RVE
Most popular questions and responses by RVE
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 
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 
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 
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 
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 
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 
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 sixsided die (with sides labelled 1,2,…,6)
asked on May 14, 2015 
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 
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 
MathematicalModels
Consider the BlackScholesMerton 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 
MathematicalModels
A defaultfree 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 3month defaultfree zerocoupon bond is traded at 99,
asked on May 18, 2015 
MathematicalModels
Consider a BlackScholesMerton model with r=0.1, T=0.5 years, S(0)=100. Suppose the BlackScholes 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 
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 
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 
MathematicalModels
Consider the Vasicek model for the short rate dr(t)=(b−ar(t))dt+γdW1(t) and the BlackScholesMerton 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 riskneutral probability, and they have correlation ρ.
asked on May 18, 2015 
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 
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 
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=3X1 , 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^21. For y>=1, Y = 3X^2  1. For y
asked on March 17, 2015 
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

Probability
x_1 minXi i c=0.29957
posted on May 26, 2015 
Probability
true value for c = 0.29957
posted on May 26, 2015 
Maths Probability
The true value for c = 0.29957
posted on May 26, 2015 
Probability
1 7/9 0.40343
posted on May 26, 2015 
Math Probability
p(XY >= 1) = 0.40343
posted on May 26, 2015 
Maths Probability
please solve P(XY≥1)= 0.40343
posted on May 26, 2015 
Probability
3. 1.6 6.0.283333
posted on May 20, 2015 
probability
answers to "Problem 1: Steadystate 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 
probability
answers to "Problem 1: Steadystate 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 
probability
5. 2*LW*LE/((LW+LE)^2) 6.a k1 5.b LE^7*LW^(k7)/(LE+LW)^k
posted on May 13, 2015 
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 k1 5.b LE^7*LW^(k7)/(LE+LW)^k
posted on May 13, 2015 
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 
Probability
5. 2*20^n/(22^(n+1))
posted on May 13, 2015 
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 
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 
Probability
a=(lambda*s)^(mn)*e^(lambda*s) b= mn c= lambda^m*s^(mn)*t^n*e^(lambda*(s+t)) d= mn f= (s^(mn)*t^n)/((s+t)^m) g= m h= mn E[NM]= (lambda*t)*(lambda*s)+lambda*t+(lambda*t)^2
posted on May 11, 2015 
Probability
b) 41/17
posted on May 11, 2015 
probability
17/10 41/17 b/a
posted on May 11, 2015 
probability
1) 2*lambda*exp(2*lambda*t) 2) lambda*e^(lambda*x) 3) yes 4. 2*lambda*exp(lambda*t)*(1exp(lambda*t)) 5) 3/(2*lambda)
posted on May 5, 2015 
Probability
1.1. h 1.2. 2.25/n 2. 22500 3. 90000 4. H1.96*(3^0.5)/((4*n)^0.5), H+1.96*(3^0.5)/((4*n)^0.5)
posted on April 30, 2015 
Probability
1.1. h 1.2. 2.25/n 2. 22500 3. 90000
posted on April 30, 2015 
Probability
1.1. h 1.2. 2.25/n 2. 22500 3. 90000
posted on April 30, 2015 
probability
5. H1.96*(3^0.5)/((4*n)^0.5), H+1.96*(3^0.5)/((4*n)^0.5)
posted on April 25, 2015 
Probability
5. H1.96*(3^0.5)/((4*n)^0.5), H+1.96*(3^0.5)/((4*n)^0.5)
posted on April 25, 2015 
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 H1.96*(3^0.5)/((4*n)^0.5), H+1.96*(3^0.5)/((4*n)^0.5)
posted on April 25, 2015 
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 
probability
1. alpha = ln(mu/(2*lambda))/(mulambda)
posted on April 15, 2015 
Probability
the maximum number of tickets to be sold: 320
posted on April 15, 2015 
Probability
E[XY] = 3 var(X+Y)= 33 checked!!
posted on April 15, 2015 
Probability
2. P(X≤U)= 0.25
posted on April 15, 2015 
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 
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 
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 
Probability
1. p*2^(3k)  p*2^(3k) + (1p)*2^k 2. k ≤ 3/2 + 1/2*log_2(p/(1p)) 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 
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 
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 
Probability
Let N^=c1A+c2 be the LLMS estimator of N given A. Find c1 and c2 in terms of p. c1= 1p PLEASE, could you help out by giving away the answer for c2 ???
posted on April 3, 2015 
Probability
3. second choice q = k/sum(k, i=1) t_i
posted on April 3, 2015 
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 
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 
probability
1. 2*z
posted on April 3, 2015 
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 
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 
probability
1. a= 0.4286 2. For 0≤y≤1, fY(y)= 0.6429 For 1
posted on April 3, 2015 
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 
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 
Probability
2. 0.01622
posted on April 3, 2015 
Probability
1. p*(1p) 2. n*p*(1p) 3. p*(1p) 4. 0 5. p^2*(1p)^2 6. 57/64
posted on April 3, 2015 
probability
c= 5/64 P(Y
posted on April 3, 2015 
Probability
1. 1/n! 2.(nm)!/n! 3. 1/(n m) 4. (1p)^m 5. (n m)*(1p)^m * p^(nm)
posted on April 3, 2015 
Probability
4. c1=3, c2= 4, c3=6
posted on April 3, 2015 
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 
Probability
2. 1/13*4*(18 12)/(52 13)
posted on April 3, 2015 
probability
1. 0.6667 3.0.2963
posted on April 3, 2015 
Probability
1. 0.46091 2. 0.8
posted on April 3, 2015 
probability
(4*n6)/(n*(n1))
posted on April 3, 2015 
probability
0.125 0.125 0.375 0.5
posted on April 3, 2015 
probability
0.4 0.75 0.3
posted on April 3, 2015 
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 
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 
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 
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 
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 
probability
could you help me out with some of the other answers???
posted on March 20, 2015 
Probability
3. Official answer 1/(2*(2p))
posted on March 12, 2015 
Probability
2. official answer: 1(1(X)*(1Y)*(1Z))*(1(1X)*(Y)*(1Z))*(1(1X)*(1Y)*(Z))
posted on March 12, 2015 
Probability
2. 155/7776 3. 25/108
posted on March 9, 2015 
Probability
1. (1X)*(1Z) be aware your keyboard is set to US language not US international or other language
posted on March 9, 2015 
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 
Probability
Is Y2+Y3 independent of Y1? NO! Is Y2−Y3 independent of Y1? YES!
posted on March 8, 2015 
Probability
1. 6 different rolls = 6/6*5/6*4/6*3/6*2/6*1/6
posted on March 8, 2015 
Probability
1.a. 0.25*p*(1p)^(x1) 1.b. p*(1p)^(x1) 2. 0.5*p*(1p)^(x1) 3. ¿ 0*p ? (not sure about this one) 4. 2
posted on March 8, 2015 
Probability
1. a= 0.4286 2. For 0≤y≤1, fY(y)= 0.6429 For 1
posted on February 28, 2015 
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 
Math
P(−1≤Y≤2)= 0.3413
posted on February 28, 2015 
Probability
c= 5/64 P(Y
posted on February 21, 2015 
probability
c= 5/64 P(Y
posted on February 21, 2015 
Probability
a. False b. False c. True d. False e. True
posted on February 21, 2015 
probability
1. p*(1p) 2. n*p*(1p) 3. p*(1p) 4. 0 5. p^2*(1p)^2 6. 57/64
posted on February 21, 2015 
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 
Probability
1. 2/3 2. No 3. 0.2963
posted on February 14, 2015