Lab 2 - Decision theory

Transcription

1 Lab 2 - Decision theory Edvin Listo Zec edvinli@student.chalmers.se September 29, 2014 Co-worker: Jessica Fredby

2 Introduction The goal of this computer assignment is to analyse a given set of data of seven different stocks: AstraZenica, Electrolux, Ericsson, Gambio, Nokia, Swedish Match and Svenska Handelsbanken. The records are in form of time series from to Our main objective is to do a portfolio optimisation of these stocks; in other words to divide our capital between the stocks so that the difference between two points next to each other in time of the portfolio is maximised, while also taking the risk into account using a utility function. 1 Assignment 1 - Data exploration 1.1 Assignment Assumptions about the data The task in this assignment is to see if the given data of seven different stocks are normal distributed. All calculations are done in Matlab. We started by loading the data from stockdata.tsv and calculating the log-returns for each stock, see figure 1.1, according to the definition: X(t) = log(s(t)) log(s(t 1)), where S(t) is the price of a stock at time t and X(t) is the log-return. The log-returns are assumed to be i.i.d. normal in time (X 1 (t) independent of X 1 (t 1)). Out of interest we also plotted the actual stocks against time as seen in figure 1.2. The seven given stocks are: AstraZenica, Electrolux, Ericsson, Gambio, Nokia, Swedish Match and Svenska Handelsbanken. The records are in form of time series, S(t), from to We also plotted Q-Q-plots and histograms of the logreturns: see figures 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9. Two goodness-of-fit tests were made, a χ 2 test and a Kolmogorov Smirnov test. The χ 2 test returned a 1 for all stocks except for the fourth one, meaning we should reject the hypothesis that the other stocks are N (0, 1). The Kolmogorov Smirnov test however rejected all seven stocks. As seen in the Q-Q-plots, no log-return follow a normal distribution completely; they all seem to be skewed in some way. Nothing so far seems to be indicating that the log-returns are N (0, 1). 1

3 AstraZenica Electrolux Ericsson Gambio Nokia Swedish Match Svenska Handelsb AstraZenica Electrolux Ericsson Gambio Nokia Swedish Match Svenska Handelsb x x 10 4 Figure 1.1: The log-returns of the stocks plotted against time. Figure 1.2: The stocks plotted against time. Figure 1.3: QQ-plot and histogram of Stock 1. 2

4 Figure 1.4: QQ-plot and histogram of Stock 2. Figure 1.5: QQ-plot and histogram of Stock 3. 3

5 Figure 1.6: QQ-plot and histogram of Stock 4. Figure 1.7: QQ-plot and histogram of Stock 5. 4

6 Figure 1.8: QQ-plot and histogram of Stock 6. Figure 1.9: QQ-plot and histogram of Stock 7. To go further, we thought that maybe the mean and standard deviation could be different than 0 and 1. Therefore we estimated these parameters by maximum likelihood estimation. Using Matlab s mle function, we got the following results: 5

7 Table 1.1: Maximum likelihood estimations for each stock. Stock 1 Stock 2 Stock 3 Stock 4 Stock 5 Stock 6 Stock 7 µ σ Doing the χ 2 and Kolmogorov-Smirnov tests again yields the same results: the χ 2 rejected all stocks except stock 4, and the Kolmogorov-Smirnov test rejected all seven stocks. All tests of normal distribution are thus inconclusive. The next part of this assignment was to estimate the auto-correlation function (ACF) of the logreturns and of the absolute values of the log-returns. The ACF is defined as: r X (t, t + h) = r X (h) = Cov{X(t), X(t + h)}, Var{X(t)} In figures 1.10, 1.11, 1.12, 1.13, 1.14, 1.15 and 1.16 we ve plotted the ACF of each log-return and the respective absolute value, using the autocorr function in Matlab to estimate the ACF. Autocorrelation is used to check randomness of processes. It is the tendency for observations made next to each other in time to be related. If a process is random the auto-correlation should be near zero. In the plots below the lag (time span between observations) is shown on the x-axis and the auto-correlation is shown on the y-axis. The blue lines represent a 95% confidence interval. If the sample is inside the confidence interval we don t have auto-correlation and if the sample is outside of the confidence interval we have auto-correlation. We can see that in our plots each log-return becomes more independent with time. This is also shown in the right hand side plots where we plotted the absolute values of each log-return. We chose to plot 200 lags so we could see how the dependence varied over time. From theory, we know that we can assume linear independence if ˆr X (h) 1.96 (1) n where ˆr X is the ACF. We therefore made calculations in Matlab that checked for each log-return and lag if equation 1 was true, which it was for all log-returns and all lags. We therefore conclude that the log-returns are independent. 6

