Homework about Financial Risk Management

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timer Asked: Feb 5th, 2019
account_balance_wallet $9.99

Question Description

Task 1

Go to Yahoo Finance and download historical data for SPY ETF over the last two years, starting from 2017 to today. Compute the daily return using the adjusted close price column. This should result in a new time series (column).

Task 2

Using the daily returns of the SPY ETF, calibrate a Geometric Brownian Motion process. Report the corresponding mu and sigma.

Task 3

Given the calibrated model, simulate the SPY price path over the next three months, i.e. 256/4=64 trading days. Provide a couple of plots to demonstrate this.

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Task 1 Go to Yahoo Finance and download historical data for SPY ETF over the last two years, starting from 2017 to today. Compute the daily return using the adjusted close price column. This should result in a new time series (column). Task 2 Using the daily returns of the SPY ETF, calibrate a Geometric Brownian Motion process. Report the corresponding mu and sigma. Task 3 Given the calibrated model, simulate the SPY price path over the next three months, i.e. 256/4=64 trading days. Provide a couple of plots to demonstrate this. FE535-A: Introduction to Financial Risk Management Week 2 Majeed Simaan1 1 Assistant Professor of Finance School of Business Stevens Institute of Technology February 4, 2019 1/31 Spring 2019 (Stevens) Financial Risk Management FE535 1 / 31 Agenda Introduction to Monte Carlo Methods I I Law of Large Numbers Pricing using Numerical Methods Univariate Stochastic Process I I Brownian Motion Simulating Stock Prices Application to Risk Management I I Value-at-Risk Simulating Portfolio Value/Returns 2/31 Spring 2019 (Stevens) Financial Risk Management FE535 2 / 31 Monte Carlo Simulations: Overview Monte Carlo (henceforth MC) Simulations are central to financial engineering and risk management In particular, MC methods allow risk managers to I I I avoid complicated analytical solutions price complex instruments, e.g. derivatives derive complex portfolio distribution Today, MC methods have become widespread thanks to technological advancement 3/31 Spring 2019 (Stevens) Financial Risk Management FE535 3 / 31 Monte Carlo Simulations: Overview Monte Carlo (henceforth MC) Simulations are central to financial engineering and risk management In particular, MC methods allow risk managers to I I I avoid complicated analytical solutions price complex instruments, e.g. derivatives derive complex portfolio distribution Today, MC methods have become widespread thanks to technological advancement Nonetheless, there are also drawbacks of MC methods MC relies heavily on the model’s assumptions, such as I I I Distribution shape Underlying parameters Pricing functions Risk managers should be aware of possible errors in such assumptions I recall Model Risk? 3/31 Spring 2019 (Stevens) Financial Risk Management FE535 3 / 31 Game 1 By flipping a coin, let’s consider the following game I I earn a $1 for heads (H) lose a $1 for tails (T) How much would you pay to participate in such a game? 4/31 Spring 2019 (Stevens) Financial Risk Management FE535 4 / 31 Game 1 By flipping a coin, let’s consider the following game I I earn a $1 for heads (H) lose a $1 for tails (T) How much would you pay to participate in such a game? The answer solely depends on the probability of H, denoted by π If it is a fair coin, then π = 0.5 and 0.5 × (−1) + 0.5 × 1 = 0 But what if the coin is unfair? I I If π > 0.5, would pay a premium to participate? If π < 0.5, would require a premium to participate? 