Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science6.041/6.431: Probabilistic Systems Analysis(Spring 2006)Problem Set 5Due: March 22, 20061. Consider an exponentially distributed random variable X with parameter λ. Let F be thedistribution function. Find the real number µ that satisfies: F(µ) =12. This number µ is calledthe median of the random variable.2. One of two wheels of fortune, A and B, is selected by the flip of a fair coin, and the wheel chosenis spun once to determine the experimental value of random variable X. Random variable Y ,the reading obtained w ith wheel A, and random variable W , the reading obtained with wheel B,are described by the PDFsfY(y) =1, 0 < y ≤ 1;0, otherwise,and fW(w) =3, 0 < w ≤13;0, otherwise.If we are told the experimental value of X was less than14, what is the conditional probabilitythat wheel A was the one selected?3. A 6.041 graduate opens a new casino in Las Vegas and decides to make the games more challengingfrom a probabilistic point of view. In a new version of roulette, each contestant spins the followingkind of roulette wheel. The wheel has radius r and its perimeter is divided into 20 intervals,alternating red and black. The red intervals (along the perimeter) are twice the width of theblack intervals (also along the perimeter). The red intervals all have the same length and theblack intervals all have the same length. After the wheel is spun, the center of the ball is equallylikely to settle in any position on the edge of the wheel; in other words, the angle of the finalball position (marked at the ball’s center) along the wheel’s perimeter is distributed uniformlybetween 0 and 2π radians.(a) What is the probability that the center of the ball settles in a red interval?(b) Let B denote the event that the center of the ball settles in a black interval. Find theconditional PDF fZ|B(z), where Z is the d istance, along the perimeter of the roulette wheel,between the center of the ball and the edge of the interval immediately clockwise from thecenter of the ball?(c) What is the unconditional PDF fZ(z)?Another attraction of th e casino is the Gaussian slot machine. In this game, the machine pro-duces independent identically distributed (IID) numbers X1, X2, ... that have normal distributionN(0, σ2). For every i, when the number Xiis positive, the player receives from the casino a sumof money equal to Xi. When Xiis negative, the player pays the casino a sum of money equal to|Xi|.(d) What is the standard deviation of the net total gain of a player after n plays of the Gaussianslot machine?(e) What is the probability that the absolute value of the net total gain after n plays is greaterthan 2√nσ?4. Let a continuous random variable X be un iform ly distributed over the interval [−1, 1]. Derivethe PDF fY(y) where:Page 1 of 4Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science6.041/6.431: Probabilistic Systems Analysis(Spring 2006)(a) Y = sin(π2X)(b) Y = sin(2πX)5. A Communication Example During lecture, we looked at a simple point-to-point communi-cation example. The task was to transmit messages from one point to another, by usin g a noisychannel. We modeled the problem probabilistically as follows:• A binary message source generates independent successive messages M1, M2, ···: each mes-sage is a discrete random variable that takes the value 1 with probability p and the value 0with probability 1 − p. For a single message, we write the P MF as:pM(m) =1 − p If m = 0p If m = 1• A binary symmetric channel acts on an input bit to produce an output bit, by flipping theinput with “crossover” probability e, or transmitting it correctly with probability 1 −e.For a single tran s mission, this channel can be modeled using a conditional PMF pY |X(y|x),where X and Y are random variables for the channel input and output bits respectively.We then have:pY |X(y|x) =1 − e If y = x.e If y 6= x.Next, we make a series of assumptions that m ake the above single-transmission descriptionenough to describe multiple transmissions as well. We say that transmissions are:(a) Independent: Outputs are conditionally independent from each other, given the inputs.(b) Memoryless: Only the current in put affects the current output.(c) Time-invariant: We always have the same conditional PMF.We write:pY1,Y2,···|X1,X2,···(y1, y2, ···|x1, x2, ···)(a)= pY1|X1,X2,···(y1|x1, x2, ···) · pY2|X1,X2,···(y2|x1, x2, ···) ···(b)= pY1|X1(y1|x1) · pY2|X2(y2|x2) ···(c)= pY |X(y1|x1) · pY |X(y2|x2) ···Page 2 of 4Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science6.041/6.431: Probabilistic Systems Analysis(Spring 2006)Any transmission scheme through this channel must transform messages into channel inputs(encoding), and transform channel outputs into estimates of the transmitted messages (decoding).For this problem, we will encode each message separately (more elaborate schemes look at blocksand trails of messages), and generate a sequence of n channel input bits. T he encoder is thereforea map:{0, 1} 7−→ {0, 1}nM −→ X1, X2, ··· , XnThe decoder is simply a revers e map:{0, 1}n7−→ {0, 1}Y1, Y2, ··· , Yn−→ˆMNote that we use the “hat” notation to indicate an estimated quantity, but bare in mind thatM andˆM are two distinct random variables. The complete communication problem looks asfollows:Finally, to measure the performance of any transmission scheme, we look at the probability oferror, i.e. the event that the estimated message is different than the transmitted message:P(error) = P(ˆM 6= M).(a) No encoding:The simplest encoder send s each message bit directly through the channel, i.e. X1= X =M. Then, a reasonable d ecoder is to use the output channel directly as message estimate:ˆM = Y1= Y . What is the probability of error in this case?(b) Repetition code with majority decoding rule:The next thing we attempt to do is to send each message n > 1 times through this channel.On the decoder end, we do what seems natural: decide 0 when there are more 0s than 1s,and decide 1 otherwise. This is a “majority” decoding rule, and we can wr ite it as follows(making the dependence ofˆM on the channel outputs explicit):ˆM(y1, ··· , yn) =0 If y1+ ··· + yn< n/2.1 If y1+ ··· + yn≥ n/2.Analytical results:i. Find an expression of the probability of error as a
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