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# part15 - EE 5375/7375 Random Processes Homework #2...

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Unformatted text preview: EE 5375/7375 Random Processes Homework #2 Solutions September 9, 2003 Problem 1. textbook problem 3.12 Let U be a uniform random variable in the interval [-1,1]. Find the probabilities (a) P (U > 0) (b) P (U < 5) (c) P (|U | < 1/3) (d) P (1/3 < U < 1/2) (e) P (|U | ≥ 3/4). (a) The pdf for U is fU (u) = 1/2 for −1 ≤ u ≤ 1. P (U > 0) = 1 (b) P (U < 5) = −1 1 du = 1. 2 (c) P (|U | < 1/3) = P (−1/3 < U < 1/3) = (d) P (1/3 < U < 1/2) = (e) P (|U | ≥ 3/4) = 1/2 1 du = 1 /3 2 −3/4 1 1 2 du + 3/4 −1 1 12 . 1 2 du 1/3 1 du −1/3 2 1 = 3. 11 du 02 = 1. 2 = 1 8 + 1 8 = 1. 4 Problem 2. textbook problem 3.15 A random variable Y has the PDF FY (y ) = 0 1 − y −n if y < 1 if y ≥ 1 where n is a positive integer. (a) Plot the PDF of Y . (b) Find the probability P (k < Y ≤ k + 1) for a positive integer k . (a) The PDF is shown in Fig. 1. (b) P (k < Y ≤ k + 1) = FY (k + 1) − FY (k ) = 1 − (k + 1)−n − (1 − k −n ) = k −n − (k + 1)−n . Problem 3. textbook problem 3.16 A continuous random variable X has the PDF 0 FX (x) = c(1 + sin(x)) 1 (a) Find c. (b) Plot FX (x). (a) A continuous random variable must have a smooth PDF without discontinuities. Therefore c must satisfy (i) c(1 + sin(−π/2)) = 0 and (ii) c(1 + sin(π/2)) = 1. Condition (i) is satisﬁed for every c because 1 + sin(−π/2) = 0. Condition (ii) is c(1 + sin(π/2)) = 2c = 1, so c = 1/2. (b) The PDF is shown in Fig. 2. Problem 4. textbook problem 3.18 Let X be an exponential random variable with parameter λ. (a) For d > 0 and a positive integer k , ﬁnd the probabilities P (X ≤ d), P (kd ≤ X ≤ (k + 1)d), and P (X > kd). (b) Segment the positive real line into 5 equiprobable disjoint intervals. (a) The exponential pdf is fX (x) = λe−λx and PDF is FX (x) = 1 − e−λx for x ≥ 0. Hence P (X ≤ d) = FX (d) = 1 − e−λd P (kd ≤ X ≤ (k + 1)d) = FX ((k + 1)d) − FX (kd) = 1 − e−λ(k+1)d − (1 − e−λkd ) = e−λkd (1 − e−λd ) P (X > kd) = 1 − FX (kd) = e−λkd 1 if x ≤ −π/2 if −π/2 < x ≤ π/2 if π/2 ≤ x (b) The problem is to ﬁnd {X1 , X2 , X3 , X4 } where FX (X1 ) = 1/5, FX (X2 ) = 2/5, FX (X3 ) = 3/5, and FX (X4 ) = 4/5. In general, FX (Xi ) = i/5 for i = 1, 2, 3, 4. These points will divide the real line into 5 equiprobable intervals. We have FX (Xi ) = 1 − e−λXi = i/5, therefore 1 − e−λXi = 1− i 5 i = e−λXi 5 1 i Xi = − ln 1 − λ 5 3 5 1 , X3 = − ln λ 2 5 1 , X4 = − ln λ 1 5 That is, 1 X1 = − ln λ 4 5 1 , X2 = − ln λ Problem 5. textbook problem 3.19 A random variable X has the pdf fX (x) = cx(1 − x) 0 if 0 ≤ x ≤ 1 otherwise (a) Find c. (b) Find P (1/2 ≤ X ≤ 3/4). (c) Find FX (x). (a) The pdf must integrate to 1: 1 1 1 cx(1 − x)dx = 0 0 cxdx − 0 cx2 dx = c c c − = =1 23 6 +2 13 2 so c = 6. 2 3/4 32 (b) P (1/2 ≤ X ≤ 3/4) = 1/2 6x(1 − x)dx = 3 4 − 3 1 − 2 2 x x x (c) FX (x) = 0 6y (1 − y )dy = 0 6ydy − 0 6y 2 dy = 3x2 − 2x3 . 33 4 = 27 16 − 3 4 − 54 64 + 2 8 = 11 32 . Problem 6. textbook problem 3.21 A random variable X has the pdf shown in Fig. P3.2. (a) Find fX (x). (b) Find the PDF of X . (c) Find b such that P (|X | < b) = 1/2. (a) The pdf has a symmetric triangular shape: fX (x) = c 1− 0 |x| a if |x| ≤ a if |x| > a The area of the triangle is c(2a)/2 but this must equal 1, so c = 1/a. (b) The PDF is found by integration of the pdf. First, FX (x) = 0 for x < −a. Second, for −a ≤ x ≤ 0, x FX (x) = −a y 11 1 1+ dy = + a a 2a x 0 x+ x2 2a x2 2a For 0 ≤ x ≤ a, FX (x) = Finally, for all x > a, FX (x) = 1. (c) We have P (|X | < b) = 1 + 2 1 y 11 1− dy = + a a 2a x− 1 2 = FX (b) − FX (−b) = 2 a 2 b− b2 2a If we solve for b, then we ﬁnd b = a(1 − 1 √ ). 