4/31 Spring 2019 (Stevens) Financial Risk Management FE535 4 / 31 Game 2 Let’s consider a little more complicated game Assume we have a fair dice, such that result i has a probability of 1/6 ∀i = 1, .., 6 The rules of the game go as follows I I I I I Roll the dice 3 times, giving three results X1 , X2 , and X3 Let Xmax = max(X1 , X2 , X3 ) You earn a $1 if Xmax = 6 Otherwise, you get nothing You need to pay p dollars to play this game The question is how much would you pay for this game? 5/31 Spring 2019 (Stevens) Financial Risk Management FE535 5 / 31 Fair Price With a loss of generality, let’s ignore the time value of money, such that the fair price of a single period game should take into account all payoffs and their respective likelihood, i.e. p= S X CFs × P(s) (1) s=1 where CFs denotes the cash-flows from state s and P(s) is the probability of s 6/31 Spring 2019 (Stevens) Financial Risk Management FE535 6 / 31 Fair Price With a loss of generality, let’s ignore the time value of money, such that the fair price of a single period game should take into account all payoffs and their respective likelihood, i.e. p= S X CFs × P(s) (1) s=1 where CFs denotes the cash-flows from state s and P(s) is the probability of s In game 2, there are multiple states in which one gets paid, i.e. the different permutations of {X1 , X2 , X3 } in which Xmax = 6 takes place In other words, p =P(Xmax = 6) (2) To know p, we need to know the probability P(Xmax = 6) To know P(Xmax = 6), we need to know the distribution of Xmax Spring 2019 (Stevens) Financial Risk Management FE535 6/31 6 / 31 We can find the distribution of Xmax (hence the price p) in two different ways I I Analytically Numerically (MC) 7/31 Spring 2019 (Stevens) Financial Risk Management FE535 7 / 31 We can find the distribution of Xmax (hence the price p) in two different ways I I Analytically Numerically (MC) Analytically We know that Xi ∼ U(1, 6), for i = 1, 2, 3, such that P(Xi ≤ i) = i/6 It also can be shown that P(Xmax ≤ i) = (i/6)3 (3) P(Xmax = 6) = 1 − P(Xmax ≤ 5) = 1 − (5/6)3 = 0.4213 (4) Hence, 7/31 Spring 2019 (Stevens) Financial Risk Management FE535 7 / 31 Numerically Start with n = 1 Generate three random variables from Xj,n ∼ U(1, 6) for j = 1, 2, 3 Compute Xmax,n = max(X1,n , X2,n , X3,n ), if Xmax,n = 6 return 1 and zero otherwise Repeat the above N = 105 times Finally, compute how many times out of N, we have Xmax,n = 6 8/31 Spring 2019 (Stevens) Financial Risk Management FE535 8 / 31 Numerically Start with n = 1 Generate three random variables from Xj,n ∼ U(1, 6) for j = 1, 2, 3 Compute Xmax,n = max(X1,n , X2,n , X3,n ), if Xmax,n = 6 return 1 and zero otherwise Repeat the above N = 105 times p 0.425 0.420 0.415 0.410 0.405 0.400 In other words... > N <- 10^5 > I_seq <- numeric() > for (n in 1:N) { + X_1 <- sample(1:6,1) + X_2 <- sample(1:6,1) + X_3 <- sample(1:6,1) + X_max_n <- max(c(X_1,X_2,X_3)) + I_n <- (X_max_n == 6)*1 + I_seq <- c(I_seq,I_n) + } > round(mean(I_seq),4) 0.430 Finally, compute how many times out of N, we have Xmax,n = 6 0e+00 2e+04 4e+04 6e+04 8e+04 1e+05 N [1] 0.4227 Spring 2019 (Stevens) 8/31 Financial Risk Management FE535 8 / 31 Law of Large Numbers To put formally, if we have a sequence of independent identically distributed (iid) Xn for i = 1, .., N with mean µ and variance σ 2 < ∞, then PN n=1 Xn →µ (5) lim N→∞ N 9/31 Spring 2019 (Stevens) Financial Risk Management FE535 9 / 31 Law of Large Numbers To put formally, if we have a sequence of independent identically distributed (iid) Xn for i = 1, .., N with mean µ and variance σ 2 < ∞, then PN n=1 Xn →µ (5) lim N→∞ N In game 2, the price of the game is equivalent to the expected payoff At each time the payoff of the game is either 1 or zero depending on the result of Xmax,n If we denote ( 1 P(Xmax = 6) In = (6) 0 otherwise As a result, it follows that PN lim N→∞ n=1 In N → E[In ] = P(Xmax,n = 6) = p (7) 9/31 Spring 2019 (Stevens) Financial Risk Management FE535 9 / 31 Game 3: MC Application to Asset Prices Game 2 resembles what is known as a binary option A binary option returns $1 in case some event takes place and zero otherwise Let’s consider the following game now Game 3 The current stock price is $100 If the stock price goes beyond $110 next month, you get paid $1 If not, you are paid zero and you lose the down payment of p What is the fair price of p? To answer the above, we need to know the behavior (distribution) of the stock price I I What is the potential growth of the stock? What is the volatility of the stock? 