# EE 505 - Robert Marks.org by ert554898

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```									            EE 505
Random Processes - Example Random Processes

Example RP’s
Example Random Processes
 Gaussian

Recall Gaussian pdf
1                 
          1           ( x  m )T K 1 ( x  m )
fX (x)                  e 2
2 
n/2    1/ 2
|K|

Let Xk=X(tk) , 1  k  n. Then if, for all n, the
corresponding pdf’s are Gaussian, then the
RP is Gaussian.
The Gaussian RP is a useful model in signal
processing.
Flip Theorem
Let A take on values of +1 and -1 with equal
probability
Let X(t) have mean m(t) and
autocorrelation RX
Let Y(t)=AX(t)
Then Y(t) has mean zero and
autocorrelation RX
Multiple RP’s
X(t) & Y(t)
   Independence
(X(t1), X(t2), …, X(tk ))
is independent to
(Y(1), Y( 2), …, Y( j ))

…for All choices of k and j and
all sample locations

Multiple RP’s
X(t) & Y(t)
   Cross Correlation
RXY(t,  )=E[X(t)Y()]
   Cross-Covariance
CXY(t,  )= RXY(t,  ) - E[X(t)] E[Y()]
   Orthogonal: RXY(t,  ) = 0
   Uncorrelated: CXY(t,  ) = 0
   Note: Independent Uncorrelated, but not
the converse.
Example RP’s
Multiple Random Process
Examples
 Example

X(t) = cos(t+), Y(t) = sin(t+),
Both are zero mean.
Cross Correlation=?
p.338

Example RP’s
Multiple Random Process Examples
   Signal + Noise
X(t) = signal, N(t) = noise
Y(t) = X(t) + N(t)
If X & N are independent,RXY=?               p.338
Note: also, var Y = var X + var N

var X
SNR 
var N

Example RP’s
Multiple Random Process Examples (cont)
 Discrete time RP’s
X[n]
Mean
Variance
Autocorrelation
Autocovariance
   Discrete time i.i.d. RP’s
 Bernoulli RP’s Binomial RP’s                   p.340
   Binary vs. Bipolar
 Random Walk       p.341-2

Autocovariance of Sum Processes
n
Sn   X [ k ]
k 1
X[k]’s are iid.

E [ S n ]  nX

var[ S n ]  n var( X )

Autocovariance=?

Autocovariance of Sum Processes
                      
CS (n, k )  E ( S n  S n )(S k  S k )
 E ( S n  nX )(S k  kX )
 n
             k
             

 E  ( X i  X ) ( X j  X )
 i 1
              j 1
             

When i=j, the answer is var(X). Otherwise, zero.
How many cases are there where i = j?
min( n, k )  CS ( n , k )  min( n , k ) var( X )

Autocovariance of Sum Processes
For Bernoulli sum process,
var( X )  pq
CS ( n , k )  min( n , k ) pq

For Bipolar case
var( X )  4 pq
CS ( n , k )  4 min( n , k ) pq

Continuous Random Processes
Poisson Random Process
   Place n points randomly on line of length T
t

T
   Choose any subinterval of length t.
   The probability of finding k points on the
subinterval is
 n  k nk     t
Pr[ k po int s ]    p q
k         ;p
              T
Continuous Random Processes
Poisson Random Process (cont)
   The Poisson approximation: For k big and p small…

 n  k nk     np (np)
k
Pr[k points]    p q
k         e
                    k!
(nt / T ) k
 e  nt / T
k!

Continuous Random Processes
The Poisson Approximation…
   For n big and p small (implies k << n since p k/n<<1)
 n  k nk     np (np)
k
 p q
k         e
                    k!
Here’s why…
n       n!        n(n  1)(n  2)...(n  k  1) n k
 
 k  k!(n  k )!                               
                               k!                k!

q n k  (1  p)nk  (1  p)n  (e p )n
Continuous Random Processes
Poisson Random Process (cont)
 n  k nk          (np) k
Pr[k points]    p q
k          e  np
                     k!
k
 nt / T (nt / T )
e
k!
   Let n  such that =n/T = frequency of points
remains constant.

 t (t )
k
Pr[ k points on interval t ]  e
k!
Continuous Random Processes
Poisson Random Process (cont)
 t   ( t )   k
Pr[ k points on interval t ]  e
k!
   This is a Poisson process with parameter
 occurrences per unit time
   Examples: Modeling
 Popcorn
 Rain (Both in space and time)
t
 Passing cars
 Shot noise
 Packet arrival times

Continuous Random Processes
Poisson Counting Process
X(t )

Poisson Points
 t   ( t )   k
Pr[ X ( t )  k ]  e
k!
Continuous Random Processes
Recall for Poisson RV with parameter a
k
a   (a)
Pr[ X  k ]  e                           X  var( X )  a
k!
Poisson Counting Process Expected Value is
thus

E [ X ( t )]  t

Continuous Random Processes
The Poisson Counting Process is independent
increment process. Thus, for   t and j  i,

Pr[ X (t )  i, X ( )  j ]
 Pr[X (t )  i, X ( )  X (t )  j  i ]
 Pr[X (t )  i ] Pr[ X ( )  X (t )  j  i ]


t i et   (  t ) j i e ( t )
i!                        ( j  i )!

