# 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

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.

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

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

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

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,

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)

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

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

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

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,

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

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

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

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.

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

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

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

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

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

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

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

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

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

posted on March 12, 2015
66. ## Probability

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