10/31 Spring 2019 (Stevens) Financial Risk Management FE535 10 / 31 Let Pt denote the price in month t Game 3 pays a $1, if I I the price goes up P1 > 110, where P0 = 100 or the return on stock goes up by 9.53%, i.e. R1 > log (110/100) = 9.53% (8) Hence, if we know the distribution of the monthly return Rt , then similar to (7), the option price would be P(R1 > 9.53%) (9) 11/31 Spring 2019 (Stevens) Financial Risk Management FE535 11 / 31 Let Pt denote the price in month t Game 3 pays a $1, if I I the price goes up P1 > 110, where P0 = 100 or the return on stock goes up by 9.53%, i.e. R1 > log (110/100) = 9.53% (8) Hence, if we know the distribution of the monthly return Rt , then similar to (7), the option price would be P(R1 > 9.53%) (9) For instance, if R1 ∼ N(0.02, 0.042 ), then P(R1 > 0.0953) = 1 − Φ  0.0953 − 0.02 0.04  = 1 − 0.9701 ≈ 3% (10) If we assume a discount rate of zero over one month period, then the price of the binary option today is $0.03 I I An increase of 9.53% over one month seems very unlikely For this reason the option is cheap 11/31 Spring 2019 (Stevens) Financial Risk Management FE535 11 / 31 If we were to find the price of the binary option numerically we should do the following steps 1 2 3 4 5 Start with n = 1 Generate one random variable from R1,n ∼ N(0.02, 0.042 ) Check whether R1,n > 0.0953 and assign In = 1 if true and zero otherwise Repeat the above N = 105 times Finally, the average In = 1 over the N iterations should converge to P(R1,n > 0.1) = 0.0228 12/31 Spring 2019 (Stevens) Financial Risk Management FE535 12 / 31 If we were to find the price of the binary option numerically we should do the following steps 5 > > > + + + + > N <- 10^5 I_seq <- numeric() for (n in 1:N) { R_1 <- rnorm(1,0.02,0.04) I_n <- (R_1 > log(110/100))*1 I_seq <- c(I_seq,I_n) } mean(I_seq) 0.040 4 0.035 3 Start with n = 1 Generate one random variable from R1,n ∼ N(0.02, 0.042 ) Check whether R1,n > 0.0953 and assign In = 1 if true and zero otherwise Repeat the above N = 105 times Finally, the average In = 1 over the N iterations should converge to P(R1,n > 0.1) = 0.0228 p 2 0.030 1 0.025 [1] 0.02998 > # alternative fast solution > mean(rnorm(N,0.02,0.04) > log(110/100)) 0e+00 2e+04 4e+04 6e+04 8e+04 1e+05 N [1] 0.03045 Spring 2019 (Stevens) 12/31 Financial Risk Management FE535 12 / 31 Complicated Games So far, the problems we have talked about all have analytical solutions However, in practice, things can be too complex to price analytically I analytical solution may not exist at all Relying on numerical solution is inevitable Examples of More Complicated Options European Option - has a closed form solution American Option - no analytical solution Asian Option - no analytical solution what makes the American and Asian complicated is the fact that both are path-dependent 13/31 Spring 2019 (Stevens) Financial Risk Management FE535 13 / 31 Simulating Price Path Under market efficiency, financial prices should exhibit a random walk Prices are assumed to follow what it is known as a Markov Process I the next period price depends on today’s alone The future prices are stochastic and obey certain motion It is common to represent the price over time using a number of components: I I I I Growth or expected return - µ Volatility - σ Time - t Stochastic Component - Zt 14/31 Spring 2019 (Stevens) Financial Risk Management FE535 14 / 31 Standard Brownian Motion The stochastic component Zt is a specific Markov Process known as a standard Brownian Motion (BM) or Weiner process Zt has a number of main properties 1 2 Z0 = 0 It has a normal (Gaussian) distribution Zt ∼ N(0, t) 3 (11) Its increments are independent and also follow