i am trying to run a python code for ARCH model
np.random.seed(13)# a0 and a1 are constant and should be greater than 0
a0 = 2
a1 = .5# Random number generator with size 1000:
y = w = np.random.normal(size=1000)
Y = np.empty_like(y)for t in range(len(y)):
Y[t] = w[t] * np.sqrt((a0 + a1*y[t-1]**2)) #Formula of ARCH(1)# simulated ARCH(1) series, looks like white noise
tsplot(Y, lags=30)
however the output is giving error as
name 'tsplot' is not defined
I am trying to get graphs time series analysis ARCH model.
Expecting graphs
Related
I am not familiar with nonlinear regression and would appreciate some help with running an exponential decay model in R. Please see the graph for how the data looks like. My hunch is that an exponential model might be a good choice. I have one fixed effect and one random effect. y ~ x + (1|random factor). How to get the starting values for the exponential model (please assume that I know nothing about nonlinear regression) in R? How do I subsequently run a nonlinear model with these starting values? Could anyone please help me with the logic as well as the R code?
As I am not familiar with nonlinear regression, I haven't been able to attempt it in R.
raw plot
The correct syntax will depend on your experimental design and model but I hope to give you a general idea on how to get started.
We begin by generating some data that should match the type of data you are working with. You had mentioned a fixed factor and a random one. Here, the fixed factor is represented by the variable treatment and the random factor is represented by the variable grouping_factor.
library(nlraa)
library(nlme)
library(ggplot2)
## Setting this seed should allow you to reach the same result as me
set.seed(3232333)
example_data <- expand.grid(treatment = c("A", "B"),
grouping_factor = c('1', '2', '3'),
replication = c(1, 2, 3),
xvar = 1:15)
The next step is to create some "observations". Here, we use an exponential function y=a∗exp(c∗x) and some random noise to create some data. Also, we add a constant to treatment A just to create some treatment differences.
example_data$y <- ave(example_data$xvar, example_data[, c('treatment', 'replication', 'grouping_factor')],
FUN = function(x) {expf(x = x,
a = 10,
c = -0.3) + rnorm(1, 0, 0.6)})
example_data$y[example_data$treatment == 'A'] <- example_data$y[example_data$treatment == 'A'] + 0.8
All right, now we start fitting the model.
## Create a grouped data frame
exampleG <- groupedData(y ~ xvar|grouping_factor, data = example_data)
## Fit a separate model to each groupped level
fitL <- nlsList(y ~ SSexpf(xvar, a, c), data = exampleG)
## Grab the coefficients of the general model
fxf <- fixed.effects(fit1)
## Add treatment as a fixed effect. Also, use the coeffients from the previous
## regression model as starting values.
fit2 <- update(fit1, fixed = a + c ~ treatment,
start = c(fxf[1], 0,
fxf[2], 0))
Looking at the model output, it will give you information like the following:
Nonlinear mixed-effects model fit by maximum likelihood
Model: y ~ SSexpf(xvar, a, c)
Data: exampleG
AIC BIC logLik
475.8632 504.6506 -229.9316
Random effects:
Formula: list(a ~ 1, c ~ 1)
Level: grouping_factor
Structure: General positive-definite, Log-Cholesky parametrization
StdDev Corr
a.(Intercept) 3.254827e-04 a.(In)
c.(Intercept) 1.248580e-06 0
Residual 5.670317e-01
Fixed effects: a + c ~ treatment
Value Std.Error DF t-value p-value
a.(Intercept) 9.634383 0.2189967 264 43.99329 0.0000
a.treatmentB 0.353342 0.3621573 264 0.97566 0.3301
c.(Intercept) -0.204848 0.0060642 264 -33.77976 0.0000
c.treatmentB -0.092138 0.0120463 264 -7.64867 0.0000
Correlation:
a.(In) a.trtB c.(In)
a.treatmentB -0.605
c.(Intercept) -0.785 0.475
c.treatmentB 0.395 -0.792 -0.503
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-1.93208903 -0.34340037 0.04767133 0.78924247 1.95516431
Number of Observations: 270
Number of Groups: 3
Then, if you wanted to visualize the model fit, you could do the following.
## Here we store the model predictions for visualization purposes
predictionsDf <- cbind(example_data,
predict_nlme(fit2, interval = 'conf'))
## Here we make a graph to check it out
ggplot()+
geom_ribbon(data = predictionsDf,
aes( x = xvar , ymin = Q2.5, ymax = Q97.5, fill = treatment),
color = NA, alpha = 0.3)+
geom_point(data = example_data, aes( x = xvar, y = y, col = treatment))+
geom_line(data = predictionsDf, aes(x = xvar, y = Estimate, col = treatment), size = 1.1)
This shows the model fit.
I'm trying to plot the Yukawa Potential in Matlab and I want to have my program go through user inputs for the values alpha (called alph in my program) and l. The values I need to use are 0.1, 0.2 and 0.3 for alpha with values of 0, 1 and 2 of l for each value of alpha. I know I could set up a loop for this but it doesn't have to be pretty and I want to test the values one at a time. Anyway I keep getting an error after I input the values for alpha, the error I keep getting is in my function, saying that I don't have enough input arguments. The output should be the T matrix, the Hamiltonian matrix and a plot of the first 10 eigenfunctions.
