EE 193 - Department of Electrical & Computer Engineering - Tufts ...

EE 193 - Applied Probability and Statistics for EngineersDepartment of Electrical and Computer EngineeringTufts University Fall 2007Problem Set #4 SolutionsReading: Yates and Goodman, Chapter 3Problem 1• 3.4.2 From the Appendix in the text, the variance of an exponential RV is 1/λ 2 . Since thevariance of Y is 25, we conclude that for this RV, λ = 1/5.– The PDF isp Y (y) ={15 ey/5 y > 00 else– Since the variance satisfies σ 2 Y = E[Y 2 ] − (E[Y ]) 2 we have E[Y 2 ] = σ 2 Y + (E[Y ])2 =1/λ 2 + 1/λ 2 = 2/λ 2 = 2/25– The required probability isP [Y > 5] =∫ ∞515 e−y/5 dy = e −1• 3.4.9Let T be the time for a call. In terms of T , the costs for plans A and B are:Now we computed expected values:C A = 10T{C B =99 T < 2099 + 10(T − 20) T > 20∫ ∞E[C A ] =10 τ 10te−t/τ dt= 10E[T ] = 10τE[C B ] =∫ 20099 1 τ e−t/τ dt += 99 − 10τe −20/τ∫ ∞20(99 − 10(t − 20)) 1 τ e−t/τ dtPlaying around a bit with Matlab, gives us that for τ greater than about 12.34 minutes,calling plan B is a better deal than calling plan A.• 3.5.3 We have that A ∼ (N(0, σ X ) meaning[ |X|P [|X| ≤ 10] = P ≤ 10 ]= Φσ X σ X( 10σ X)− Φ(− 10 ) ( ) 10= 2Φ − 1 = 0.1σ X σ Xwhere I have used the method of Example 3.17. Now, solving the above for σ X gives[ ( )] 1.1 −1σ X = 10 Φ −1 ≈ 10/0.13 ≈ 7721

• 3.5.5 We have T ∼ N(µ, 15). From the problem statement we know[ t − µP [T > 10] = 0.5 → P15 > 10 − µ ] ( ) 10 − µ= 1 − Φ = 1 1515 2which means Φ((10 − µ)/15) = 1/2. Since Φ −1 (0.5) = 0 we conclude that µ = 10. Hence[ ]T − 10 32 − 10P [T > 32] = P >15 15= 1 − Φ(22/15) = 1 − Φ(1.47) ≈ 1 − 0.9292 = .0708[ T − 10P [T < 0] = P < 0 − 10 ]15 15= Φ(−0.66) = 1 − Φ(0.66) ≈ 1 − 0.7454 = 0.2546[ ]T − 10 60 − 10P [T > 60] = P 84] =23, 000100, 000, 000 .[ ] [ ]X − 70 84 − 7014> = 1 − Φ =σ X σ X σ X23100, 000 .σ X =14) ≈ 4.00 using norminv in MatlabΦ(1 −1 − 23100,000(b) We have[ X − 70P [X > 96] = P >4]96 − 70= 1 − Φ (6.5)4(c) The probability that any individual is over 90 inches is( ) 90 − 70P [X > 90] = 1 − Φ= 1 − Φ(5) ≈ 2.87e − 7 ≡ p.4Now, for this problem, there are 100,000,000 men in the U.S. If each man is a Bernoullitrial with p = 2.87e − 7, then the probability that there are no men over 90 inches inheight is (1 − p) 100,000,000 ≈ 3.56e − 13. This is a very small number indeed.(d) Here we are asking for the expected number of “successes” in 100,000,000 trials of abinomial RV with p = 2.87e − 7. This is just 100, 000, 000 × p ≈ 28.7 or roughly 30 men.2