a normal distribution Zt − Zs ∼ N(0, t − s) (12) which is independent of any past values Zu for u < s It is common to represent the change in the process over small time increment as ∆Zt , i.e. that is the change in the value of Zt over ∆t period Given the above properties it follows that Zt+∆t − Zt = ∆Zt ∼ N(0, ∆t) (13) 15/31 Spring 2019 (Stevens) Financial Risk Management FE535 15 / 31 Simulating Standard Brownian Motion Assume we have t units of time We like to simulate the BM between t = 0 and t = 1 The simulated process is observed over ∆t = 0.01 time increments At time t = 0, the process is Z0 = 0 The process moves with ∆Zt increments over t = 0.01, 0.02, ..., 1 Z0.01 = Z0 + ∆Z0.01 , where ∆Z0.01 ∼ N(0, 0.01) (14) For 0.02, the same applies Z0.02 = Z0.01 + ∆Z0.02 , where ∆Z0.02 ∼ N(0, 0.01) (15) Generalizing, in d steps from t = 0, the process value is given by Z = Z0 + d 100 d X ∆Z (16) i 100 i=1 Since Z0 = 0 and ∆Z i 100 ∼ N(0, 0.01) is iid, it follows that Z d 100 ∼ N(0, d ) 100 (17) which is in line with Equation (11) 16/31 Spring 2019 (Stevens) Financial Risk Management FE535 16 / 31 0.5 0.2 0.0 0.0 Z −0.2 Z −0.5 −0.4 −1.0 −0.6 0 20 40 60 80 100 0 20 40 60 80 100 60 80 100 Index Z −0.6 Z −0.4 −0.5 −0.2 0.0 0.0 0.2 Index −1.0 −0.8 −1.0 −1.2 BM_path <- function(n) { d <- 100 t <- 1 dt <- t/d Z <- 0 for (i in 1:d) { dZ <- rnorm(1,0,sqrt(dt)) Z_i <- Z[i] + dZ Z <- c(Z,Z_i) } return(Z) } Z <- BM_path() plot(Z, type = "l") abline(h = 0,lty = 2) −1.5 > + + + + + + + + + + + > > > 0 20 40 60 80 100 Index 0 20 40 Index > Z_mat <- sapply(1:10^4,BM_path) > dim(Z_mat) > Z_1 <- Z_mat[nrow(Z_mat),] > mean(Z_1);sd(Z_1) [1] [1] -0.01323055 [1] 1.006404 101 10000 17/31 Spring 2019 (Stevens) Financial Risk Management FE535 17 / 31 General Brownian Motion Similar to standard BM, the general BM has the following properties ∆Xt = µ∆t + σ∆Zt (18) ∆Xt ∼ N(µ∆t, σ 2 ∆t) (19) where Illustration Let t refer to annual frequency with 252 trading days a year Let ∆t denote the change in time over one day, i.e. ∆t = 1/252 Assume the price at time 0 is X0 = 100, with µ = 0.10 and σ = 0.2 We can simulate the price in the next day as X where ∆X 1 252 = 0.1 × 1 252 = X0 + ∆X (20) 1 252 1 + 0.2 × ∆Z 1 ∼ N 252 252  0.1 0.2 , 252 252  (21) with ∆Z1 ∼ N(0, 1/252) 18/31 Spring 2019 (Stevens) Financial Risk Management FE535 18 / 31 If we repeat the previous procedure multiple times, we can simulate the process over number of periods using today’s price To illustrate this, the price in d periods ahead is given by X d 252 = X0 + d X ∆X (22) i 252 i=1 Since the increments ∆Xi are iid, then we can simulate d random numbers from 0.1 0.22 N( 252 , 252 ) X 99.80 99.85 99.90 99.95 100.00 To implement, > X <- 100 > mu <- 0.1; sig <- 0.2 > dt <- 1/252 > X <- 100 > for(i in 1:126) { + dX <- rnorm(1,mu*dt,sig*sqrt(dt)) + X_1 <- X[i] + dX + X <- c(X,X_1) + } > plot(X,type = "l") > abline(h = 100, lty = 2) 0 20 40 60 80 100 120 Index 19/31 Spring 2019 (Stevens) Financial Risk Management FE535 19 / 31 The problem with the previous BM is that prices could take negative values For instance, if the current price were $1 while σ = 0.5, by simulating the price 104 times, we get the following > > > > > + + + > prices <- numeric() N <- 10^4 X <- 1 sig <- 0.5 for(n in 1:N) { X_end <- 1 + sum(rnorm(126,mu*dt,sig*sqrt(dt))) prices <- c(prices,X_end) } summary(prices) Min. 1st Qu. -0.3484 0.8218 Median 1.0572 Mean 3rd Qu. 1.0566 1.2945 Max. 2.5832 Additional problems with the this process is that fails to mimic other aspects of prices, e.g. 1 2 Prices change relative to the previous levels Prices exhibit positive skewness - which is not the case for normal 20/31 Spring 2019 (Stevens) Financial Risk Management FE535 20 / 31 Geometric Brownian Motion The most common process to simulate stock prices is the Geometric Brownian Motion (GBM) In this case, ∆St = St µ∆t + St σ∆Zt (23) ∆St = µ∆t + σ∆Zt St (24) alternatively, Note that ∆St St resembles the stock return