EE 302 Division 1. Homework 6 Solutions.

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1 EE 3 Division. Homework 6 Solutions. Problem. A random variable X has probability density { C f X () e λ,,, otherwise, where λ is a positive real number. Find (a) The constant C. Solution. Because of the normalization aiom, the PDF should integrate to. On the other hand, we showed in class that the second moment of an eponential random variable with parameter λ is equal to /λ in other words, λe λ is equal to /λ. We therefore have: f X ()d C λ C λ λ C λ3. (b) The cumulative distribution function of X. Solution. By definition, F X () f X (u)du λe λ d From the epression of PDF we can see that if <, the integral is equal to. When, we use integration by parts twice: F X () λ3 λ3 λ3 λ3 u e λu du u d ( λ ) e λu [ u λ ] e λu + λ e λu d(u ) [ λ3 λ e λ + ud ( λ )] λ e λu [ λ e λ λ u λ e λu + ] λ λ e λu du [ λ3 λ e λ λ e λ ] λ λ e λu [ λ3 λ 3 +λ + λ ] λ 3 e λ (+λ + λ ) e λ.

2 λ () f X F X ().5 / λ 4/ λ 6 / λ 8 / λ. / λ 4/ λ 6 / λ 8/ λ (a) PDF of X (b) CDF of X Figure : PDF and CDF of X in Problem The PDF and CDF are depicted in Figure. (c) The probability P( X /λ). Solution. By definition of CDF, we have: P( X /λ) F X (/λ) F X (). By plugging in /λ and into the epression of CDF obtained in Part(b), we get: F X (/λ) (++/)e 5 e, F X (), P( X /λ) F X (/λ) 5 e.83. Problem. A random variable X has probability density f X () Ce µ, where µ is a real number and is a positive real number. Find (a) The constant C. Solution. Denoting by X the random variable from the end-of-chapter Problem, and substituting λ /, we see that our random variable is X X + µ. (To see this, we use the same argument as the one we used for the shifted eponential random variable in Homework 5). According to Problem from Chapter 3, the constant should be C λ/ /( ) in order to make the PDF correctly normalized. Similarly to Problem (a), we can also get the value of C directly, by applying the normalization aiom:

3 .7 f X () µ Figure : PDF of X in Problem C f X ()d C µ C C C. e µ d e ( µ) d (due to symmetry about µ) ( µ) e u du (u ) (b) The mean of X. Solution. Since X X + µ, E[X] E[X ]+µ µ. This makes sense, since, as shown in Figure, the PDF of X is symmetric about µ. Alternatively, we can derive this by applying the definition of epectation: E[X] f X ()d Making a change of variable u µ, wehave: E[X] (u + µ)ce u C ue u du }{{} µ. Ce µ d. du +µ Ce u du } {{ } (c) The variance of X. Solution. According to the end-of-chapter Problem, we have: var(x) /λ. 3

4 Again, we can derive this directly by applying the definition of variance: var[x] ( E[X]) f X ()d ( µ) e µ u e u du u u e du the second moment of an eponential r.v. with λ ( ). d Problem 3. A speedskating race has eactly three participants:,, and 3. They finish the race with respective times of X, X, and X 3, where the X i are uniformly distributed from 5 to 6 minutes. Find the probability that skater came in second. Solution. Since there is no difference among the three skaters, each one is equally likely to be the second one. The answer is therefore /3. We can also arrive at this by the following calculations, which are not necessary for this problem, but constitute an important technique for many other problems. Given that X i (i,, 3) are uniformly distributed on [5, 6], their PDF s are : f Xi ( i ) { 5 i 6 otherwise. (i,, 3) We assume the three participants are independent. The joint PDF of X,X,X 3 is the product of the three PDF s, i.e.: f XX X 3 (,, 3 ) f X ( )f X ( )f X3 ( 3 ) { 5 6, 5 6, otherwise. Let C {(,, 3 ):5 6, 5 6, 5 3 6} to be the sample space of the result of the race. We can see that the outcome of the race is uniformly distributed in the unit cube C. Let M {(,, 3 ) C : < < 3 } M {(,, 3 ) C : > > 3 } The probability that skater came in second is equal to P(M )+P(M ). Because of symmetry, P(M )P(M ) and we only need to calculate one of them. Let s do it for M. As depicted in Figure 3, the area above the plane EDAC corresponds to < and the area below BCFD corresponds to 3 >. The intersection of the two, tetrahedron ABCD, is equal to M. Because of uniformity, the probability of M is equal to its volume. Remember that the volume of a tetrahedron is equal to /3 times the area of the base times the height. 4

5 6 E B C D 5 F 6 3 A 6 Figure 3: The sample space of Problem 3 Therefore P(M )/6. It can also be obtained by integration as follows: P(M ) f XX X 3 (,, 3 )d d d 3 (,, 3) M d d d 3 (,, 3) M d 3 d d (6 )d d 6 ( 3 + )d 6. 5 (6 )( 5)d Therefore, The probability that skater came in second is equal to P(M )/3. In general, as long as their distributions of finishing time are identical, the answer should be /3. Problem 4. Let X be a normal random variable with mean µ and variance. (a) Compute the probabilities P(X µ + ) P(X µ ) P(X µ +) Solution. If we define random variable Y (X µ)/, Y is a standard normal random variable 5