8 Figure 1.10: ACF plot of log-return (left) and absolute values of log-return (right) of Stock 1. Figure 1.11: ACF plot of log-return (left) and absolute values of log-return (right) of Stock 2. 7

9 Figure 1.12: ACF plot of log-return (left) and absolute values of log-return (right) of Stock 3. Figure 1.13: ACF plot of log-return (left) and absolute values of log-return (right) of Stock 4. 8

10 Figure 1.14: ACF plot of log-return (left) and absolute values of log-return (right) of Stock 5. Figure 1.15: ACF plot of log-return (left) and absolute values of log-return (right) of Stock 6. 9

11 Figure 1.16: ACF plot of log-return (left) and absolute values of log-return (right) of Stock 7. Finally, we were supposed to calculate to covariance matrix for the log-returns. The diagonal is the variance for each log-return. 1 Cov(logreturns) = We see here that the covariance is almost zero in all cases. Therefore, as expected, we can draw the conclusion that the log-returns are independent. 1.2 Assignment Utility and expected utility In this assignment we were to explore the utility function 1 e kx for different k. This function describes a so-called risk averse utility, which always is concave. For example, it favours gaining a small amount of money with certainty. It also puts a large negative weight on loosing your capital. A risk-averse person might prefer to choose to put his or her money into a bank account that yields a low but guaranteed interest rate, rather than to invest in a stock that might have high expected returns but also has a chance of losing value. We plotted the function for different values of k in figure We see that the graph moves closer to the y-axis as k grows larger. We therefore interpret k as a measure of risk aversiveness. The larger the k, the less likely one is to take risks, due to the increase of concavity. 10

12 Figure 1.17: Utility function 1 e kx plotted for different values of k. We now consider each stock at a time and calculate the expected utility for a few different k. We used the estimated parameters from table 1.1. To calculate the expected utility we solved the integral in equation 2 numerically in Matlab using the quad function. We however chose to integrate from 0.2 to 0.2 since integrating from to didn t give us result. We chose the interval by looking at the graph of the function to see what values it lied between. U π (w) = (1 e kx 1 ) e (x z)2 2σ 2 dx (2) 2πσ In table 1.2 one can see the expected utility we calculated for each stock at a time. Table 1.2: Expected utility when investing everything in a single stock, with their respective estimated parameters. k Stock 1 Stock 2 Stock 3 Stock 4 Stock 5 Stock 6 Stock From this table one can see that stock 4 gives the maximum expected utility for k = 0.5 and k = 1. For k = 2, however, stock 6 was the most advantageous. 11

13 2 Assignment 2 In this assignment we were to show that maximising equation 2 is the same as maximising the following equation: under the constraints µ T w k 2 wt Σw. w 1,..., w n 0 and n w i = 1 Using that z = µ T w and σ 2 = w T Σw we get the following expression: i=1 z k 2 σ2 (3) 2.1 Proof We begin by multiplying the utility function with the distribution function and receive the following: U π (w) = 1 e (x z)2 2σ 2 dx 2πσ 1 2πσ e kx e (x z)2 2σ 2 dx (4) The first integral in this expression is equal to 1 since it is the integral over a normal distributed function. In the other integral we rewrite the exponent as: (x z)2 kx 2σ 2 = (x2 + 2kxσ 2 2xz + z 2 ) 2σ 2 (5) Completing the square now yields: (x (z kσ 2 )) 2 2kzσ 2 + k 2 σ 2 2σ 2 = (x (z kσ2 )) 2 2σ 2 k (z k2 ) σ2 (6) We then land at the following expression: U π (w) = 1 e k(z k 2 σ2 ) 1 2πσ e (x (z kσ2 )) 2 2σ 2 dx (7) This integral is also an integral over a normal distributed function, the only difference being the µ = z kσ 2, so this integral evaluates to 1 as well. This results in: 1 e k(z k 2 σ2 ) (8) It is now clear that maximising equation 3 minimises the exponential term, which finally maximises equation 8. 12

14 3 Assignment 3 - Optimisation with two stocks In this assignment the task was to optimise the portfolio using two stocks: Ericsson (stock 3) and Gambio (stock 4). We started by estimating the mean vector and the covariance matrix. With these estimates, and k = 1, we calculated the and plotted the expected utility for w 1 [0, 1], w 2 = 1 w 1. Here w 1 is the weight of Ericsson and w 2 the weight for Gambio. The results can be seen in figure 3.1. Figure 3.1: Expected utility, k = 1. The maximum can approximately be located at w 1 = To find the optimum more accurately we used the fmincon function in Matlab. This yielded the result w 1 = for k = 1. The expected utility is then We repeated this for several different values of k. The results are seen in table 3.1. We see that the maximum value for w 1 increases with an increasing k. For small k (when we are being more risk-neutral) it seems best to invest mostly in Gambio. However, as k increases we become more risk-aversive and then judging from the results it s better to invest a little more in Ericsson. k max w Table 3.1: Optimal weight for different k. 13