between t and t + ∆t. To see this, St+∆t − St ∆St = ≈ log St St  St+∆t St  = ∆log (St ) (25) In fact, the solution to (23) or (24), requires the solution to the stochastic differential equation (SDE) ∆log (St ) 21/31 Spring 2019 (Stevens) Financial Risk Management FE535 21 / 31 This class doesn’t require knowledge about SDEs, but it follows that the solution for the GBM is     St+∆t σ2 log = µ− ∆t + σ∆Zt (26) St 2 which is the same as Equation (4.6) from the Textbook To simplify the notation, let ∆Rt denote the return of the stock over ∆t ∆Rt = log  St+∆t St  (27) In fact, it follows that the ∆Rt is a general BM, such that  ∆Rt ∼ N µ− σ2 2  ∆t, σ 2 ∆t  (28) is an iid process 22/31 Spring 2019 (Stevens) Financial Risk Management FE535 22 / 31 This class doesn’t require knowledge about SDEs, but it follows that the solution for the GBM is     St+∆t σ2 log = µ− ∆t + σ∆Zt (26) St 2 which is the same as Equation (4.6) from the Textbook To simplify the notation, let ∆Rt denote the return of the stock over ∆t ∆Rt = log  St+∆t St  (27) In fact, it follows that the ∆Rt is a general BM, such that  ∆Rt ∼ N µ− σ2 2  ∆t, σ 2 ∆t  (28) is an iid process Therefore, to simulate the price at time t + ∆t, one needs 1 2 3 the price at time t, St estimate µ and σ Finally, simulate ∆Rt , i.e. draw a random number from the normal distribution described in (28) In other words St+∆t = St × exp(∆Rt ) (29) 22/31 Spring 2019 (Stevens) Financial Risk Management FE535 22 / 31 Let’s consider again the same example as before Implementation of GBM Simulation Let ∆t = 1/252 The current price at time t = 0 is S0 = 100 Given µ and σ, draw a random number from (28) denoted by ∆R 1 252 The price next day is S 1 252 = S0 × exp(∆R 1 252 ) (30) To simulate the second day price, draw another random number from (28) denoted by ∆R 2 , such that 252 S 2 = S 1 × exp(∆R 2 ) (31) 252 252 252 To generalize it follows that the next d days price is given by S d 252 = S0 d Y exp(∆R i 252 ) = S0 × exp d X ! ∆R (32) i 252 i=1 i=1 23/31 Spring 2019 (Stevens) Financial Risk Management FE535 23 / 31 106 100 S 100 102 104 95 S 90 98 85 96 80 0 20 40 60 80 100 120 0 20 40 60 80 100 120 80 100 120 Index S 100 105 90 95 S 95 110 100 115 Index 85 S <- 100 dt <- 1/252 mu <- 0.1 sig <- 0.2 for(i in 1:126) { dR <- rnorm(1, dt*(mu - 0.5*sig^2), sig*sqrt(dt) ) S_dt <- S[i]*exp(dR) S <- c(S,S_dt) } plot(S,type = "l") abline(h = S[1], lty = 2) 80 > > > > > + + + + + + > > 108 Let’s demonstrate how to implement GBM 0 20 40 60 Index 80 100 120 0 20 40 60 Index 24/31 Spring 2019 (Stevens) Financial Risk Management FE535 24 / 31 Application to Portfolio Let’s denote today’s portfolio value by F0 We are interested in evaluating the future portfolio performance at t, i.e. Ft which is unknown MC simulations allow us to generate different scenarios as a robustness check, for instance I I With 95% confidence, what’s the maximum loss on the portfolio? What happens to the underlying portfolio if the market premium (volatility) drops (increases) by x%? 25/31 Spring 2019 (Stevens) Financial Risk Management FE535 25 / 31 Application to Portfolio Let’s denote today’s portfolio value by F0 We are interested in evaluating the future portfolio performance at t, i.e. Ft which is unknown MC simulations allow us to generate different scenarios as a robustness check, for instance I I With 95% confidence, what’s the maximum loss on the portfolio? What happens to the underlying portfolio if the market premium (volatility) drops (increases) by x%? Assuming the portfolio obeys to GBM as in (29), then its future price can be described as   σp2 Ft = F0 × exp (µp − )t + σp Zt (33) 2 where µp and σp denote the portfolio mean return and volatility, respectively Hence, if one knows µp and σp , then the future price is determined by the stochastic component Zt 25/31 Spring 2019 (Stevens) Financial Risk Management FE535 25 / 31 Valu ...
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Tutor Answer

