Stochastic Processes Lecture Notes PDF

Stochastic Processes CSE-5605 Dr. Safdar Nawaz Khan Marwat DCSE, UET Peshawar Lecture 3 [email protected]  Expected Value of Discrete RV Entire pmf required for completely describing RV behavior In some cases, interest in parameters summarizing pmf Expected value or...

Stochastic Processes CSE-5605 Dr. Safdar Nawaz Khan Marwat DCSE, UET Peshawar Lecture 3 [email protected]  Expected Value of Discrete RV Entire pmf required for completely describing RV behavior In some cases, interest in parameters summarizing pmf Expected value or mean of discrete RV X defined by m X = EX  =  xp X ( x) xS X [email protected] 2  Expected Value of Discrete RV (cont.) The “expected value” does not mean expected outcome E[X] not necessarily an outcome ❑ E.g. the expected value of Bernoulli RV is p ❑ But outcomes are always 0 or 1 E[X] corresponds to “average of X” ❑ In large number of observations of X [email protected] 3 Key.txt Expected Value of Functions of RV Function g(X) of RV X can be denoted by Z ❑ Expected value of Z would be E[ Z ] = E[ g ( X )] =  g ( xk )p X ( xk ) k Or simply multiply each value of Z with its probability and add products for each k ❑ For more than one value of X mapped to one value of Z E[ Z ] =  g (xk )p X (xk ) =  z j pZ (z j ) k j Property (see other properties in Garcia 2008): E[ag ( X ) + c] = aE[ g ( X )] + c [email protected] 4  Variance of Discrete RV Expected value provides limited information Interest also in the variation about expected value X – E[X] Squaring the variations gives positive values (X – E[X])2 Variance defined as the expected value of this square  2 X = VARX  = E[( X − E[ X ]) ] 2 [email protected] 5  Variance of Discrete RV (cont.)   2 X =  ( x − E[ X ]) xS X 2 p X ( x) =  ( xk − E[ X ]) p X ( xk ) k =1 2 The square root of variance is standard deviation  X = STD[ X ] = VAR[ X ] Variance also expressed as E[( X − E[ X ]) ] = E[ X − 2 E[ X ] X + E [ X ]] 2 2 2 = 𝐸 𝑋 2 − 2𝐸 𝑋 𝐸 𝑋 + 𝐸 2 𝑋 = 𝐸 𝑋 2 − 𝐸 2 𝑋 E[X ] is 2nd moment of X, similarly E[X ] the nth moment 2 n [email protected] 6  Important Discrete RVs Bernoulli Random Variable SX = {0, 1} p0 = q = 1 – p, p1 = p E[X] = p, VAR[X] = pq Binomial Random Variable SX = {0, 1, 2,... , n} n Pk = C pk qn-k k E[X] = np, VAR[X] = npq [email protected] 7  Important Discrete RVs (cont.) Geometric Random Variable SX = {1, 2, 3,...} Pk = qk-1p E[X] = 1/p, VAR[X] = q/p2 Uniform Random Variable SX = {1, 2, 3,... , L} Pk = 1/L E[X] = (L+1)/2, VAR[X] = (L2-1)/12 [email protected] 8  Important Discrete RVs (cont.) Poisson Random Variable SX = {0, 1, 2,...} Pk = (αk/k!)e-α E[X] = α, VAR[X] = α Counting number of occurrences of an event in a time period Arises in situations where events occur at random ❑ E.g. counts of emissions from radioactive substances Here, α is number of arrivals in time interval of length t ❑ α is unitless and given as α = λt ❑ λ is arrival rate with unit of jobs/time (e.g. packets/sec) [email protected] 9  Examples The number N of queries arriving in t seconds at a call centre is a Poisson random variable with α = λt where λ is the average arrival rate in queries/second. Assume that the arrival rate is 4 queries per minute. Find the probability of the following events: ❑ 4 queries in 10 seconds ❑ More than 4 queries in 10 seconds ❑ Less than or equal to 5 queries in 2 minutes [email protected] 10  Examples (cont.) The number N of packet arrivals in t seconds at a multiplexer is a Poisson random variable with α = λt where λ is the average arrival rate in packets/second. Find the probability that there are no packet arrivals in t seconds. [email protected] 11