Gamma(n+\frac{1}{2})\\ &=\frac{(2n)!}{2^nn!}\nu^n=C_n\nu^n \end{split} \tag{21.19} \end{equation}$$

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 $C_n=\frac{(2n)!}{2^nn!}$.