2 Problem 7. textbook problem 3.27 Let X be the exponential random variable with the pdf fX (x) = λe−λx for x ≥ 0. (a) Find and plot FX (x|X > t). How does FX (x|X > t) diﬀer from FX (x)? (b) Find and plot fX (x|X > t). (c) Show that P (X > t + x|X > t) = P (X > x). Explain why this is called the memoryless property. (a) By deﬁnition of conditional probability, FX (x|X > t) = P (X ≤ x|X > t) P (X ≤ x and X > t) = P (X > t) 0 if x < t = FX (x)−FX (t) if x ≥ t 1−FX (t) = 0 e−λx −e−λt e−λt if x < t if x ≥ t A plot of FX (x|X > t) is shown in Fig. 3. It appears like a delayed version of FX (x). (b) To ﬁnd fX (x|X > t), we have to diﬀerentiate FX (x|X > t) found in part (a): fX (x|X > t) = d dx e−λx − e−λt e−λt = λe−λx = λe−λ(x−t) e−λt for x ≥ t. A plot of fX (x|X > t) is shown in Fig. 4. (c) We want to show P (X > t + x|X > t) = 1 − P (X ≤ t + x|X > t) =1− = e−λ(t+x) − e−λt e−λt e−λ(t+x) e−λt −λx =e = P (X > x) Thus, the probability of waiting an additional x does not depend on the previous waiting time t. The additional waiting time is the same as if the waiting just started. The waiting time is memoryless in the sense that the previous waiting time has no eﬀect on future waiting. Problem 8. textbook problem 3.34 Let N be a geometric random variable with PMF PN (x) = p(1 − p)x for x = 0, 1, . . .. (a) Find P (N > k ). (b) Find the PDF of N . (c) Find P (N is an even number). (d) Find P (N = k |N ≤ m). (a) P (N > k ) = ∞ j =k+1 p(1 − p)j = p(1 − p)k+1 (b) FN (x) = P (N ≤ x) = 1 − P (N > x) = 1 − (1 − p) where [x] is the integer part of x. ∞ p (c) P (N is an even number) = P (N = 0) + P (N = 2) + · · · = j =0 p(1 − p)2 j = 1−(1−p)2 = 3 ∞ j =0 (1 − [x]+1 p)j = p(1−p)k+1 1−(1−p) = (1 − p)k+1 1 2−p (d) P (N = k |N ≤ m) = P (N =k and N ≤m) P (N ≤m) = P (N =k) P (N ≤m) = p(1−p)k 1−(1−p)m+1 for 1 ≤ k ≤ m. Problem 9. textbook problem 3.35 Prove the memoryless property of the geometric random variable: P (M ≥ k + j |M > j ) = P (M ≥ k ) for all j, k > 0. In what sense is M memoryless? We can show the memoryless property by P (M ≥ k + j |M > j ) = P (M ≥ k + j and M > j ) P (M > j ) P (M ≥ k + j ) = for k ≥ 1 P (M > j ) ∞ i−1 i=k+j p(1 − p) =∞ i−1 i=j +1 p(1 − p) = (1 − p)k+j −1 (1 − p)j = (1 − p)k−1 = P (M ≥ k ) This says that the probability of k additional trials until the ﬁrst success is independent of how many failures have already occurred (i.e., the future number of trials does not depend on the past). Problem 10. textbook problem 3.42 The rth percentile, Pr , of a random variable X is deﬁned by P (X ≤ Pr ) = r/100. (a) Find the 90%, 95%, and 99% percentiles of the exponential random variable with parameter λ. (b) Repeat part (a) for the Gaussian random variable with parameters m = 0 and σ . (a) The exponential random variable has PDF P (X ≤ Pr ) = 1 − e−λPr = Pr , we have 1 r Pr = − ln 1 − λ 100 For the 90%, 95%, and 99% percentiles, P90 ≈ 3 4.6 23 , P95 ≈ , P99 ≈ λ λ λ r 100 . Solving for the rth percentile, (b) If X is a normal random variable with mean m = 0 and variance σ 2 , the PDF can be found from the x standard normal distribution by P (X ≤ x) = Φ σ . We can look up the percentiles from any table of x x Φ(x). The 90% percentile Φ σ = 0.9 occurs when σ = 1.28, so