mikewinter3
School: Cornell University

Attached.

1

Geometric Brownian motion Process for SPY ETF Stock Price

Student name:
Institutional affiliation:

2
Geometric Brownian motion Process for SPY ETF Stock Price
The attached word document and excel file addresses the question “Homework about Financial
Risk Management” by answering the following:
Task 1
Go to Yahoo Finance and download historical data for SPY ETF over the last two years, starting
from 2017 to today. Compute the daily return using the adjusted close price column. This should
result in a new time series (column).
Task 2
Using the daily returns of the SPY ETF, calibrate a Geometric Brownian Motion process. Report
the corresponding mu and sigma.
Task 3
Given the calibrated model, simulate the SPY price path over the next three months, i.e. 256/4=64
trading days. Provide a couple of plots to demonstrate this.


1

Running Head: Geometric Brownian motion

Geometric Brownian motion Process for SPY ETF Stock Price

Student name:
Institutional affiliation:

2

Geometric Brownian motion

Geometric Brownian motion Process for SPY ETF Stock Price
Task 1
See the attached excel file for the data I downloaded from Yahoo Finance.
Task 2
Using the daily returns of the SPY ETF, calibrate a Geometric Brownian Motion process. Report
the corresponding mu and sigma.
See the attached excel file the daily returns of the SPY ETF and calibrated Geometric Brownian
motion process:
Mean (mu)
Std deviation( sigma)

-0.03912%
0.008507644

Task 3
See the excel file for the simulated SPY price path over the next three months, that is between 6 th
February 2017 to 8th May 2017 ( a total of 64 trading days). The following graphs demonstrate the
simulation:

3

Geometric Brownian motion

4

Geometric Brownian motion
References
Jenkins, M. S. (1992 ). The geometry of stock market profits : a guide to professional trading for
a living. Greenville, SC : Traders Press.
Lawrence, K. D., Klimberg, R. K., & Lawrence, S. M. (2009). Fundamentals of forecasting
using Excel. New York, N.Y.: Industrial Press.