15 4 Assignment 4 - Optimisation with seven stocks In this task we repeated the optimisation in the previous assignment, but using all seven stocks simultaneously. Using fmincon we now got the results seen in table 4.1. We also calculated the expected utility for the naive way of investing equally in all seven stocks. When using a risk aversive utility function there is a large difference in expected utility, implying the naive way of investing is too risky for a risk aversive investor. We see, however, that the smaller k is the smaller the difference is. Since the function becomes more risk neutral when k decreases, this indicates that the naive way of investing is pretty risk-neutral for small k. Furthermore, the expected utility decreases with an increasing k. This coincides with our discussion of k from earlier assignments: an increase in k implies that one is less likely to take risks due to the increase in concavity. Analysing the results further, it seems that stock 4 and stock 6 are the most advantageous ones to invest in, no matter the value of k. Since the 4th and 6th stocks are so advantageous, it could be of interest to investigate things further: what is the expected utility if we only invest in stock 4 or 6? This is also seen in table 4.1. It seems like it s almost always better to invest everything in either stock 4 or 6, than to divide everything equally between the seven stocks. In conclusion, it is worth noting that the analysis done here is not perfect, due to the assumption that the stocks follow a normal distribution even though that the statistical tests in the beginning suggested otherwise. Also, worth mentioning is a possible source of error. In assignment 1 we calculate the expected utility for each stock at a time using the integral, and in assignment 4 we do it with the other function seen in assignment 2. We get different values for the expected utility when doing these two calculations, when in theory they should be equal. This may be due to the bounds we choose for the integration, meaning that Matlab s way of calculating the integral numerically may be too sensitive. Table 4.1: Optimal values of w and values of the expected utility function for different k. k Optimal w E{U(w optimal )} E{U(w naive )} E{U(w 4 )} E{U(w 6 )} 0.1 (0.0000, , , , , , ) (0.0000, , , , , , ) (0.0000, , , , , , ) (0.0000, , , , , , ) (0.0000, , , , , , ) (0.0646, , , , , , )

16 A Matlab code A.1 Assignment %%Lab set (0, DefaultFigureWindowStyle, docked ) %dock all the figures 3 clc ; clf ; 4 load stockdata.tsv %load the given data 5 logs = log(stockdata (:,2:8) ); %log the data 6 logreturns = logs(2:end,:) logs(1:end 1,:); %calculate the logreturn 7 figure (1) 8 clf 9 plot(stockdata(1:length(stockdata) 1,1), logreturns); %logreturns against time 10 figure (2) 11 clf 12 plot(stockdata(1:length(stockdata),1),stockdata (:,2:8) ) %stock against time 13 %% 14 clc ; clf ; 15 chi=zeros(1,7); isnormal=zeros(1,7); 16 kolm=zeros(1,7); mu=zeros(1,7); sigma=zeros(1,7); 17 kolm2=zeros(1,7); chi2=zeros(1,7); 18 for i=1:7 19 figure (i) 20 clf 21 subplot (1,2,1) 22 qqplot(logreturns (:, i)) %QQ plot 23 subplot (1,2,2) 24 hist (logreturns (:, i)) %histogram 25 chi( i)=chi2gof(logreturns (:, i)); %chi2 goodness of fit 26 kolm(i)=kstest(logreturns (:, i)); %kolmogorov goodness of fit phat=mle(logreturns(:,i)); %maximum likelihood estimates since it s not standard normal distr. 29 mu(i)=phat(1); %mean of each stock 30 sigma(i)=phat(2); %variance (or stand. dev.?) of each stock 31 end 32 chi 33 kolm 34 mu 35 sigma 36 %% 37 %second try with new estimated paramters 38 mu_hat=mean(mu); 39 sigma_hat=mean(sigma); 40 for i=1:7 41 figure (i); 42 clf ; 43 qqplot(logreturns (:, i),probdistunivparam( normal, [mu_hat sigma_hat])) %QQ plot w/ new parameters 44 kolm2(i) = kstest(logreturns (:, i),probdistunivparam( normal, [mu_hat sigma_hat])); %kolmogorov w/ new parameters 45 end 46 kolm %% 49 clc ; 50 lag=1; 51 acf=zeros(lag+1,7); 15

