RVE

Most popular questions and responses by RVE
  1. probability

    This figure below describes the joint PDF of the random variables X and Y. These random variables take values in [0,2] and [0,1], respectively. At x=1, the value of the joint PDF is 1/2. (figure belongs to "the science of uncertainty) 1. Are X and Y

    asked on February 28, 2015
  2. Probability

    The joint PMF, pX,Y(x,y), of the random variables X and Y is given by the following table: (see: the science of uncertainty) 1. Find the value of the constant c. c = 0.03571428571428571428 2. Find pX(1). pX(1)= 1/2 3. Consider the random variable Z=X2Y3.

    asked on February 25, 2015
  3. probability

    Tossing a pair of coins We have a white coin, for which P(Heads)=0.4 and a black coin for which P(Heads)=0.6. The flips of the same or of different coins are independent. For each of the following situations, determine whether the random variable N can be

    asked on April 11, 2015
  4. probability

    Problem 2: Oscar's running shoes Oscar goes for a run each morning. When he leaves his house for his run, he is equally likely to use either the front or the back door; and similarly, when he returns, he is equally likely to use either the front or the

    asked on May 14, 2015
  5. probability

    Determine whether each of the following statement is true (i.e., always true) or false (i.e., not always true). 1. Let X be a random variable that takes values between 0 and c only, for some c≥0, so that P(0≤X≤c)=1. Then, var(X)≤c2/4. TRUE 2. X and

    asked on February 28, 2015
  6. probability

    For each one of the following figures, identify if it is a valid CDF. The value of the CDF at points of discontinuity is indicated with a small solid circle. (original images belonging to: "The science of uncertainty") 1. No, it is not a valid CDF 2. Yes,

    asked on February 28, 2015
  7. probability

    Problem 3: Checking the Markov property For each one of the following definitions of the state Xk at time k (for k=1,2,…), determine whether the Markov property is satisfied by the sequence X1,X2,…. A fair six-sided die (with sides labelled 1,2,…,6)

    asked on May 14, 2015
  8. probability

    Let K be a discrete random variable with PMF pK(k)=⎧⎩⎨⎪⎪1/3,2/3,0if k=1,if k=2,otherwise. Conditional on K=1 or 2, random variable Y is exponentially distributed with parameter 1 or 1/2, respectively. Using Bayes' rule, find the conditional PMF

    asked on February 28, 2015
  9. MathematicalModels

    At time zero you enter a long position in a forward contract on 1 share of the stock XYZ at the forward price of 10.00. Moreover, you buy one exotic derivative, with the same maturity as the forward contract, which pays to the holder exactly one share of

    asked on May 18, 2015
  10. MathematicalModels

    Consider the Black-Scholes-Merton model for two stocks: dS1(t)=0.1S1(t)dt+0.2S1(t)dW1(t) dS2(t)=0.05S2(t)dt+0.1S2(t)dW2(t) Suppose the correlation between W1 and W2 is 0.4. Consider the dynamics of the ratio S1/S2, where A,B,C,D,F,G,I,J,K,L are constants

    asked on May 18, 2015
  11. MathematicalModels

    A default-free coupon bond maturing in 6 months, that pays a coupon of 2.00 after 3 months and makes a final payment of 102.00 (the last coupon and the principal), trades at 101.00 today. Moreover, a 3-month default-free zero-coupon bond is traded at 99,

    asked on May 18, 2015
  12. MathematicalModels

    Consider a Black-Scholes-Merton model with r=0.1, T=0.5 years, S(0)=100. Suppose the Black-Scholes price of the digital option that pays one dollar if S(T)≥100 and zero otherwise, is equal to 0.581534. Enter the value of volatility σ (hint: it is one of

    asked on May 18, 2015
  13. MathematicalModels

    Suppose you have written a derivative that pays the squared value of the stock price at maturity T=1; that is, it pays S2(1). The stock currently trades at S(0)=100. Your model is a single period binomial tree with up value for the stock equal to 102 and

