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10.1.6 Solved Problems

Problem

Let $Y_1$, $Y_2$, $Y_3$, $\cdots$ be a sequence of i.i.d. random variables with mean $EY_i=0$ and $\textrm{Var}(Y_i)=4$. Define the discrete-time random process $\big\{X(n), n \in \mathbb{N}\big\}$ as \begin{align*} X(n)=Y_1+Y_2+\cdots+Y_n, \quad \textrm{for all }n \in \mathbb{N}. \end{align*} Find $\mu_X(n)$ and $R_X(m,n)$, for all $n,m \in \mathbb{N}$.

  • Solution
    • We have \begin{align*} \mu_X(n) &=E[X(n)] \\ &=E[Y_1+Y_2+\cdots+Y_n]\\ &=E[Y_1]+E[Y_2]+\cdots+E[Y_n]\\ &=0. \end{align*} Let $m \leq n$, then \begin{align*} R_X(m,n) &=E[X(m)X(n)] \\ &=E\left[X(m)\big(X(m)+Y_{m+1}+Y_{m+2}+\cdots+Y_n\big)\right]\\ &=E[X(m)^2]+E[X(m)]E[Y_{m+1}+Y_{m+2}+\cdots+Y_n]\\ &=E[X(m)^2]+0\\ &=\textrm{Var}\big(X(m)\big)\\ &=\textrm{Var}\big(Y_1\big)+\textrm{Var}\big(Y_2\big)+\cdots+\textrm{Var}\big(Y_m\big)\\ &=4m. \end{align*} Similarly, for $m \geq n$, we have \begin{align*} R_X(m,n) &=E[X(m)X(n)] \\ &=4n. \end{align*} We conclude \begin{align*} R_X(m,n) &=4 \min (m,n). \end{align*}

Problem

For any $k \in \mathbb{Z}$, define the function $g_k(t)$ as \begin{align} \nonumber g_k(t) = \left\{ \begin{array}{l l} 1 & \quad k \lt t \leq k+1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align} Now, consider the continuous-time random process $\big\{X(t), t \in \mathbb{R}\big\}$ defined as \begin{align*} X(t)=\sum_{k=-\infty}^{+\infty} A_k g_k(t), \end{align*} where $A_1$, $A_2$, $\cdots$ are i.i.d. random variables with $EA_k=1$ and $\textrm{Var}(A_k)=1$. Find $\mu_X(t)$, $R_X(s,t)$, and $C_X(s,t)$ for all $s,t \in \mathbb{R}$.

  • Solution
    • Note that, for any $k \in \mathbb{Z}$, $g(t)=0$ outside of the interval $(k,k+1]$. Thus, if $k \lt t \leq k+1$, we can write \begin{align*} X(t)=A_k. \end{align*} Thus, \begin{align*} \mu_X(t) &=E[X(t)] \\ &=E[A_k]=1. \end{align*} So, $\mu_X(t)=1$ for all $t \in \mathbb{R}$. Now consider two real numbers $s$ and $t$. If for some $k \in \mathbb{Z}$, we have \begin{align*} k \lt s,t \leq k+1, \end{align*} then \begin{align*} R_X(s,t) &=E[X(s)X(t)] \\ &=E[A_k^2]=1+1=2. \end{align*} On the other hand, if $s$ and $t$ are in two different subintervals of $\mathbb{R}$, that is if \begin{align*} k \lt s\leq k+1, \quad \textrm{and} \quad l \lt t \leq l+1, \end{align*} where $k$ and $l$ are two different integers, then \begin{align*} R_X(s,t) &=E[X(s)X(t)] \\ &=E[A_k A_l]=E[A_k]E[A_l]=1. \end{align*} To find $C_X(s,t)$, note that if \ \begin{align*} k \lt s,t \leq k+1, \end{align*} then \begin{align*} C_X(s,t) &=R_X(s,t)-E[X(s)]E[X(t)] \\ &=2-1\cdot 1=1. \end{align*} On the other hand, if \begin{align*} k \lt s\leq k+1, \quad \textrm{and} \quad l \lt t \leq l+1, \end{align*} where $k$ and $l$ are two different integers, then \begin{align*} C_X(s,t) &=R_X(s,t)-E[X(s)]E[X(t)] \\ &=1-1\cdot 1=0. \end{align*}

Problem

Let $X(t)$ be a continuous-time WSS process with mean $\mu_X=1$ and \begin{align} \nonumber R_X(\tau) = \left\{ \begin{array}{l l} 3-|\tau| & \quad -2 \leq \tau \leq 2 \\ & \quad \\ 1 & \quad \text{otherwise} \end{array} \right. \end{align}

