Random Processes in Communication and Control/M-Sep14

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PMF $P[X=x]=P[s:X(s)=x] \!\,$

a) $P_X(x) \geq 0 \forall x \in S_x \!\,$

b) $\sum_{x \in S_X} P_X(x)=1 \!\,$

c) event $B \subset S_X \!\,$

$P[B]= \sum_{x\in B} P_X(x) \!\,$

Some Useful Random Variables

Bernoulli R.V

$P_X(x)= \begin{cases} 1-p & x=0 \\ p & x=1 \\ 0 & o.w. \end{cases} \!\,$

$0 \leq p \leq 1 \!\,$ success probability

Example

1) Flip a coin $X= \!\,$# of H

2) Manufacture a Chip $X=\!\,$# of acceptable chips

3) Bits you transmit successfully by a modem

Geometric Random Variable

Number of trails until (and including) a success for an underlying Bernoulli

$P_X(x)=p(1-p)^{x-1} \quad x=1,2,3,\ldots \!\,$

Example

1) Repeated coin flips $X= \!\,$# of tosses until H

2) Manufacture chips $X= \!\,$3 of chips produced until an acceptable time

Binomial R.V

"# of successes in n trials"

$P_x(x)= {n \choose x} p^x (1-p)^{n-x} \quad x=0,1,\ldots n \!\,$

Example

1) Flip a coin n times. $X= \!\,$ # of heads.

2) Manufacture n chips. $X= \!\,$ # of acceptable chips.

Note: Binomial $X=Y_1+ Y_2+Y_3 + \ldots+Y_n \!\,$ where $Y_1+Y_2+Y_3 \ldots+ Y_n \!\,$ are independent Bernoulli trials

Note: n=1; Binomial=Bernoulli; $X=Y_1 \!\,$

Pascal R.V

"number of trials until (and including) the kth success with an underlying Bernoulli"

$P_X(x)= {x-1 \choose k-1} p^k (1-p)^{x-k} \quad x=k,k+1,k+2,\ldots \!\,$

where ${x-1 \choose k-1} \!\,$ is $k-1 \!\,$ successes in $x-1\!\,$ trials

Note: Pascal $X=Y_1+Y_2+\ldots+Y_k \!\,$ where $Y_1,Y_2,\ldots Y_k \!\,$ are geometric R.V.

Note: K=1 Pascal=Geometric

Example

$X=\!\,$# of flips until the kth H

Discrete Uniform R.V.

$P_X(x)= \begin{cases} \frac{1}{b-a+1} & x=a,a+1,\ldots,b \\ 0 & \mbox{otherwise} \end{cases} \!\,$

Example

1) Rolling a die. $a=1, b=6 \!\,$

$P_X(x)= \begin{cases} \frac16 & x=1,\ldots,6 \\ 0 & \mbox{otherwise} \end{cases} \!\,$

2) Flip a fair coin. $X \!\,$=# of H

$P_X(x)= \begin{cases} \frac12 & x=0 \\ \frac12 & x=1 \\ 0 & \mbox{ otherwise } \end{cases} \!\,$

Poisson R.V.

$P_X(x)= e^{-\alpha} \frac{\alpha^x}{x!} \quad x=0,1,2\ldots \!\,$

(Exercise) limiting case of binomial with $n \rightarrow \infty, p \rightarrow 0, np=\alpha \!\,$

PMF is a complete model for a random variable

Cumulative Distribution Function

$F_X(x)= P[X \leq x] = \sum_{x' \leq x} P_X(X=x') \!\,$

Like PMF, CDF is a complete description of random variable.

Example

Flip the coins $X= \!\,$# of H

$P_X(x)= \begin{cases} \frac14 & x=0 \\ \frac12 & x=1 \\ \frac12 & x=2 \\ 0 & \mbox{otherwise} \end{cases} \!\,$

$\begin{align} P[X \leq 0]&=P[X=0]\\ P[X \leq 0.5] &= P[X=0] \\ P[X \leq 1 ] &= \underbrace{P[X=0]}_{\frac14} + \underbrace{P[X=1]}_{\frac12} \\ \end{align} \!\,$

Properties of CDF

• a) $F_X(-\infty)=0 \Leftarrow F_X(-\infty)=P[X \leq -\infty] = 0 \!\,$

$F_X(\infty)=1 \Leftarrow F_X(\infty)=P[X \leq \infty]=1 \!\,$

"starts at 0 and ends at 1"

• b) For all $x' \geq x \!\,$, $F_X(x') \geq F_X(x) \!\,$

"non-decreasing in x"

$\begin{align} F_X(x') &\quad F_X(x) \\ P[X \leq x'] &\quad P[X \leq x ] \\ P[s: X(s) \leq x'] &\quad P[s:X(s) \leq x] \\ \{s: X(s) \leq x \} &\subset \{s: X(s) \leq x' \} \\ P[X \leq x] &\leq P[X \leq x']\\ F_X(x) &\leq F_X(x') \end{align} \!\,$