    asked on May 18, 2015
  14. probability

    The PDF of exp(X) Let X be a random variable with PDF f_X. Find the PDF of the random variable Y=e^X for each of the following cases: For general f_X, when y>0, f_Y(y)= f_X(ln y) --------- y When f_X(x) = {1/3,0,if −2

    asked on March 26, 2015
  15. MathematicalModels

    Consider the Vasicek model for the short rate dr(t)=(b−ar(t))dt+γdW1(t) and the Black-Scholes-Merton model for a stock S dS(t)=r(t)S(t)dt+σS(t)dW2(t) where W1 and W2 are Brownian motions under the risk-neutral probability, and they have correlation ρ.

    asked on May 18, 2015
  16. MathematicalModels

    The price of a US stock is given by dS(t)/S(t)=μdt+σdW1(t) The exchange rate Dollar/Euro is given by dQ(t)/Q(t)=βdt+δdW2(t) where W1 has correlation ρ with W2. >> (i) Select the Brownian motions W∗1 and W∗2 such that the discounted dollar value of

    asked on May 18, 2015
  17. MathematicalModels

    Assume that the future dividends on a given stock S are known, and denote their discounted value at the present time t by D¯(t). For American call and put options values C(t) ,P(t), suppose we have that P(t)−D¯(t)−K>C(t)−S(t) Suppose you sell the

    asked on May 18, 2015
  18. probability

    FUNCTIONS OF A STANDARD NORMAL The random variable X has a standard normal distribution. Find the PDF of the random variable Y, where: 1. Y=3X-1 , Y = 3X - 1 answer: fY(y)=1/3*fX*(y+1/3) f_ Y(y)=1/3*f_ X*(y+1/3) 2. Y=3X^2-1. For y>=-1, Y = 3X^2 - 1. For y

    asked on March 17, 2015
  19. probability

    Exercise: Sections of a class A class consists of three sections with 10 students each. The mean quiz scores in each section were 40, 50, 60, respectively. We pick a student, uniformly at random. Let X be the score of the selected student, and let Y be the

    asked on March 13, 2015
  1. Probability

    x_1 minXi i c=0.29957

    posted on May 26, 2015
  2. Probability

    true value for c = 0.29957

    posted on May 26, 2015
  3. Maths Probability

    The true value for c = 0.29957

    posted on May 26, 2015
  4. Probability

    1 7/9 0.40343

    posted on May 26, 2015
  5. Math Probability

    p(XY >= 1) = 0.40343

    posted on May 26, 2015
  6. Maths Probability

    please solve P(XY≥1)= 0.40343

    posted on May 26, 2015
  7. Probability

    3. 1.6 6.0.283333

    posted on May 20, 2015
  8. probability

    answers to "Problem 1: Steady-state convergence" . Belonging to this very same problem set.... 1.a. False 1.b. False 2.a. False 2.b. False 3.a. True 3.b. True

    posted on May 18, 2015
  9. probability

    answers to "Problem 1: Steady-state convergence" , belonging to this very same problem set.... 1.a. False 1.b. False 2.a. False 2.b. False 3.a. True 3.b. True

    posted on May 14, 2015
  10. probability

    5. 2*LW*LE/((LW+LE)^2) 6.a k-1 5.b LE^7*LW^(k-7)/(LE+LW)^k

    posted on May 13, 2015
  11. Probability

    1. LW/(LW+LE) 2. LE*exp(-LE*x) 3. exp(-2*LW*t) 4. LE^7*v^6*exp(-LE*v)/720 5. 2*LW*LE/((LW+LE)^2) 6.a k-1 5.b LE^7*LW^(k-7)/(LE+LW)^k

    posted on May 13, 2015
  12. probability

    1. lambda 2. Yes, it is a Poisson process. 3. lambda+mu 4. 1/mu 5. 2*20^n/(22^(n+1))

    posted on May 13, 2015
  13. Probability

    5. 2*20^n/(22^(n+1))

    posted on May 13, 2015
  14. Probability

    1. lambda 2. Yes, it is a Poisson process. 3. lambda+mu 4. 1/mu 5. 2*20^n/(22^(n+1))