P90 = 1.28σ . Similarly, the 95% percentile x x x x Φ σ = 0.95 occurs when σ = 1.5, so P95 = 1.5σ . The 99% percentile Φ σ = 0.99 occurs when σ = 2.33, so P99 = 2.33σ . Problem 11. textbook problem 3.45 A communication channel accepts an arbitrary voltage v and outputs a voltage Y = v + N where N is a Gaussian random variable with mean 0 and variance σ 2 = 1. Suppose that the channel is used to transmit binary information in this way: an input of -1 causes a transmission of 0, and an input of +1 causes a transmission of 1. The receiver decides a 0 was sent if the voltage is negative and a 1 otherwise. Find the probability of the receiver making an error if a 0 was sent; and if a 1 was sent. 4 An error is made if the input was -1 but the output is positive (then the receiver will decide the transmission was a 1). The conditional probability is P (error|v = −1) = P (Y ≥ 0|v = 1) = P (v + N ≥ 0|v = −1) = P (N ≥ 1) = 1 − Φ(1) = 0.159 Similarly, an error is made if the input was 1 but the output is negative (then the receiver will decide the transmission was a 0). The conditional probability is P (error|v = 1) = P (Y < 0|v = 1) = P (v + N < 0|v = 1) = P (N < −1) = Φ(−1) = 1 − Φ(1) = 0.159 Problem 12. textbook problem 3.47 Messages arrive at a center at a rate of one message per second. Let X be the time for the arrival of 5 messages. Find the probability that X < 6; and that X > 8. Assume that message interarrival times are exponential random variables. When interarrival times are exponentially distributed, then the number of arrivals within a certain interval is a Poisson random variable. Suppose the interarrival times are exponential with parameter λ, then the number of arrivals N during an interval (0, t) has the Poisson PMF P (N = k ) = e−λt (λt)k k! The event {X < 6} means that the 5 arrivals have occurred within the interval (0,6), or there are at least 5 arrivals within (0,6). The probability of this event can also be found as P (5 or more arrivals in (0,6)) = 1 − P (up to 4 arrivals in (0,6)) 4 =1− k=0 e−λ6 (λ6)k k! Since the arrival rate λ = 1, this probability becomes 4 P (5 or more arrivals in (0,6)) = 1 − k=0 e−6 6k = 0.7149 k! The event {X > 8} corresponds to the event that the 5 arrivals took longer than (0,8), or there are 4 or less arrivals within (0,8). The probability of this event is 4 P (4 or less arrivals in (0,8)) = k=0 e−8 8k = 0.09963 k! Problem 13. exponential The PDF of a random variable X is given by FX (x) = 1 − e−x for x ≥ 0. The probability of the event {X < 1 or X > 2} is (a) 0.914 (b) 0.767 (c) 0.632 (d) 0.135 (e) 0.041. 5 The answer is (b). P (X < 1 or X > 2) = P (X < 1) + P (X > 2) = (1 − e−1 ) + (1 − P (X < 2)) = (1 − e−1 ) + (1 − (1 − e−2 )) = 1 − e−1 + e−2 = 1 − 0.368 + 0.135 = 0.767 Problem 14. normal (optional for EE 5375) A radar scans the skies for UFOs (unidentiﬁed ﬂying objects). Let M be the event that a UFO is present and M c the complementary event that a UFO is absent. Let fX |M (x|M ) = 1 −(x−r)2 /2 e 2π be the conditional pdf of the radar return signal X when a UFO is actually there and let fX |M c (x|M c ) = 1 −x2 /2 e 2π be the conditional pdf of the radar return signal X whenthere is no UFO. To be speciﬁc, let r = 1 and let the alert level be xA = 0.5. Let A denote the event of an alert, i.e., {X > xA }. Compute P (A|M ), P (Ac |M ), P (A|M c ), and P (Ac |M c ). P (A|M ) = P (X > 0.5|M ) ∞ 1 −(x−1)2 /2 = e dx 2π 0 .5 ∞ 1 −y2 /2 = e dy −0.5 2π 0 .5 = −∞ 