I've tried going in and simply defining alpha as the numbers I want to look at and the program works fine with displaying the output I'm looking for. I just want to be able to change the values for alpha without having to change the program itself. I haven't had any problems with the l inputs.
r = linspace(0.05,19.95,1999)
n = 1999
dr = 0.05
a = full(gallery("tridiag",n,1,-2,1))
T = -0.5*a/(dr^2)
l = input('Input a value for l.')
alph = input('Input a value for alpha.')
v = arrayfun(#(r) yuk_pot(r,l),r);
V = diag(v)
H = T + V
[O,D] = eig(H);
plot(r,O(:,1),r,O(:,2),r,O(:,3),r,O(:,4),r,O(:,5),r,O(:,6),r,O(:,7),r,O(:,8),r,O(:,9),r,O(:,10))
function v = yuk_pot(r,alph,l)
v = (-exp(-alph*r)/r) + 0.5*(l*(l+1)/(r^2));
end
your function function v = yuk_pot(r,alph,l) has 3 input arguments.
you call it with 2 arguments (r and l)
v = arrayfun(#(r) yuk_pot(r,l),r);
what about the second alph argument?
I am using Support Vector Regression(SVR) in libsvm package to predict outputs. Kernel : RBF
Train set size : 729x40
Test set size : 137x40
The output of train set seems fine when measured against ground truth. But the predictions on test set are all the same. It outputs same values.
After checking the related posts, I normalized the data and played with the values of gamma(10-100000) but still the problem persists.
trainGT=games(((games(:,46)>=2010) & (games(:,46)<2015) & (games(:,1)~=8)),43);
featuresTrain=lastGame(games,true,1);
testGT=games((games(:,46)>=2015 & (games(:,1)~=8)),43);
featureTest=lastGame(games,false,1);
eval(['model = svmtrain( trainGT, featuresTrain,''-s 4 -t 2 -c 10 -g 10 ' ''');']);
w = (model.sv_coef' * full(model.SVs));
b = -model.rho;
predictionsTrain = svmpredict(trainGT, featuresTrain,model);
predictionsTest = svmpredict(zeros(length(testGT),1), featureTest, model);
My output is as follows
optimization finished, #iter = 1777
epsilon = 0.630588
obj = -19555.036253, rho = -17.470386
nSV = 681, nBSV = 118
Mean squared error = 305.214 (regression)
Squared correlation coefficient = -1.#IND (regression)
All my predictionTest values are 17.4704 (which is same as the rho value in the output). Can someone please help me on this? Thanks.
When running my MATLAB script below, I keep getting an error that states:
Error using spa (Line 147)
The value of the window
size must be an integer greater than 2.
Error in "projectName" G = spa(xFunction2, x)
I've tried putting in multiple types of arguments into "spa" (data, windowsize, frequency) but it still yields the same error(s). Help?
n = 1:1024;
%Signal Function
xFunction = sqrt(10)*exp(j*2*pi*0.10*n)+ sqrt(20)*exp(j*2*pi*0.20*n) + sqrt(625);
%Complex Noise Function
zFunction = 0.707*randn(size(n)) + j*0.707*randn(size(n));
%Computing the value of x(n) + sqrt(625)*z
xFunction2 = xFunction + sqrt(625)*zFunction;
G = spa(xFunction2,51);
figure(1);
plot(w, 10*log10(G));
Acording the documentation of spa the first argument is iddata type. In addition, the time serie has to be a column vector.
So, to make it works change G = spa(xFunction2,51); for G = spa(xFunction2(:),51);. To do it the correct way, convert your time serie to iddata:
Ts = 0.1; % what ever is your sampling time.
myiddata = iddata(xFunction2(:),[],Ts);
G = spa(myiddata,51);
In addition, you should use spectrum(G) or bode(G)to plot the result.
Is there some simple way of calculating of p-value of t-Test in MATLAB.
I found something like it however I think that it does not return correct values:
Pval=2*(1-tcdf(abs(t),n-2))
I want to calculate the p-value for the test that the slope of regression is equal to 0. Therefore I calculate the Standard Error
$SE= \sqrt{\frac{\sum_{s = i-w }^{i+w}{(y_{s}-\widehat{y}s})^2}{(w-2)\sum{s=i-w}^{i+w}{(x_{s}-\bar{x}})^2}}$
where $y_s$ is the value of analyzed parameter in time period $s$,
$\widehat{y}_s$ is the estimated value of the analyzed parameter in time period $s$,
$x_i$ is the time point of the observed value of the analysed parameter,
$\bar{x}$ is the mean of time points from analysed period and then
$t_{score} = (a - a_{0})/SE$ where $a_{0}$ where $a_{0} = 0$.
I checked that p values from ttest function and the one calculated using this formula:
% Let n be your sample size
% Let v be your degrees of freedom
% Then:
pvalues = 2*(1-tcdf(abs(t),n-v))
and they are the same!
Example with Matlab demo dataset:
load accidents
x = hwydata(:,2:3);
y = hwydata(:,4);
stats = regstats(y,x,eye(size(x,2)));
fprintf('T stat using built-in function: \t %.4f\n', stats.tstat.t);
fprintf('P value using built-in function: \t %.4f\n', stats.tstat.pval);
fprintf('\n\n');
n = size(x,1);
v = size(x,2);
b = x\y;
se = diag(sqrt(sumsqr(y-x*b)/(n-v)*inv(x'*x)));
t = b./se;
p = 2*(1-tcdf(abs(t),n-v));
fprintf('T stat using own calculation: \t\t %.4f\n', t);
fprintf('P value using own calculation: \t\t %.4f\n', p);