    posted on May 13, 2015
  15. probability

    the right answers are: 1)1/8 2)1/8 3)1/8 4)1/6 5)5/16 6)10/3

    posted on May 12, 2015
  16. Probability

    a=(lambda*s)^(m-n)*e^(-lambda*s) b= m-n c= lambda^m*s^(m-n)*t^n*e^(-lambda*(s+t)) d= m-n f= (s^(m-n)*t^n)/((s+t)^m) g= m h= m-n E[NM]= (lambda*t)*(lambda*s)+lambda*t+(lambda*t)^2

    posted on May 11, 2015
  17. Probability

    b) 41/17

    posted on May 11, 2015
  18. probability

    17/10 41/17 b/a

    posted on May 11, 2015
  19. probability

    1) 2*lambda*exp(-2*lambda*t) 2) lambda*e^(-lambda*x) 3) yes 4. 2*lambda*exp(-lambda*t)*(1-exp(-lambda*t)) 5) 3/(2*lambda)

    posted on May 5, 2015
  20. Probability

    1.1. h 1.2. 2.25/n 2. 22500 3. 90000 4. H-1.96*(3^0.5)/((4*n)^0.5), H+1.96*(3^0.5)/((4*n)^0.5)

    posted on April 30, 2015
  21. Probability

    1.1. h 1.2. 2.25/n 2. 22500 3. 90000

    posted on April 30, 2015
  22. Probability

    1.1. h 1.2. 2.25/n 2. 22500 3. 90000

    posted on April 30, 2015
  23. probability

    5. H-1.96*(3^0.5)/((4*n)^0.5), H+1.96*(3^0.5)/((4*n)^0.5)

    posted on April 25, 2015
  24. Probability

    5. H-1.96*(3^0.5)/((4*n)^0.5), H+1.96*(3^0.5)/((4*n)^0.5)

    posted on April 25, 2015
  25. Probability

    4. Assume that X is uniformly distributed on [0,3]. Using the Central Limit Theorem, identify the most appropriate expression for a 95% confidence interval for h H-1.96*(3^0.5)/((4*n)^0.5), H+1.96*(3^0.5)/((4*n)^0.5)

    posted on April 25, 2015
  26. Probability

    The above answers are wrong, here is the official indictment: 1. 1/(theta*ln(2)) 3. x/(2*ln(2)) 4 . c1= 0.06452 c2= 0.58065

    posted on April 15, 2015
  27. probability

    1. alpha = ln(mu/(2*lambda))/(mu-lambda)

    posted on April 15, 2015
  28. Probability

    the maximum number of tickets to be sold: 320

    posted on April 15, 2015
  29. Probability

    E[XY] = 3 var(X+Y)= 33 checked!!

    posted on April 15, 2015
  30. Probability

    2. P(X≤U)= 0.25

    posted on April 15, 2015
  31. Probability

    some answers P(X≥8)= 0.054799 var (X+Y) = 33 ?? (not really sure about this one) fellow classmate: help us out with E[XY] if you are reading this....Feel free to provide your answer to the whole TEST

    posted on April 15, 2015
  32. probability

    1. mean = 40, variance = 24 2. mean = 50 3. mean = 50, variance = 24 4. mean = 0, variance = 0 the variance for question 2 is missing...Help us out, please!!

    posted on April 11, 2015
  33. probability

    2.) 1/3*e^(-μ*α) + 2/3*(1-(e^-λ*α)) please, share what you have got for the whole problem set!!

    posted on April 7, 2015
  34. Probability

    1. p*2^(3-k) ------------------------ p*2^(3-k) + (1-p)*2^k 2. k ≤ 3/2 + 1/2*log_2(p/(1-p)) 4. it increases or stays the same please, if you are reading these answers ...Dont be selfish and share the other answers to the whole problem set

    posted on April 6, 2015
  35. Probability

    For 0≤q≤1, fQ∣A(q)= 4*q^3 P(B∣A) = 0.8 if you are reading these answers. Don't be selfish! you should share the other answers to the whole problem set....