• 3.7.2 We start by finding the CDF for Y :F Y (y) = P [Y < y] = P [ √ X < y] = P [X < y 2 ]=∫ y 20λe −λx dx= 1 − e −λy2which holds for y ≥ 0. Differentiating with repsect to y gives the PDF:{f Y (y) = dF Y (y) 2λye −λy2 y ≥ 0=dy 0 elseThe form of a Rayleigh distribution isf X (x) =The two are the same if a 2 = 2λ or a = √ 2λ.• 3.7.8{a 2 xe −x2 a 2 /2x ≥ 00 else(a) First, the radius, R, of a circle with unit circumference satisfies 2πR = 1 → R = 1/2π.Now, if we let A denote the angle subtended by X we know from elementary analyticgeometry that A = 2πX. Next, Y , the area of the circle with radius 1/2π enclosed in asector with angle A is, using integration in polar coordinates:Y =but since A = 2πX we end up with∫ 1/2π0rdr∫ A0dθ =A8π 2Y = X 4π(b) Using the definition of F Y (y) we have[ ] XF Y (y) = P [Y < y] = P4π < y⎧⎪⎨ 0 y < 0∫= P [X < 4πy] = 4πy01dx = 4πy 0 < y 14π(c) Taking derivatives we getf Y (y) = dF Y (y)dy={4π 0 < y < 14π0 else3

(d) This is a uniform RV with parameters a = 0 and b = 1 . Hence the expected value is• 3.8.118π .(a) The probability of B isSo the conditional PDF isP [B] =f X |B(x|B) = f X(x)P [B]∫ 3−34π1dX = 6/10.10x ∈ B ={ 1/106/10 = 1 6|x| < 30 else(b) Since the conditional PDF is uniform with a = −3 and b = 3, the conditional expectedvalue is E[X|B] = (a + b)/2 = 0(c) From the a and b parameters identified in part (b), σ 2 X|B = (b − a)2 /12 = 36/12 = 3• 3.8.5 Since E[T ] = .01 = 1/λ, the PDF for T is{100e −100t t > 0f T (t) =0 elseand the probability of the event A = {T > .02} isHence, the conditional PDF isf T |A (t|A) =P [T > .02] =∫ ∞.02100e −100t dt = e −2 .{100e −2 e −100t = 100e −100(t−.02) t > 0.020 elseBut this is just a “shifted” exponential random variable. Hence, it’s mean (i.e., the conditionalexpected value) will be the mean of the original,0.01, plus 0.02 which is 0.03. Since we stillhave a decay paramater, λ = 100 (i.e., since the shape of the random variance is the sameregardless of the shift along the T axis) the conditional variance will be the same as theoriginal, 1/λ 2 = 10 −4 .• 3.8.7(a) Given that a person is healthy, the PDF is Gaussian with mean 90 and standard deviation20. Hence the PDF is1f X|H (x|H) = √ e −(x−90)/800800π(b)[ X − 90P [T + |H] = P [X ≥ 140|H] = P20[ X − 90P [T − |H] = P [X ≤ 110|H] = P20≥≤]140 − 90= 1 − Φ(5/2) = 0.00620]110 − 90= Φ(1) = 0.841204

(c) Start from the definition of conditional probabilityP [H|T − ] = P [HT − ]P [T − ]=P [T − |H]P [H]P [T − |H]P [H] + P [T − |D]P [D] .We have all of the pieces of the formula on the right hand side except for P [T − |D] whichis[ ]X − 160P [T − 110 − 160|D] = P [X ≤ 110|D] = P ≤ = 1 − Φ(5/4) = 0.106.40 40Putting everything together gives P [H|T − ] = 0.986.(d) Note that P [T 0 |H] = 1 − P [T + |H] − P [T − |H] = 0.153. We can think of this need torepeat tests as a Bernoulli process where a “failure” is given my an ambiguous test andhas probability of p = 0.153 while a success is an unambiguous result and occurs withprobability 1−p = 0.847. The number number of tests until the first unambiguous resultgiven that a person is healthy is a geometric random variable{(1 − p)p n n = 1, 2, 3, . . .p N|H (n|H) =0 else5