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    Devry
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    Mar-2010 - Oct-2016

Category > Programming Posted 22 May 2017 My Price 9.00

Generate a stream of i.i.d. random variables

Problem 1
Generate a stream of i.i.d. random variables U = {Un | 1 ≤ n ≤ N} which are

uniformly distributed over [0, 1], for each N = 1000, 5000, 10000.

(a) Use the resulting streams to draw the probability density histograms of U over

the intervals (0.1i, 0.1i + 0.1], for i = 0, 1, .. . , 9. Also, compare the histogram

levels with those derived analytically.

(b) Compute the sample means and variances for each of the three runs. Then,

compare the results with the mean and variance values derived analytically.

– when the variables {Xn | 1 ≤ n ≤ N} are observed, N > 1, the sample

mean µN and sample variance σN 2 are calculated as follows:

µN =

PN Xi
N
, σN 2 =

PN (Xi - µN)2

N - 1

.
Problem 2
Generate a stream of i.i.d. random variables T = {Tn | 1 ≤ n ≤ N} which

are exponentially distributed with parameter (intensity) λ = 0.1, for each N =

1000, 5000, 10000.

(a) Use the resulting streams to compute and exhibit the probability density histograms for T, over the intervals (i, 1 + i], for i = 0, 1, .. . , 19. Also, compare

the histogram levels with those derived analytically.

(b) Compute the sample means and variances, for each one of the three runs. Then,

compare the results with the analytically derived mean and variance values

– Note: The theoretical mean and variance of Tn are λ 1 and λ 12, respectively.

(c) Show that if Un is uniformly distributed over (0, 1], and the random variable

Tn
is calculated as Tn = - ln(λ Un), then Tn is exponentially distribution with

intensity λ.

Problem 3
Consider a traffic stream represented by the stochastic point process A = {An |
n ≥ 0}, where An denotes the time of occurrence of the n-th message arrival and

A0 = 0. The n-th inter-arrival time is Tn = An-An-1, for n ≥ 1, where An = Pn Ti.

Assume that the inter-arrival time sequence T = {Tn | n ≥ 1} consists of i.i.d.
exponentially distributed random variables with parameter λ. Then, the traffic stream

A is said to be a Poisson point process. Use the method described in Problem 2 to

generate a sequence T = {Tn | 1 ≤ n ≤ N} that consists of i.i.d. exponentially

distributed random variables with λ = 0.1, for each N = 1000, 5000, 10000. Then,

 


write a program to generate the corresponding Poisson point process streams A =
{An | n ≥ 0}, where A0 = 0, for each N = 1000, 5000, 10000. Plot realizations for
these point process streams, for N = 1000, 5000, 10000 (y-axis is the arrival time and
x-axis is indices of the arrival events).
Problem 4
Let N(t) be the counting variable associated with the point process A. N(t)
represents the number of arrivals (events) occurring during the interval (0, t] where
N(0) = 0. The process M = {N(t) | t ≥ 0} is the associated counting process.
Consider a Poisson point process A with intensity λ. The associated counting
process M is then called a Poisson Counting Process. For this process, we note the
following properties:
(P1) N(t) follows a Poisson distribution with parameter λ:
P(N(t) = n) = e-λt(λt)n
n! , ∀n = 0, 1, . . . ,
E[N(t)] = λt; V ar[N(t)] = λt.
(P2) The probability of an event occurrence in an interval (t, t + h] of small duration
h (so that λh << 1) is approximately equal to λh:
P(N(t + h) - N(t) = 1) = λh + o(h),
P(N(t + h) - N(t) = 0) = 1 - λh + o(h),
where limh→0 o(h h) = 0.
(P3) The number of events occurring over an interval is statistically independent of
the number of events occurring over any other disjoint interval.
(a) Write a computer program that generates and plots a realization of a Poisson counting process M until N(t) = N, by starting with the generated Poisson point process A (y-axis is N(t) and x-axis is index of t), for each N =
1000, 5000, 10000.
(b) For λ = 0.1, t = 30, generate the Poisson counting variable N(30) with K
independent realizations of N(30). Then, use the results to draw a probability
mass function by using histogram expressing P(N(30) = n), for n ≥ 0. Also use
the results to compute the mean and variance realized values of N(30). Assume
the following values: K = 1000, 5000, 10000. Compare these results with the
analytically derived probability distribution values, and with the analytically
computed mean and variance values.
(c) For λ = 0.1 and h = 1, verify the above mentioned property (P2), using K
independent realizations, for each K = 1000, 5000, 10000. You can consider any
time period of [t, t + h), for any time t (e.g., by choosing t = 29)

 

Problem 1Generate a stream ofi.i.d.random variablesU={Un|1≤ n ≤ N}which areuniformly distributed over [0,1], for eachN= 1000,5000,10000.(a) Use the resulting streams to draw the probability density histograms ofUoverthe intervals (0.1i,0.1i+ 0.1], fori= 0,1, .. . ,9. Also, compare thehistogramlevels with those derived analytically.(b) Compute the sample means and variances for each of the three runs.Then,compare the results with the mean and variance values derived analytically.–when the variables{Xn|1≤ n ≤ N}are observed,N >1, the samplemeanµNand sample varianceσN2are calculated as follows:µN=PN i=1XiN, σN2=PN i=1(Xi- µN)2N -1.Problem 2Generate a stream ofi.i.d.random variablesT={Tn|1≤ n ≤ N}whichare exponentially distributed with parameter (intensity)λ= 0.1, for eachN=1000,5000,10000.(a) Use the resulting streams to compute and exhibit the probability densityhistograms forT, over the intervals (i,1 +i], fori= 0,1, .. . ,19. Also,comparethe histogram levels with those derived analytically.(b) Compute the sample means and variances, for each one of the threeruns. Then,compare the results with the analytically derived mean and variance values–Note: The theoretical mean and variance ofTnareλ1andλ12, respectively.(c) Show that ifUnis uniformly distributed over (0,1], and the randomvariableTnis calculated asTn=-ln(λ Un), thenTnis exponentially distribution withintensityλ.Problem 3Consider a traFc stream represented by the stochastic point processA={An|n ≥0}, whereAndenotes the time of occurrence of then-th message arrivaland

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Status NEW Posted 22 May 2017 08:05 AM My Price 9.00

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