Date
2/6/2017
2/7/2017
2/8/2017
2/9/2017
2/10/2017
2/13/2017
2/14/2017
2/15/2017
2/16/2017
2/17/2017
2/21/2017
2/22/2017
2/23/2017
2/24/2017
2/27/2017
2/28/2017
3/1/2017
3/2/2017
3/3/2017
3/6/2017
3/7/2017
3/8/2017
3/9/2017
3/10/2017
3/13/2017
3/14/2017
3/15/2017
3/16/2017
3/17/2017
3/20/2017
3/21/2017
3/22/2017
3/23/2017
3/24/2017
3/27/2017
3/28/2017
3/29/2017
3/30/2017
3/31/2017
4/3/2017
4/4/2017
4/5/2017
4/6/2017
4/7/2017
4/10/2017
4/11/2017

Open
High
Low
Close
Adj Close Volume Daily Returns
228,87
229,33
228,54
228,93 220,3782 57790100
-0,00436%
229,38
229,66
228,72
228,94 220,3878 57931200
-0,13088%
228,94
229,39
228,31
229,24 220,6766 51566200
-0,58974%
229,24
230,95
229,24
230,6 221,9857 65955200
-0,39309%
231
231,77
230,62
231,51 222,8618 66015900
-0,54130%
232,08
233,07
232,05
232,77 224,0747 55182100
-0,39794%
232,56
233,71
232,16
233,7
224,97 67794500
-0,51933%
233,45
235,14
233,39
234,92 226,1444 86785800
0,08520%
234,95
235,16
233,85
234,72 225,9519 84722400
-0,15738%
233,95
235,09
233,93
235,09 226,308 77204100
-0,59199%
235,52
236,69
235,51
236,49 227,6557 88946100
0,08887%
236,02
236,54
235,83
236,28 227,4536 62115200
-0,06767%
236,88
236,9
235,56
236,44 227,6076 74615900
-0,12673%
235,46
236,79
235,41
236,74 227,8964 82381600
-0,15604%
236,64
237,31
236,35
237,11 228,2526 56515400
0,27065%
236,67
236,95
236,02
236,47 227,6365 96961900
-1,38043%
238,39
240,32
238,37
239,78 230,8228 1,49E+08
0,63372%
239,56
239,57
238,21
238,27 229,3693 70246000
-0,06291%
238,17
238,61
237,73
238,42 229,5136 81974300
0,29867%
237,5
238,12
237,01
237,71 228,8302 55391500
0,29960%
237,36
237,77
236,76
237 228,1467 65103700
0,18599%
237,34
237,64
236,4
236,56 227,7231 78168800
-0,12666%
236,7
237,24
235,74
236,86 228,0119 90683900
-0,34920%
237,97
238,02
236,59
237,69 228,8109 81991700
-0,05046%
237,62
237,86
237,24
237,81 228,9265 57256800
0,38414%
237,18
237,24
236,19
236,9 228,0504 59880800
-0,85793%
237,56
239,44
237,29
238,95 230,0239 96081800
0,19709%
239,11
239,2
238,1
238,48 229,5714 78344000
0,17592%
237,75
237,97
237,03
237,03 229,1682 89002100
0,10981%
237,03
237,36
236,32
236,77 228,9169 52537000
1,30065%
237,47
237,61
233,58
233,73 225,9777 1,32E+08
-0,23476%
233,77
234,61
233,05
234,28 226,5094 97569200
0,10682%
234,28
235,34
233,6
234,03 226,2677
1E+08
0,07270%
234,38
235,04
232,96
233,86 226,1034 1,13E+08
0,10273%
231,93
233,92
231,61
233,62 225,8713 87454500
-0,72242%
233,27
235,81
233,14
235,32 227,515 93483900
-0,09341%
234,99
235,81
234,73
235,54 227,7277 61950400
-0,31740%
235,47
236,52
235,27
236,29 228,4528 56737900
0,23331%
235,9
236,51
235,68
235,74 227,921 73733100
0,17421%
235,8
236,03
233,91
235,33 227,5246 85546500
-0,06369%
235
235,58
234,56
235,48 227,6696 56466200
0,29815%
236,26
237,39
234,54
234,78 226,9929 1,09E+08
-0,28033%
234,94
236,04
234,43
235,44 227,631 69135800
0,10205%
235,15
236
234,64
235,2 227,3989 74412300
-0,05950%
235,36
236,26
234,73
235,34 227,5343 67615300
0,11913%
234,9
235,18
233,34
235,06 227,2636 88045300
0,44011%