17 52 lindep=zeros(lag+1,7); 53 n=size(logreturns,2) ; 54 for i=1:7 55 figure (i) 56 clf ; 57 subplot (1,2,1) 58 acf (:, i) = autocorr(logreturns (:, i), lag); %calculate the ACF for each logreturn 59 autocorr(logreturns (:, i), lag); %plot the ACF for each logtreturn 60 subplot (1,2,2) 61 autocorr(abs(logreturns (:, i)), lag) %plot the ACF of each abs(logreturn) 62 lindep (:, i)=(abs(autocorr(logreturns(:, i), lag))<1.96/sqrt(n)); % 1 = true, 0 = false 63 end 64 %assume linear indepedence if autocorr < 1.96/sqrt(n) 65 lindep %here we see independence for h>1, but dependence for h=1 for all stocks 66 %% 67 %covariance is 0 if x and y are independent 68 C=cov(logreturns) %covariance matrix, variance on diag A.2 Assignment clc ; clf ; 2 x=linspace(0,2,1000); 3 col=[ b, r, g, magenta ]; 4 for k=0.5:0.5:2 5 u1=1 exp( k x); %risk averse 6 plot(x,u1,col(k 2)); hold on; 7 end 8 legend( k=0.5, k=1, k=1.5, k=2, Location, NorthWest ) 9 %we see that the graph moves closer to the y axis as k grows 10 %=> k can be interpreted as a measure of risk aversiveness 11 %larger k => less likely to take risks due to concavity grid on 14 xlabel( x ); ylabel( Utility ); 15 %% 16 clf ; clc ; 17 mu_hat=zeros(1,7); s_hat = zeros(1,7); 18 k=1; 19 x=linspace( 10,10,10000); 20 q=zeros(1,7); 21 for i=1:7 22 phat=mle(logreturns(:,i)); %maximum likelihood estimates since it s not standard normal distr. 23 mu_hat(i)=phat(1); 24 s_hat(i)=phat(2); 25 figure (i) 26 plot(x, exutility (x,k,mu_hat(i),s_hat(i))) % 27 q(i)=quad(@(x)exutility(x,k,mu_hat(i),s_hat(i)), 0.2,0.2); %calculating the expected utility, 28 %each stock at a time 29 end 30 q 16

18 A.3 Assignment 3 & 4 1 clc ; clf ; 2 load stockdata.tsv 3 logs = log(stockdata (:,2:8) ); 4 logreturns = logs(2:end,:) logs(1:end 1,:); 5 mu_hats=zeros(1,2); s_hats=zeros(1,2); 6 w1=linspace(0,1,100); 7 w2 = 1 w1; 8 w=[w1; w2]; 9 ericsson = logreturns (:,3) ; gambio = logreturns(:,4) ; % ericsson and gambio stocks 10 phat1 = mle(ericsson); phat2 = mle(gambio); %mean vector estimator 11 mu_hats(1) = phat1(1); mu_hats(2) = phat2(1); 12 C = cov(ericsson,gambio); %sigma estimator in diagonal of covariance matrix k=1; Y=zeros(1,100); 17 for i=1: Y(i)=exu2(w(:,i),k,mu_hats,C); %calculating expected utility 19 end plot(w(1,:),y) 23 %maximum at approx (0.1515, ) options = optimset( Algorithm, Interior point ); 27 [max_wei,max_exp] = fmincon(@(w) exu2(w,k,mu_hats,c),[0.1;0.9],[],[],[1 1],1,[0;0],[1;1],[], options); 28 %optimising with fmincon, finds the real optimum for w1 29 %and the respective expected utility 30 max_wei 31 max_exp= max_exp/ %% lab 2 exercise 4 33 clc ; 34 mu_hats = zeros(1,7); 35 s_hats = zeros(1,7); 36 for i=1:7 37 phat=mle(logreturns(:,i)); 38 mu_hats(i) = phat(1); 39 end 40 C=cov(logreturns); %checking for optimal weights, now including all seven stocks, for different k 43 k=1; 44 options = optimset( Algorithm, Interior point ); 45 [max_wei,max_exp] = fmincon(@(w) exu2(w,k,mu_hats,c),[0.1;0.9;0;0;0;0;0],[],[],[ ],1, [0;0;0;0;0;0;0],[1;1;1;1;1;1;1],[], options); 47 max_wei 48 max_exp= max_exp/ w2=(1/7) ones(1,7); 51 exp_naive = exu2(w2,k,mu_hats,c); %naive way of w equally divided between the stocks 52 exp_naive = exp_naive/ exp_single=zeros(1,7); %investing everything in a single stock 17

19 55 for i=1:7 56 w3=zeros(1,7); 57 w3(i)=1; 58 exp_single(i) = exu2(w3,k,mu_hats,c); 59 end 60 exp_single=exp_single/