Expected values of discrete Random Variables

The function that maps S into S X in R and which is denoted by X(.) is called a random variable. The name random variable is a poor one in that the function X(.) is not random but a known one and usually one of our own choosing. Example: In phase-shift keyed (PSK) digital system a ”0” or “1” is communicated to a receiver by sending A 1 or a 0 occurs with equal probability so we model the choice of a bit as a fair coin tossing experiment. random variable Definition of Discrete RV (Review)

Last lecture overview: Example Servicing Customers A supermarket has one express lane open from 5 to 6PM on weekdays. No more than 10 items are allowed, thus average time is 1 min. The supermarket wants that no more that one person waiting in line 95% of the time. 3 How many lanes should be opened?

Example: Servicing Customers 4 Since the processing time is 1 min, there can be at most two arrivals in each time slot of 1 min length. (One is serviced and the other is waiting for a min). Assuming no two arrivals per second is possible we have a sequence of independent Bernoulli trials 0 – no arrivals 1 - arrival

Example: Servicing Customers Observing the total arrivals, the average is 70 arrivals per min or 70 per 3600 seconds. Thus probability of arrival is p = 70 / 3600 = We have the Binomial PMF, with p = and M = 60 seconds. Approximating it with Poisson PMF, the probability of k arrivals in 60 seconds is 5 Where is expected number of customers arriving in the 1 minute interval.

Example: Servicing Customers There should be at most 2 customers arriving 95% of the time. Substituting Poisson PMF Hence, the probability of 2 or fewer customers arriving at the express lane is not grater than Adding one more lane, we have 70/2=35 customers per minute. Therefore. 6

Example: Servicing Customers Since the arrivals are modeled as independent Bernoulli trials, so that 7

Probability Mass Functions PMF is a complete description of a discrete random variable. It allows to determine probabilities of any event. We might be interested in a rainfall between 8 and 12 inches to grow a particular crop. Estimated PMF Is this adequate or should the probability be higher? 9.76

Determining Averages from the PMF Rather it might be more useful to know the average rainfall, since it is close to the requirement of an adequate amount of rainfall. Two methods to estimate average

Determining Averages from the PMF Example: A barrel is filled with equal number of US dollar bills $1,5,10,20. A person playing the game gets to choose a bill from the barrel, but must do so blindfolded. He pays 10$ to play the game, which consists of a single draw with replacement from the barrel. He wins that amount of money. Will he make a profit by playing this game? The average of winning per play

Determining Averages from the PMF The number of times a player wins k dollars N 1 = 13,N 2 = 13,N 10 = 10,N 20 = 14 If he were to play the game a large number of times, then N  ∞ we would have N k /N  p X [k]. p X [k] is PMF of choosing a bill with k$. Proportion of bills On average he will lose $1 per play

Determining Averages from the PMF The expected value is also called the expectation of X, the average of X, and the mean of X. The expected value of a discrete RV X is defined as For all nonzero p X [x i ]. Expected value The expected value may be interpreted as the best prediction of the outcome of a random experiment for a single trial. The expected value is analogous to the center of mass of a system of linearly arranged masses.

Expected values of Important Random Variables. Bernoulli: If X ~ Ber(p), then the expected value is Binomial X: If X ~ bin(M, p), then the expected value is Setting M / = M – 1, k / = k -1, this becomes

Expected values of Important Random Variables. Geometric: if X ~ geom(p), the expected value is Let’s modify the summand to be a PMF by letting q = 1 – p and then differentiating the expression Since 0 < q < 1 we have Σ k=1..∞ q k = q/(1 - q ) If p = 1/10, then on average it takes 10 trials for a success.

Expected values of Important Random Variables. Poisson: If X ~ Pois(p), then it can be shown that E[X] = λ. This result is consistent with the Poisson approximation to the binomial PMF since the approximation constrains Mp to λ. Not all PMFs have expected values. Discrete RVs with a finite number of values always have expected values. Discrete RVs with countably infinite number of values may not have an expected value. Consider the PMF Attempting to find the expected value produces It is valid PMF since it can be show to sum to one

Expected values of Important RV It is possible for a sum to produce different results depending upon the order in which the terms are added up. To avoid these difficulties, we require the sum to be absolutely summable.

