RVE

This page lists questions and answers that were posted by visitors named RVE.

Questions

The following questions were asked by visitors named RVE.

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,...

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 indep...

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 th...

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...

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<x≤1,otherwise, we have f_Y(y) = {g(y)...

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...

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 ba...

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) is ro...

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...

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,...

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 put opti...

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...

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...

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 t...

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 the eur...

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. Let...

Answers

The following answers were posted by visitors named RVE.

part (a) 1 and 2 No. part (b) 1. No. 2. Looks like conditional probability (i failed this question. This answer was not used back in the day) scenarios: (the only possibility for an 8) 3,5 =0.2*0.2 4,4=0.2*0.2 5,3=0.2*0.2 at least one of the dices was a 3...

(48) (4) ----- (52) (8)

question #1. Easy! 4/52

it is all narrowed down to: 29/89

1. (6)(1/4)^2 *(3/4)^4 (2) 2. 1/3 3. 1/20

a) Forest A b) 1/3 c)0.5263 d)1/20 e)0.9722 f)0.1333

1. 2/3 2. No 3. 0.2963

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

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

c= 5/64 P(Y<X)= 83/128 P(Y=X)= 1/32 P(X=1)= 10/64 P(X=2)= 17/64 P(X=3)= 0 P(X=4)= 37/64 E[X]= 3 E[XY]= 227/32 var(X)= 3/2

c= 5/64 P(Y<X)= 83/128 P(Y=X)= 1/32 P(X=1)= 10/64 P(X=2)= 17/64 P(X=3)= 0 P(X=4)= 37/64 E[X]= 3 E[XY]= 227/32 var(X)= 3/2

P(−1≤Y≤2)= 0.3413

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

1. a= 0.4286 2. For 0≤y≤1, fY(y)= 0.6429 For 1<y≤2, fY(y)= 0.2142*(4-y^2) 3. E[1/X∣Y=3/2]= 0.5714

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

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

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

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

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

2. 155/7776 3. 25/108

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

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

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

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

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 t...

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

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...

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 o...

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

1. 0.46091 2. 0.8

1. 0.6667 3.0.2963

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

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

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

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

c= 5/64 P(Y<X)= 83/128 P(Y=X)= 1/32 P(X=1)= 10/64 P(X=2)= 17/64 P(X=3)= 0 P(X=4)= 37/64 E[X]= 3 E[XY]= 227/32 var(X)= 3/2

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

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

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

1. a= 0.4286 2. For 0≤y≤1, fY(y)= 0.6429 For 1<y≤2, fY(y)= 0.2142*(4-y^2) 3. E[1/X∣Y=3/2]= 0.5714

1. Are X and Y independent? NO 2. Find fX(x). Express your answers in terms of x using standard notation . If 0<x<1, fX(x)= x/2 If 1<x<2, fX(x)= -3*x/2+3 Find fY|X(y∣0.5). If 0<y<1/2, fY|X(y∣0.5)= 2 3. Find fX|Y(x∣0.5). If 1/2<x<1, fX|Y(x∣0.5)= 0.5 If 1<x...

For general fX, when y>0, fY(y)= Solution: f_x(ln(y))/y When fX(x) = {1/3,0,if −2<x≤1,otherwise, we have fY(y) = {g(y),0,if a<y≤b,otherwise. Give a formula for g(y) and the values of a and b using standard notation . (In your answers, you may use the symb...

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 WHOLE PROBLE...

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

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

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 ???

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

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....

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

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

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!!

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

2. P(X≤U)= 0.25

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

the maximum number of tickets to be sold: 320

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

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

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)

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

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

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

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

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)

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)

17/10 41/17 b/a

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

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

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

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

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

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

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

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

3. 1.6 6.0.283333

please solve P(XY≥1)= 0.40343

p(XY >= 1) = 0.40343

The true value for c = 0.29957

true value for c = 0.29957

x_1 minXi i c=0.29957