    posted on April 6, 2015
  36. Probability

    2. Find the MAP estimate of Θ based on the observation X=x and assuming that 0≤x≤1. Express your answer in terms of x. For 0≤x≤1,θ^MAP(x)= x/2 PLEASE, if you are bugged with this problem set. Do not be mean and provide the other answers

    posted on April 3, 2015
  37. Probability

    Let N^=c1A+c2 be the LLMS estimator of N given A. Find c1 and c2 in terms of p. c1= 1-p PLEASE, could you help out by giving away the answer for c2 ???

    posted on April 3, 2015
  38. Probability

    3. second choice q = k/sum(k, i=1) t_i

    posted on April 3, 2015
  39. Probability

    c2 = n*ln(1) please! if you are reading this, be generous and provide your answers for the whole problem set

    posted on April 3, 2015
  40. probability

    1.Carry out this minimization and choose the correct formula for the MAP estimate, θ^1, from the options below. (second choice) θ^1=∑ni=1ti(yi−θ0−θ2t2i)σ2+∑ni=1t2i 4. t1 and t2 = 10 IF YOU ARE READING THIS. PLEASE, PROVIDE YOUR ANSWERS TO THE

    posted on April 3, 2015
  41. probability

    1. 2*z

    posted on April 3, 2015
  42. probability

    For general fX, when y>0, fY(y)= Solution: f_x(ln(y))/y When fX(x) = {1/3,0,if −2

    posted on April 3, 2015
  43. Probability

    1. Are X and Y independent? NO 2. Find fX(x). Express your answers in terms of x using standard notation . If 0

    posted on April 3, 2015
  44. probability

    1. a= 0.4286 2. For 0≤y≤1, fY(y)= 0.6429 For 1

    posted on April 3, 2015
  45. Probability

    1 P(X>0.75)= 0.2266 2 P(X≤−1.25)= 0.1056 Let Z=(Y−3)/4. Find the mean and the variance of Z. 3. E[Z]= -0.25 4. var(Z)= 0.5625 5. P(−1≤Y≤2)= 0.3413

    posted on April 3, 2015
  46. Probability

    1 P(X>0.75)= 0.2266 2 P(X≤−1.25)= 0.1056 Let Z=(Y−3)/4. Find the mean and the variance of Z. 3. E[Z]= -0.25 4. var(Z)= 0.5625 5. P(−1≤Y≤2)= 0.3413

    posted on April 3, 2015
  47. Probability

    2. 0.01622

    posted on April 3, 2015
  48. Probability

    1. p*(1-p) 2. n*p*(1-p) 3. p*(1-p) 4. 0 5. p^2*(1-p)^2 6. 57/64

    posted on April 3, 2015
  49. probability

    c= 5/64 P(Y

    posted on April 3, 2015
  50. Probability

    1. 1/n! 2.(n-m)!/n! 3. 1/(n m) 4. (1-p)^m 5. (n m)*(1-p)^m * p^(n-m)

    posted on April 3, 2015
  51. Probability

    4. c1=3, c2= 4, c3=6

    posted on April 3, 2015
  52. Probability

    1. (6 2)*(1/4)^2*(3/4)^4 3. 0.05 4. c1=3, c2= 4, c3=6

    posted on April 3, 2015
  53. Probability

    2. 1/13*4*(18 12)/(52 13)

    posted on April 3, 2015
  54. probability

    1. 0.6667 3.0.2963

    posted on April 3, 2015
  55. Probability

    1. 0.46091 2. 0.8

    posted on April 3, 2015
  56. probability

    (4*n-6)/(n*(n-1))

    posted on April 3, 2015
  57. probability

    0.125 0.125 0.375 0.5

    posted on April 3, 2015
  58. probability

    0.4 0.75 0.3

    posted on April 3, 2015
  59. probability

    At least two of the events A, B, C occur. Event E6 Regions: 2 4 5 6 At most two of the events A, B, C occur. Event E2 Regions: 1 2 3 5 6 7 8 None of the events A, B, C occurs. Event E5 Region: 8 All three events A, B, C occur. Event E1 Region: 4 Exactly