Continuous Random Processes
t           
Autocorrelation: If  > t
R X ( t , )  E X ( t ) X (  )

 E X ( t ) X (  )  X ( t )  X 2 ( t )             
 E X ( t ) X (  )  X ( t )  E X 2 ( t )            
 E X ( t )E X (  )  X ( t )  E X ( t )         2


 t  (   t )  t  t          2

  t  t
2
RX ( t , )  2t   min( t , )
Continuous Random Processes
Autocovariance of a Poisson sum process

C X ( t , )  R X ( t , )  E  X ( t )E  X (  )
  t   min( t , )  t  
2

  min( t , )

Continuous Random Processes
Other RP’s related to the Poisson process
   Random telegraph signal

X(t )

Poisson Points
Poisson Random Processes
   Random telegraph signal
2 |t |
E[ X (t )]  e
2|t  |
C X ( t , )  e

PROOF…

Poisson Random Processes
   Random telegraph signal. For t>0,

E[ X (t )]  1 PrX (t )  1  (1)  Pr[X (t )  1]
 Prnumber of points on ( 0,t) is even
 Prnumber of points on ( 0,t) is odd

Poisson Random Processes
   Random telegraph signal. For t>0,

Prnumber of points on ( 0,t) is even
 t
 (t ) 2 (t ) 4      
 e 1                   ...

   2!      4!         
 e t cosht 

Poisson Random Processes
 Random telegraph signal. For t>0.
Similarly…

Prnumber of points on ( 0,t) is odd

 t       (t )3 (t )5      
e        t                ...
        3!     5!        
                         
 e  t sinh(t )

Poisson Random Processes
Random telegraph signal. For t>0.
Thus
E[ X (t )]  1 PrX (t )  1  (1)  Pr[ X (t )  1]
 Prnumber of points on ( 0 ,t) is even
 Prnumber of points on ( 0 ,t) is odd
 e  t cosh(t )  sinh(t )
 e  2t ; t  0
2 |t |
For all t…       E[ X (t )]  e
Poisson Random Processes
   Random telegraph signal. For t > ,
X()
1

-1                   1     X(t)

-1

Pr[ X (t ) X ( )  1]  PrX (t )  1, X ( )  1
 PrX (t )  1, X ( )  1
 PrX (t )  1 | X ( )  1Pr X ( )  1
 PrX (t )  1 | X ( )  1Pr X ( )  1
Poisson Random Processes
   Random telegraph signal. For t > ,

PrX (t )  1 | X ( )  1
 PrX (t )  1 | X ( )  1
 Prnumber of points on ( , t ) is even
 e  ( t  ) cosh (t   ) 
Thus…
PrX (t )  1, X ( )  1
 PrX (t )  1 | X ( )  1Pr X ( )  1
 cosh (t   ) e  (t  ) cosh( )e 

Poisson Random Processes
   Random telegraph signal. For t > ,
PrX (t )  1, X ( )  1
 PrX (t )  1 | X ( )  1Pr X ( )  1
X()
 cosh (t   ) e t cosh( )                        1

And…
-1              1   X(t)
PrX (t )  1, X ( )  1
-1
 PrX (t )  1 | X ( )  1Pr X ( )  1
 cosh (t   ) e t sinh( )

Poisson Random Processes
   Random telegraph signal. For t > .
Onward…

PrX (t )  1 | X ( )  1
 Pr X (t )  1 | X ( )  1
 Prnumber of points on ( , t ) is odd
 e  ( t  ) sinh  (t   ) 

Poisson Random Processes
   Random telegraph signal. For t > .
PrX (t )  1, X ( )  1
 PrX (t )  1 | X ( )  1Pr X ( )  1                  X()
1
 sinh  (t   ) e t cosh( )
And…
-1              1   X(t)
PrX (t )  1, X ( )  1
-1
 PrX (t )  1 | X ( )  1Pr X ( )  1
 sinh  (t   ) e t sinh( )

Poisson Random Processes
   Random telegraph signal. For t > .
PrX (t )  1, X ( )  1
 PrX (t )  1 | X ( )  1Pr X ( )  1                  X()
1
 sinh  (t   ) e t cosh( )
And…
-1              1   X(t)
PrX (t )  1, X ( )  1
-1
 PrX (t )  1 | X ( )  1Pr X ( )  1
 sinh  (t   ) e t sinh( )

Poisson Random Processes
   Random telegraph signal. For t > .                                    X()
1

-1              1   X(t)

RX (t , )  EX (t ) X ( )                                          -1

 1 PrX (t ) X ( )  1  1 PrX (t ) X ( )  1

 cosh (t   ) e t cosh( )  sinh (t   ) e t sinh( )           
 sinh (t   ) e   t
cosh( )  cosh (t   ) e t       sinh( )

In general… X (t , )  RX (t , )  X (t ) X ( )  e2 |t  |
C
Continuous Random Processes
Other RP’s related to the Poisson process
 Poisson point process, Z(t)
Let X(t) be a Poisson sum process. Then
d
Z ( t )  X ( t )    ( t  Sn )
dt         n
Z( t )

pp.352

Poisson Points
Continuous Random Processes
Other RP’s related to the Poisson process
 Shot Noise, V(t)
Z(t)                     V(t)
h(t)
V ( t )   h( t  S n )
n
V( t )

pp.352
Poisson Points
Continuous Random Processes
Wiener Process
   Assume bipolar Bernoulli sum process with jump
bilateral height h and time interval 
   E[X(t)]=0; Var X(n) = 4npqh 2 = nh 2
   Take limit as h 0 and   0 keeping  = h 2 / 
constant and t = n .
   Then Var X(t)   t
   By the central limit theorem, X(t) is Gaussian with zero
mean and Var X(t) =  t
   We could use any zero mean process to generate the
Wiener process.
p.355
Continuous Random Processes
Wiener Processes:  =1

Continuous Random Processes
Wiener processes in finance
S= Price of a Security.  = inflationary force.
If there is no risk…interest earned is proportional to investment.
dS
dS( t )  S ( t )dt      S
t          dt
Solution is S( t )  S0e
With “volatility” , we have the most commonly used model in
finance for a security:
dS( t )  S( t )dt  S( t )dV ( t )
V(t) is a Wiener process.