Times Series Analysis Assignment #1 (Jere Xu)

Part 1

(a) Create a vector of the positive odd integers less than 100. Remove the values greater than 60 and less than 80. Find the variance of the remaining set of values.

v <- 1:100
v <- v[v %% 2 != 0]
v

 [1]  1  3  5  7  9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63
[33] 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99

v <- v[v > 60]
v <- v[v < 80]
v

 [1] 61 63 65 67 69 71 73 75 77 79

variance <- var(v)
variance

[1] 36.66667

(b) What’s the difference in output between the commands 2*1:5 and (2*1):5? Why is there a difference?

2*1:5

[1]  2  4  6  8 10

(2*1):5

[1] 2 3 4 5

The first command is having a scalar be multiplied by a vector, which is 1:5. The second command is a vector with the left value being the product of 2*1, which is 2. The parentheses affected the order of operations.

(c) If you wanted to enter the odd numbers from 1 to 19 in the variable x, what command would you use?

x <- 2*(1:10)-1
x

 [1]  1  3  5  7  9 11 13 15 17 19

(d) If you create a variable using the following command y=c(-1, 2, -3, 4, -5), what command would put the positive values of y into the variable z?

y <- c(-1, 2, -3, 4, -5)
z <- y[y > 0]
z

[1] 2 4

(e) What R command would give you the 95th percentile for a chi-squared distribution with 10 degrees of freedom?

qchisq(0.95, 10)

[1] 18.30704

(f) Generate a vector of 1000 standard normal random variables using the command x=rnorm(1000). Use R to give a five number summary of your simulated data. What is the mean and variance of your x variable? Make and print a histogram for this data

x = rnorm(1000)

summary(x)

    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
-3.54719 -0.71405 -0.08119 -0.02399  0.65335  3.08018 

var(x)

[1] 0.993529

The mean is -0.02399 and the variance is 0.993529.

hist(x)

Part 2

For Stock B: 6% return with prob. 0.7; −8% return with prob. 0.3. Calculate its mean value and sample standard deviation.

s_mean <- 0.06*0.7 + -0.08* 0.3
s_variance <- (0.06**2)*0.7 + ((-0.08)**2)*0.3

s_mean

[1] 0.018

s_std <- sqrt(s_variance)


s_std

[1] 0.06663332

Part 3

Stock C: 45.42% return with prob. 0.7; −100% return with prob. 0.3. Calculate its mean value and sample standard deviation.

s_mean <- 0.4542*0.7 + -1* 0.3
s_variance <- (0.4542**2)*0.7 + ((-1)**2)*0.3

s_mean

[1] 0.01794

s_std <- sqrt(s_variance)


s_std

[1] 0.6666396

Part 4

Let X be a Bernoulli random variable which is defined as X = 1 if we get a head with probability p and X = 0 if we get a tail with probability 1 − p. Calculate E(2X), var(3X + 4), cov(3X, 5) and cov(4X, 6X + 1000).

• The expected value of a Bernoulli random variable is p. The variance of a Bernoulli random variable is p(1-p).

• The expected value of 2X is E(X)*2, which is 2p.
• Var(3X + 4) is equal to 3*Var(X), which is 3p*(1-p)
• The covariance of any random variable with a constant is zero.

• Cov(4X, 6X + 1000) is the same as Cov(4X, 6X), which is the same as 24*Cov(X, X), which is 24*Var(X), which is 24p*(1-p).

Part 5

Let X be a random variable whose mean value is 4 and variance is 5 and Y be a random variable whose mean value is 2 and variance is 10. Suppose cov(X, Y ) = 1, calculate the correlation between X and Y and var(X + Y ). If X and Y are independent, calculate var(X + Y ).

• cov(X, Y) = 1
• The standard deviation of X is sqrt(5) and the standard deviation of Y is sqrt(10). Therefore, the correlation between X and Y is 1 divided by the product of sqrt(5) and sqrt(10).

std_x <- sqrt(5)
std_y <- sqrt(10)

corr <- 1 / (std_x * std_y)

corr

[1] 0.1414214

• Variance(X + Y) = Var(X) + Var(Y) + 2cov(X, Y)

• Variance(X + Y) = 5 + 10 + 2*1 = 17

• When X and Y are independent, then var(X + Y) = Var(X) + Var(Y)

• Var(X + Y) = 5 + 10 = 15

Part 6

Prove cov(X, Y ) = cov(Y, X).

• Cov(X, Y) = E(XY) - μxμy
• Cov(Y, X) = E(YX) - μyμx
• There is the commutative property for expected value, therefore these two expressions are the same thing.