    posted on April 3, 2015
  60. probability

    Let the random variable X be uniform on [0,2] and the random variable Y be uniform on [3,4] Determine the values of a, b, c, d, and e b=3, c=4, d=5, e=6 Let W=X+Y. The following figure shows a plot of the PDF of W. Determine the values of a, b, c, d, e, f,

    posted on March 26, 2015
  61. probability

    E[X]= 3   var(X)= 6 if you are reading this answer, please help us with the whole problem set....

    posted on March 22, 2015
  62. probability

    3. -n*(p_1)*(p_2) by the book solution. Instead of p_i or p_j we are given p_1 and p_2. In any case, k is useless hi there Juan Pro and Anonymous...First and foremost if you are able to provide the answers to the rest of this problem set....you are more

    posted on March 21, 2015
  63. probability

    ρ(X−Y,X+Y)= 0 ρ(X+Y,Y+Z)= 0.5 ρ(X,Y+Z)= 0 ρ(W,V)= (b)/((b^2+2c^2)^0.5) hope it helps! I am needing the other answers to the whole problem set

    posted on March 21, 2015
  64. probability

    could you help me out with some of the other answers???

    posted on March 20, 2015
  65. Probability

    3. Official answer 1/(2*(2-p))

    posted on March 12, 2015
  66. Probability

    2. official answer: 1-(1-(X)*(1-Y)*(1-Z))*(1-(1-X)*(Y)*(1-Z))*(1-(1-X)*(1-Y)*(Z))

    posted on March 12, 2015
  67. Probability

    2. 155/7776 3. 25/108

    posted on March 9, 2015
  68. Probability

    1. (1-X)*(1-Z) be aware your keyboard is set to US language not US international or other language

    posted on March 9, 2015
  69. Probability

    1. Not always true 2. Always true 3. Not always true 4. Always true (I am taking this exam, too)

    posted on March 8, 2015
  70. Probability

    Is Y2+Y3 independent of Y1? NO! Is Y2−Y3 independent of Y1? YES!

    posted on March 8, 2015
  71. Probability

    1. 6 different rolls = 6/6*5/6*4/6*3/6*2/6*1/6

    posted on March 8, 2015
  72. Probability

    1.a. 0.25*p*(1-p)^(x-1) 1.b. p*(1-p)^(x-1) 2. 0.5*p*(1-p)^(x-1) 3. ¿ 0*p ? (not sure about this one) 4. 2

    posted on March 8, 2015
  73. Probability

    1. a= 0.4286 2. For 0≤y≤1, fY(y)= 0.6429 For 1

    posted on February 28, 2015
  74. probability

    1 P(X>0.75)= 0.2266 2 P(X≤−1.25)= 0.1056 Let Z=(Y−3)/4. Find the mean and the variance of Z. 3. E[Z]= -0.25 4. var(Z)= 0.5625 5. P(−1≤Y≤2)= 0.3413

    posted on February 28, 2015
  75. Math

    P(−1≤Y≤2)= 0.3413

    posted on February 28, 2015
  76. Probability

    c= 5/64 P(Y

    posted on February 21, 2015
  77. probability

    c= 5/64 P(Y

    posted on February 21, 2015
  78. Probability

    a. False b. False c. True d. False e. True

    posted on February 21, 2015
  79. probability

    1. p*(1-p) 2. n*p*(1-p) 3. p*(1-p) 4. 0 5. p^2*(1-p)^2 6. 57/64

    posted on February 21, 2015
  80. probability

    P(X=0)= 1/3 P(X=1)= 2/9 P(X=−2)= 1/9 P(X=3)= 0 E[X]= 0 var(X)= 4/3 P(Y=0)= 1/3 P(Y=1)= 4/9 P(Y=2)= 0

    posted on February 20, 2015
  81. Probability

    1. 2/3 2. No 3. 0.2963

    posted on February 14, 2015