4/12/2017
4/13/2017
4/17/2017
4/18/2017
4/19/2017
4/20/2017
4/21/2017
4/24/2017
4/25/2017
4/26/2017
4/27/2017
4/28/2017
5/1/2017
5/2/2017
5/3/2017
5/4/2017
5/5/2017
5/8/2017
5/9/2017
5/10/2017
5/11/2017
5/12/2017
5/15/2017
5/16/2017
5/17/2017
5/18/2017
5/19/2017
5/22/2017
5/23/2017
5/24/2017
5/25/2017
5/26/2017
5/30/2017
5/31/2017
6/1/2017
6/2/2017
6/5/2017
6/6/2017
6/7/2017
6/8/2017
6/9/2017
6/12/2017
6/13/2017
6/14/2017
6/15/2017
6/16/2017
6/19/2017

234,74
233,64
233,11
233,72
234,52
234,15
235,25
237,18
237,91
238,51
238,77
238,9
238,68
238,84
238,77
238,83
239,19
239,75
239,96
239,39
239,35
239,09
239,47
240,64
240,08
235,73
237,33
238,9
239,95
240,32
241,2
241,54
241,34
241,84
241,97
243,42
243,97
243,34
243,6
243,77
244,09
243,13
243,98
244,86
242,68
242,77
243,59

234,96
234,49
234,57
234,49
234,95
235,85
235,31
237,41
238,95
239,53
238,95
238,93
239,17
238,98
238,88
238,92
239,72
239,92
240,19
239,87
239,57
239,43
240,44
240,67
240,08
237,75
239,08
239,71
240,24
240,73
242,08
241,9
241,79
241,88
243,38
244,35
244,3
243,98
243,92
244,33
245,01
243,42
244,61
244,87
243,91
242,83
244,73

233,77
232,51
232,88
233,08
233,18
233,78
234,13
234,56
237,81
238,35
237,98
237,93
238,2
238,3
237,7
237,78
238,68
239,17
239,04
239,15
238,13
238,67
239,45
239,63
235,75
235,43
237,27
238,82
239,51
239,93
240,96
241,45
241,16
240,64
241,64
243,08
243,76
243,12
242,83
243,17
241,95
242,38
243,58
243,29
242,36
241,63
243,48

234,03
232,51
234,57
233,87
233,44
235,34
234,59
237,17
238,55
238,4
238,6
238,08
238,68
238,77
238,48
238,76
239,7
239,66
239,44
239,87
239,38
238,98
240,3
240,08
235,82
236,77
238,31
239,52
240,05
240,61
241,76
241,71
241,5
241,44
243,36
244,17
243,99
243,21
243,66
243,78
243,41
243,36
244,55
244,24
243,77
242,64
244,66

226,2677
224,7982
226,7899
226,113
225,6973
227,5343
226,8092
229,3036
230,6378
230,4928
230,6862
230,1834
230,7635
230,8505
230,5701
230,8409
231,7497
231,711
231,4983
231,914
231,4403
231,0536
232,3298
232,1171
227,9984
228,9169
230,4058
231,5757
232,0881
232,6295
233,7413
233,693
233,49
233,432
235,2883
236,0714
235,8974
235,1433
235,5784
235,6944
235,3366
235,2883
236,4388
236,1391
235,6847
235,7362
237,6987

81864400
92880400
68405400
83225800
68699900
92572200
...

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Anonymous
Tutor went the extra mile to help me with this essay. Citations were a bit shaky but I appreciated how well he handled APA styles and how ok he was to change them even though I didnt specify. Got a B+ which is believable and acceptable.

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