Properties of the expected value It is located at the “center” of the PMF if the PMF is symmetric about some point. It does not generally indicate the most probable value of RV. More than one PMF may have the same expected value.

Expected Value for a Function of RV The expected value may easily be found for a function of RV X if p X [x i ] is known. If the function of interest is Y = g(X), then by the definition of expected value or in much convenient form

Expected Value for a Function of RV A linear function If g(x) = aX + b, where a and b are constants, then If we set a = 1, then E[X + b] = E[X] + b. This allows us to set the expected value of a RV to any desired value by adding the appropriate constant to X. An extension produces For any constants a 1 and a 2 and any two functions g 1 and g 2. Thus, expectation operator E is linear.

Expected Value for a Function of RV A nonlinear function: Assume that X has a uniform PMF given by Determine E[Y] for Y = g(X) = X 2. It is not true that The expectation operator does not commute for nonlinear functions.

Variance and Moments of a RV The variability or variance is another important information about RV’s behavior and it given by Example: a uniform discrete RV is given whose PMF is mean is zero but variability of the outcomes becomes large as M increases. M = 2 M = 10

Variance and Moments of a RV It can be shown that Which yields M = 2 M = 10 Clearly, variance increases with M assume zero

Variance of Bernoulli RV If X ~ Ber(p), then since E[X] = p, we have The variance is minimized and equals zero if p = 0 or p = 1. It is maximized for p =1/2. Why? The best predictor b opt = E[X]. We wish to point out the minimum MSE. How well we can predict the outcome depends on the variance of RV.

Properties of discrete RV

An alternative expression for the variance Applying properties of expectation operator we have Hence In the case where E[X] = 0, we have the simple result that var(X) = E[X 2 ]

Properties of the variance Property 1. Alternative expression for variance Property 2. Variance for RV modified by a constant For c a constant The expectations E[X] and E[X 2 ] are called the first and second moments of X. Generally, nth moment is defined as E[X n ] and exists if E[|X| n ]. Central moments defined as E[(X – E[X]) n ]. For n = 2 we have the usual definition of the variance. Variance is a nonlinear operator

Real-world example: Data compression Data consists of a sequence of the letters A, B, C, D. Consider a typical sequence of 50 letters AAAAAAAAAAABAAAAAAAAAAAAA AAAAAACABADAABAAABAAAAAAD To encode these letters for storage we could use the two-bit code A  00 B  01 C  10 D  11 Which would then require a storage of 2 bits per letter for a total storage of 100 bits. The above sequence is characterized by a much larger probability of observing “A” as opposed to the other letters. 43 A’s, 4 B’s, 1 C, and 2D’s.

Real-world example: Data compression It makes sense then to attempt a reduction in storage by assigning shorter code words to the letters that occur more often, in this case, to the “A”. A  0 B  10 C  110 D -> 111 Using this code, would require 1     2 = 60 bits or 1.2 bits per letter. Huffman code To determine storage savings we need to determine the average length of the code word per letter. First we define a discrete RV that measures the length of the code word.

Real-world example: Data compression The probability used to generate the sequence of letters are P[A] = 7/8, P[B] = 1/16, P[C] = 1/32, P[D] = 1/32. The PMF is The average code length is give by This result in a compression ratio of 2 : 1,1875 = 1.68 or we require about 40% less storage. bits per letter

Real-world example: Data compression The average code word length per letter can be reduced even further. However it requires more complexity in coding. A fundamental theorem due to Shannon, says that the average code word length per letter can be no less than The quantity is termed the entropy of the source. In addition, he showed that a code exist that can attain, this minimum average code length. Hence, the potential compression ratio is 2: = 2.73 for a bout 63% reduction. If P[s i ] = ¼, bits per letter No compression is possible

Problems 2) 3) 4) A discrete RV X has the PMF p X [k] = 1/5 for k =0,1,2,3,4. If Y = sin[(π/2)X] find E[Y] using Which way is easier? 1)

Generate the outcomes of N trials

Estimate Means and Variances Define x i and p X using given PMF Estimate and plot means and variance

H/W problems 1) 2) (a)(b)(c)