6 with zero mean and unit variance. ( ) X µ P(X µ + ) P P(Y ) Φ() , ( ) X µ P(X µ ) P P(Y ) Φ( ) , ( ) X µ P(X µ +) P P(Y ) Φ() (b) Compute the probabilities Solution. P(µ X µ + ) P(µ X µ +) P(µ X µ + ) P(X µ + ) P(X µ ) , P(µ X µ +) P(X µ +) P(X µ ) ( ) A table of approimate values of Φ(/) /, as a function of /, is posted on the course web site. E.g., to look up the value of Φ(.3), add.5 to the number which is in row. and column.3. Do not forget that each number in the table is preceded by a decimal point, so, for eample, Φ(.3) Problem 5. A signal s 3 is transmitted from a satellite but is corrupted by noise, and the received signal is X s + W. When the weather is good, which happens with probability /3, W is normal with zero mean and variance 4. When the weather is bad, W is normal with zero mean and variance 9. In the absence of any weather information: (a) What is the PDF of X? Solution. What we can do is to first find the CDF of X and take derivative. By definition, we have: F X () P(X ) Since we don t have any weather information, if we can calculate the above probability conditioned on all the weather situations, we can apply the total probability theorem to get the final answer. Now let s suppose the weather is good, the random variable X 3+W is a Gaussian random variable with mean µ 3 and variance 4. Under this case, the probability that X is less than or equal to some is: ( ) ( ) µ 3 P(X Good) Φ Φ 6

7 Similarly, if we suppose the weather is bad, the conditional probability is: ( ) ( ) µ 3 P(X Bad) Φ Φ 3 Apply the total probability theorem to get: F X () P(X ) P(X Good)P(Good)+P(X bad)p(bad) ( ) ( ) 3 3 Φ 3 +Φ 3 3 Finally, we take derivative to get: f X () df X() e ( 3) d 3 + e ( 3) 3 π 3 π3 e ( 3) 8 + e ( 3) 8 3 π 9 π In general, we can get from Total Probability Theorem that: If B,B,,B n is a partition of the sample space, the following is true for any continuous random variable X: f X () f X B ()P(B )+f X B ()P(B )+ + f X Bn ()P(B n ), as stated on Page 3 of Chapter 3. (b) Calculate the probability that X is between and 4. Solution. By definition of CDF, we have: P( <X 4) P(X 4) P(X ) F X (4) F X () ( ) ( ) [ ( ) ( ) ] Φ 3 +Φ 3 3 Φ 3 +Φ 3 3 ( ) ( ) [ ( ) ( ) ] Φ 3 +Φ 3 3 Φ 3 +Φ Problem 6. Stations A and B are connected by two parallel message channels. A message from A to B is sent over both channels at the same time. Continuous random variables X and Y represent the message delays (in hours) over parallel channels I and II, respectively. These two random variables are independent, and both are uniformly distributed from to hours. A message is considered received as soon as it arrives on any one channel, and it is considered verified as soon as it has arrived over both channels. (a) Determine the probability that a message is received within 5 minutes after it is sent. Solution. (This problem is similar to Eample.5, the Romeo & Juliet problem in the tetbook.) Because the marginal PDF s are uniform and the X, Y are independent, the pair (X, Y ) 7

8 /4 /4 y /4 /4 y /4 /4 y (a) (b) (d) Figure 4: The events for Parts (a), (b), and (d) for Problem 6. The joint PDF of X and Y is uniform over the unit square. is uniformly distributed in the unit square as depicted in Figure 4. The probability of any event in this sample space is equal to its area. The event that a message is received within 5 minute is equivalent to X /4 ory /4, which corresponds to the shaded region in Figure 4(a). The probability of this event is equal to the area which is 7/6. (b) Determine the probability that the message is received but not verified within 5 minutes after it is sent. Solution. This event is equivalent to (X /4 and Y > /4) or (Y /4 and X>/4), which is depicted as the shaded region in Figure 4 (b). The probability of this event is equal to 3/6+3/63/8. (c) Let T represent the time (in hours) between transmission at A and verification at B. Determine the cumulative distribution function F T (t), and then differentiate it to obtain the PDF f T (t). Solution. By definition of CDF, F T (t) P(T t) P(X t and Y t) P(X t)p(y t) (due to independence), t <, F X (t)f Y (t) t, t,, t >. Differentiating, we get: f T (t) df T (t) dt { t, t,, otherwise. (d) If the attendant at B goes home 5 minutes after the message is received, what is the probability that he is present when the message should be verified? Solution. If he is present for verification, it means that the message arrived over one channel within 5 minutes of arriving over the other one: X Y /4. This is illustrated in Figure 4(d). The probability of this event is equal to minus the area of the two triangles and we get: P( X Y /4) ( /4) 7/6. 8

9 (e) If the attendant at B leaves for a 5-minute coffee break right after the message is received, what is the probability that he is present at the proper time for verification? Solution. This event is equivalent to X Y > /4. Its probability is: P( X Y /4) 9/6. (f) The management wishes to have the maimum probability of having the attendant present at both reception and verification. Would they do better to let him take his coffee break as described above or simply allow him to go home 45 minutes after transmission? Solution. If the attendant is allowed to go home 45 minutes after the transmission, the probability that he will be present at the both reception and verification is equal to the probability that both messages arrive within 45 minutes. That is: P(X 3/4 and Y 3/4) (3/4) 9/6. This probability is equal to what we obtained in Part (e). So it does not matter which decision the manager will make in terms of maimizing the probability of the attendant s presence at both reception and verification. (But note that the coffee break strategy will keep the attendant at work for a longer time on average. Indeed, when the message is received more than 45 minutes after the transmission, he will need to stay there for more than 45 minutes. In the second scheme, his working time is fied to be 45 minutes.) 9