• c) For all $x,x' \!\,$

$P[x \leq X \leq x'] = F_X(x')-F_X(x) \!\,$

"probabilities can be found by difference of the CDF"

$\{ s: X(x) \leq x' \} = \{ s: X(x) \leq x \} \cup \{s: x \leq X(s) \leq x' \} \!\,$

$P[X \leq x' ] = P[X \leq x ] + P [x \leq X \leq x']\!\,$

• d) For all $x \!\,$,

$\lim_{\epsilon \rightarrow 0} F_X(x+\epsilon)=F_X(x) \!\,$

"CDF is right continuous"

• e) For $x_i \in S_X \!\,$

$F_X(x_i)- F_X(x_i - \epsilon) = P_X(x_i) \!\,$

"For a discrete random variable, there is a jump (discontinuity) in the CDF at each value $x_i \in S_X \!\,$. This jump equals $P_X(x_i) \!\,$

$P[x_i \leq x] - P[x_i \leq x_i - \epsilon]=P[x_i- \epsilon < x < x_i] = P[X=x_i] \!\,$

• f) $F_X(x)=F_X(x_i) \!\,$ for all $x_i \leq x \leq x_{i+1} \!\,$

"Between two jumps the CDF is constant"

$\begin{align} P[X \leq x] &= P[X \leq x_i \cup x_i < X \leq x ] \\ &= P[X \leq x_i + \underbrace{P[x_i < X \leq x ]}_0 \\ &= F_X(x_i) \end{align}$

• g) $P[X > x] = 1- \underbrace{F_X(x)}_{P[X \leq x]} \!\,$

Continuous Random Variables

outcomes uncountable many

Example

T: arrival of a partical

$S_T= \{t : 0 \leq t < \infty \} \!\,$

V: voltage

$S_V=\{v : -\infty < v < \infty \} \!\,$

$\theta \!\,$: angle

$S_{\theta}=\{ \theta: 0 \leq \theta \leq 2 \pi \} \!\,$

$X \!\,$: distance

$S_X= \{ X: 0 \leq x \leq 1 \} \!\,$

$P[x \in A] = \frac{1}{n } \rightarrow 0\!\,$

No PMF, $P[X=x]=0 \forall x \!\,$

Theorem

For any random variable (continuous or discrete)

• a) $F_X(-\infty)=0 \quad F_X(\infty)=1 \!\,$

• b) $F_X(x) \!\,$ is nondecreasing in $X \!\,$

• c) $P[x < X \leq x'] = F_X(x')-F_X(x) \!\,$

• d) $F_X(x) \!\,$ is right continuous

Example

$S_X=[0,1] \!\,$

$P_[x \in A]= P[ x\ in B]\!\,$ where A, B are intervals of the same length contained in [0,1]

$P[X \leq 1] =1 \Leftrightarrow F_X(1)=1 \!\,$

$P[X \leq 0] =0 \Leftrightarrow F_X(0)=0 \!\,$

$P[x_1 < x < x_2]=P[0 < x < x_2-x_1] \!\,$

(exercise)$F_X(x_2)-F_X(x_1)=F_X(x_2-x_1) \rightarrow F_x(x)=x \!\,$

Probability Density Function (PDF)

$f_X(x)= \frac{dF_x(x)}{dx} \!\,$

discrete: PMF <--> CDF (sum/difference)

continuous <---> (derivative/integral)

Theorem: Properties of PDF

• a) $f_X(x) \geq 0 \!\,$ ($F_X(x) \!\,$ is nondecreasing)

• b) $F_X(x) = \int_{\infty}^x f_X(x) dx \quad (F_X(\infty)=0)\!\,$

• c) $\int_{-\infty}^\infty f_X(x)dx=1 \quad (F_x(\infty)=1) \!\,$

Theorem

$\begin{align} P[x_1 \leq X \leq x_2] &= F_X(x_2)-F_X(x_1) \\ &= \int_{-\infty}^{x_2} f_X(x)dx - \int_{-\infty}^{x_1} f_X(x) dx \\ &= \int_{x_1}^{x_2} f_X(x) \end{align} \!\,$

Some useful continuous Random Variables

Uniform R.V

$f_X(x)= \begin{cases} \frac{1}{b-a} & a \leq x \leq b \\ 0 & \mbox{otherwise} \end{cases} \!\,$

Exponential R.V

$f_X(x)=ae^{-ax} \quad x \geq 0\!\,$

$F_x(x)=1-e^{-ax} \!\,$

Gaussian (Normal) R.V.

$\mathcal{N} (\mu, \sigma^2) \quad \infty < x< \infty\!\,$

$f_X(x)=\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \!\,$

$F_X(x) = \int_{-\infty^1}\frac{1}{\sqrt{2 \pi \sigma^2}}e^{-\frac{x-\mu}{2\sigma^2}} \!\,$