  1. Find the expected power in $X(t)$.
  2. Find $E\left[\bigg(X(1)+X(2)+X(3)\bigg)^2\right]$.

  • Solution
      1. The expected power in $X(t)$ at time $t$ is $E[X(t)^2]$, which is given by \begin{align*} R_X(0) &=3. \end{align*}
      2. We have \begin{align*} E\left[\bigg(X(1)+X(2)+X(3)\bigg)^2\right] &=E\bigg[X(1)^2+X(2)^2+X(3)^2\\ &+2X(1)X(2)+2X(1)X(3)+2X(2)X(3)\bigg]\\ &=3R_X(0)+2R_X(-1)+2R_X(-2)+2R_X(-1)\\ &=3 \cdot 3+ 2 \cdot 2+ 2 \cdot 1 +2 \cdot 2\\ &=19. \end{align*}

Problem

Let $X(t)$ be a continuous-time WSS process with mean $\mu_X=0$ and \begin{align} \nonumber R_X(\tau) = \delta(\tau), \end{align} where $\delta(\tau)$ is the Dirac delta function. We define the random process $Y(t)$ as \begin{align*} Y(t)=\int_{t-2}^{t} X(u) du. \end{align*}

  1. Find $\mu_Y(t)=E[Y(t)]$.
  2. Find $R_{XY}(t_1,t_2)$.

  • Solution
      1. We have \begin{align*} \mu_Y(t)&=E\left[\int_{t-2}^{t} X(u) du\right] \\ &=\int_{t-2}^{t} E[X(u)] \; du\\ &=\int_{t-2}^{t} 0 \; du\\ &=0. \end{align*}
      2. We have \begin{align*} R_{XY}(t_1,t_2)&=E\left[X(t_1)\int_{t_2-2}^{t_2} X(u) du\right] \\ &=E\left[\int_{t_2-2}^{t_2} X(t_1)X(u) du\right] \\ &=\int_{t_2-2}^{t_2} R_X(t_1-u) \; du\\ &=\int_{t_2-2}^{t_2} \delta(t_1-u) \; du\\ &=\left\{ \begin{array}{l l} 1 & \quad t_2-2 \lt t_1 \lt t_2\\ & \quad \\ 0 & \quad \textrm{otherwise}\\ \end{array} \right. \end{align*}

Problem

Let $X(t)$ be a Gaussian process with $\mu_X(t)=t$, and $R_X(t_1,t_2)=1+2t_1t_2$, for all $t,t_1,t_2 \in \mathbb{R}$. Find $P\big(2X(1)+X(2) \lt 3\big)$.

  • Solution
    • Let $Y=2X(1)+X(2)$. Then, $Y$ is a normal random variable. We have \begin{align*}%\label{} EY &=2E[X(1)]+E[X(2)]\\ &=2 \cdot 1+2=4. \end{align*} \begin{align*}%\label{} \textrm{Var}(Y) &=4\textrm{Var}\big(X(1)\big)+\textrm{Var}\big(X(2)\big)+4 \textrm{Cov}\big(X(1),X(2)\big). \end{align*} Note that \begin{align*}%\label{} \textrm{Var}\big(X(1)\big)&=E[X(1)^2]-E[X(1)]^2\\ &=R_X(1,1)- \mu_X(1)^2\\ &=1+2\cdot 1 \cdot 1 -1=2. \end{align*} \begin{align*}%\label{} \textrm{Var}\big(X(2)\big)&=E[X(2)^2]-E[X(2)]^2\\ &=R_X(2,2)-\mu_X(2)^2\\ &=1+2\cdot 2 \cdot 2 -4=5. \end{align*} \begin{align*}%\label{} \textrm{Cov}\big(X(1),X(2)\big)&=E[X(1)X(2)]-E[X(1)]E[X(2)]\\ &=R_X(1,2)-\mu_X(1)\mu_X(2)\\ &=1+2\cdot 1 \cdot 2 -1 \cdot 2=3. \end{align*} Therefore, \begin{align*}%\label{} \textrm{Var}(Y) &=4\cdot 2+5+4\cdot 3=25. \end{align*} We conclude $Y \sim N(4,25)$. Thus, \begin{align*}%\label{} P\big(Y \lt 3\big)&=\Phi \left(\frac{3-4}{5} \right)\\ &=\Phi(-0.2) \approx 0.42 \end{align*}

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