# CE245 HW2

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 $$E\{x_k\}=k\ \frac{1}{6}=100*\frac{1}{6}=16.667$$.

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Figure 1. Uniform function $$p_n(n)$$