**Homework 2**

(Due on: Wed, April 15, 8:00PM)

**Problem 1**. A stochastic process is defined as \[x_{k}=x_{k-1}+m_k,\ \ \ x_{0}=0\] Find the expected value \(E\{x_k\}\) and the variance \(E\{(x_k-\overline{x}_k)^2\}\) of this process.

ANS:

equation 1. can rewrite as: \[x_{k}=x_{k-1}+m_k=x_{k-2}+m_k+m_k=x_{k-2}+2(m_k) \\
x_{k}=x_{k-3}+m_k+2(m_k)=x_{k-3}+3(m_k) \\
x_{k}=...=x_{0}+k(m_k), \ \ \ where\ x_{0}=0 \\
x_{k}=k(m_k)\]

Therefore, the expected value can compute as: \[E\{x_{k}\}=E\{x_{k-1}+m_k\}=E\{k(m_k)\}=k*E\{m_k\}=k*\int_{-\infty}^\infty m_k\ p(m_k)\ dm_k \\ E\{x_{k}\}=k\ \int_{-1/2}^{1/2} m_k\ p(m_k)\ dm_k=k\ \int_{-1/2}^{0} m_k\ 4\ (m_k+1/2)\ dm_k+\int_{0}^{1/2} m_k\ (2-4m_k)\ dm_k \\ E\{x_{k}\}=k\ (\frac{4}{3}m_k^3+m_k^2 \left| {_{-1/2}^{0} } \right.+\frac{-4}{3}m_k^3+m_k^2 \left| {_{0}^{1/2} } \right.)=k\ \frac{1}{6}\\\] And, the variance can compute as: \[E\{(x_k-\overline{x}_k)^2\}=\int_{-\infty}^\infty (x_k-\overline{x}_k)^2\ p(x)\ dx, \ \ \ where\ x_k=k\ m_k\ and\ \overline{x}_k=E\{x_{k}\}=k\ \frac{1}{6} \\ E\{(x_k-\overline{x}_k)^2\}=\int_{-\infty}^\infty k\ (m_k-\frac{1}{6})^2\ p(m_k)\ dm_k\\ E\{(x_k-\overline{x}_k)^2\}=k(\ \int_{-1/2}^{0} (m_k-\frac{1}{6})^2\ 4\ (m_k+1/2)\ dm_k+\int_{0}^{1/2} (m_k-\frac{1}{6})^2\ (2-4m_k)\ dm_k) \\ E\{(x_k-\overline{x}_k)^2\}=k\ (m_k^4+\frac{2}{9}m_k^3-\frac{5}{18}m_k^2+18m_k \left| {_{-1/2}^{0} } \right.-m_k^4+\frac{10}{9}m_k^3-\frac{7}{18}m_k^2+18m_k \left| {_{0}^{1/2} } \right.)\\ E\{(x_k-\overline{x}_k)^2\}=k\ \frac{1}{18} \\\]

**Problem 2** (only for graduate students) It is known that if \(x\) is a random variable with a pdf \(p_x(x)\), i.e., \(x \sim p_x(x)\), and \(y\) is a random variable \(y \sim p_y(y)\), then \(z=x+y\) is a random variable \(z \sim p_z(z)\), where \(p_z(z)=p_x(x)*p_y(y)\). The symbol \(*\) denotes the convolution, i.e., \[p_z(z)=p_z(x+y)=\int_{-\infty}^\infty p_y(z-x)p_x(x)dx\]

(a) If \(m_k \sim p(m_k)\), where \(p(m_k)\) is depicted in the figure, and \(m_k=n+h\), figure out \(p_n(n)\) and \(p_h(h)\).

ANS:

Becasue \(m_k=n+h\) and \(p_z(z)=p_x(x)*p_y(y)\), I could derive that \(p(m_k)=p(n+h)=p_n(n)*p_h(h)\). \(p(m_k)\) is a symmetrical triangular-shaped distribution, and, it is the result of convolution of two identical uniform distribution. Therefor, \(p_n(n)\) and \(p_h(h)\) can be two identical uniform distribution as: \[p_n(n) = \left\{
\begin{array}{l}
0, \ \ \ n < a \\
1/(b-a), \ \ \ a \le n \le b \\
0, \ \ \ \ n > b \\
\end{array} \right.
\ \ where\ a=-1/4,\ b=1/4\]

(b) Use your conclusion from (a) to write a MATLAB code that generates the sequence from Problem 1. Generate the sequence 100 (or more) times and based on these sequences, verify the results obtained in Problem 1.

ANS:

First, I use MATLAB to create uniform-shaped function \(p_n(n)\) (as figure 1). Then, I use MATLAB function (conv) to do the convultion of \(p_n(n)\) and itself (as figure 2). Finally, I generate the sequence of 100 times \(x_k=x_{k-1}+m_k\), and I got \(x_k\) about 100. And, according to the regult of Problem that \(E\{x_k\}=k\ \frac{1}{6}=100*\frac{1}{6}=16.667\).