1 −y2 /2 e dy 2π = Φ(1/2) = 0.69 c P (A |M ) = 1 − P (A|M ) = 1 − Φ(1/2) = 1 − 0.69 = 0.31 c P (A|M ) = P (X > 0.5|M c ) ∞ 1 −x2 /2 e dx = 2π 0 .5 = 1 − Φ(1/2) = 1 − 0.69 = 0.31 c c P (A |M ) = 1 − P (A|M c ) = 1 − (1 − Φ(1/2)) = Φ(1/2) = 0.69 Problem 15. uniform The arrival time of a professor to his oﬃce is a continuous random variable uniformly distributed over the hour between 8 AM and 9 AM. Deﬁne the event A as {the professor has not arrived by 8:30 AM} and B as {the professor will arrive by 8:31 AM}. Find (a) P (B |A) (b) P (A|B ). (∩ (a) By deﬁnition of conditional probability, P (B |A) = P PAAB ) . The probability P (A ∩ B ) is the probabil() ity that the professor has not arrived by 8:30 AM and the professor will arrive by 8:31 AM. This means 6 the professor will arrive between 8:30 and 8:31, which is one minute. Since the arrival time is uniformly distributed over an hour, the probability of arriving within a one-minute interval is the ratio of 1 minute to 60 minutes, or P (A ∩ B ) = 1/60. Now, P (A) is the probability that the professor will arrive between 8:30 and 9:00, which is 30/60 = 1/2. Putting these probabilities together, P (B |A) = 11/60 = 1/30. (b) /2 (∩ Similarly, P (A|B ) = P PABB ) . In this case, P (B ) is the probability of arriving between 8:00 and 8:31 which () is 31 minutes out of the 60 minutes, so the probability is P (B ) = 31/60. Putting the probabilities together, 1 P (A|B ) = 31/60 = 1/31. /60 Problem 16. Matlab (optional for EE 5375) Start Matlab and use the uniform random number generator rand to generate 300 random samples by typing: > x=rand(300,1); Take a look at the histogram of these numbers by typing: > hist(x) The histogram should look fairly ﬂat. Uniformly random numbers {X1 , X2 , . . .} can be converted to expo1 nentially distributed numbers by Yi = − λ ln(Xi ). Let us generate exponential numbers by typing: > lambda = 0.5; > y = -1 * log(x)/lambda; Take a look at the histogram of these numbers by typing: > hist(y) This histogram should appear to decrease exponentially. We can also directly generate exponentially distributed numbers by the command exprnd. The online information about exprnd can be retrieved by typing: > help exprnd We can use exprnd to generate 300 exponential numbers by typing: > y = exprnd(0.5,300,1) Take a look at the histogram of these numbers by typing: > hist(y) Again, this histogram should appear to decrease exponentially. Next, we can generate normally distributed numbers by the command normrnd. Get the online information about normrnd by typing: > help normrnd Generate some normal numbers and look at their histogram by typing: > z = normrnd(0,1,300,1); > hist(z) The shape of the histogram should look like a “bell curve.” What is the mean and variance of the distribution? The mean and variance are the ﬁrst 2 parameters of the normrnd command. In this case, the numbers were generated by normrnd(0,1,300,1) so the mean is 0 and variance is 1. 7 fx(x) -a 0 a x Fig. P3.2 for Problem 6 x Note: F( x ) = Ú -• 1 -t 2 /2 1 e dt = + erf ( x ) 2 2p † Fig. 1 for Problem 2 Fig. 2 for Problem 3 Fig. 3 for Problem 7 Fig. 4 for Problem 7 ...
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## This note was uploaded on 11/29/2009 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.

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part15 - EE 5375/7375 Random Processes Homework #2...

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