We have run an Interrupted Time Series - Poisson model on some count data.
The p-value at the point of interevntion (level change) is <0.05. We are however reproting the level change (i.e. difference between modeled data and counterfactual at time t + 8).
How would I go about deriving a separate p-value for this specific time point? And would it be different to the original level change.
Using R:
Model coded as below:
fit1a <- glm(`Subject Total` ~ Quarter + int2 + time_since_intervention2 , df, family = "poisson")
Related
I am having a hard time with the question below. I am not sure if I got it correct, but either way, I need some help futher understanding it if anyone has time to explain, please do.
Design L1 and L2 distance functions to assess the similarity of bank customers. Each customer is characterized by the following attributes:
− Age (customer’s age, which is a real number with the maximum age is 90 years and minimum age 15 years)
− Cr (“credit rating”) which is ordinal attribute with values ‘very good’, ‘good, ‘medium’, ‘poor’, and ‘very poor’.
− Av_bal (avg account balance, which is a real number with mean 7000, standard deviation is 4000)
Using the L1 distance function computes the distance between the following 2 customers: c1 = (55, good, 7000) and c2 = (25, poor, 1000). [15 points]
Using the L2 distance function computes the distance between the above mentioned 2 customers
Using the L2 distance function computes the distance between the above mentioned 2 customers.
Answer with L1
d(c1,c2) = (c1.cr-c2.cr)/4 +(c1.avg.bal –c2.avg.bal/4000)* (c1.age-mean.age/std.age)-( c2.age-mean.age/std.age)
The question as is, leaves some room for interpretation. Mainly because similarity is not specified exactly. I will try to explain what the standard approach would be.
Usually, before you start, you want to normalize values such that they are rougly in the same range. Otherwise, your similarity will be dominated by the feature with the largest variance.
If you have no information about the distribution but just the range of the values you want to try to nomalize them to [0,1]. For your example this means
norm_age = (age-15)/(90-15)
For nominal values you want to find a mapping to ordinal values if you want to use Lp-Norms. Note: this is not always possible (e.g., colors cannot intuitively be mapped to ordinal values). In you case you can transform the credit rating like this
cr = {0 if ‘very good’, 1 if ‘good, 2 if ‘medium’, 3 if ‘poor’, 4 if ‘very poor’}
afterwards you can do the same normalization as for age
norm_cr = cr/4
Lastly, for normally distributed values you usually perform standardization by subtracting the mean and dividing by the standard deviation.
norm_av_bal = (av_bal-7000)/4000
Now that you have normalized your values, you can go ahead and define the distance functions:
L1(c1, c2) = |c1.norm_age - c2.norm_age| + |c1.norm_cr - c2.norm_cr |
+ |c1.norm_av_bal - c2.norm_av_bal|
and
L2(c1, c2) = sqrt((c1.norm_age - c2.norm_age)2 + (c1.norm_cr -
c2.norm_cr)2 + (c1.norm_av_bal -
c2.norm_av_bal)2)
I use OMNeT++-4.6, sumo-0.22.0 and Veins-4a2.
I am interested to calculate the speed of the vehicle when a message is received. I used getSpeed() function to do it. But the problem is that when I calculated manually the speed basing on the time and the distance (using the formula s = d / t), the value is different.
For example, at t= 55.104470531278 s and the distance d= 29.0477 m, the speed obtained by calling the function getSpeed() is s= 3.34862 m/s = 10.8 km/h.
On the other hand the one calculated manually is s= 0.52713 m/s = 1.9 km/h.
I need help to understand why the value obtained by using getSpeed() is different please.
getSpeed() returns the current speed of the vehicle (to be precise the one in the last simulation step which is by default 1s) while your calculation gives the average speed over the last ~55s (assuming your simulation started at time 0).
I am implementing a Kalman filter for the first time to get voltage values from a source. It works and it stabilizes at the source voltage value but if then the source changes the voltage the filter doesn't adapt to the new value.
I use 3 steps:
Get the Kalman gain
KG = previous_error_in_estimate / ( previous_error_in_estimate + Error_in_measurement )
Get current estimation
Estimation = previous_estimation + KG*[measurement - previous_estimation]
Calculate the error in estimate
Error_in_estimate = [1-KG]*previous_error_in_estimate
The thing is that, as 0 <= KG <= 1, Error_in_estimate decreases more and more and that makes KG to also decrease more and more ( error_in_measurement is a constant ), so at the end the estimation only depends on the previous estimation and the current measurement is not taken into account.
This prevents the filter from adapt himself to measurement changes.
How can I do to make that happen?
Thanks
EDIT:
Answering to Claes:
I am not sure that the Kalman filter is valid for my problem since I don't have a system model, I just have a bunch of readings from a quite noisy sensor measuring a not very predictable variable.
To keep things simple, imagine reading a potentiometer ( a variable resistor ) changed by the user, you can't predict or model the user's behavior.
I have implemented a very basic SMA ( Simple Moving Average ) algorithm and I was wondering if there is a better way to do it.
Is the Kalman filter valid for a problem like this?
If not, what would you suggest?
2ND EDIT
Thanks to Claes for such an useful information
I have been doing some numerical tests in MathLab (with no real data yet) and doing the convolution with a Gaussian filter seems to give the most accurate result.
With the Kalman filter I don't know how to estimate the process and measurement variances, is there any method for that?. Only when I decrease quite a lot the measurement variance the kalman filter seems to adapt. In the previous image the measurement variance was R=0.1^2 (the one in the original example). This is the same test with R=0.01^2
Of course, these are MathLab tests with no real data. Tomorrow I will try to implement this filters in the real system with real data and see if I can get similar results
A simple MA filter is probably sufficient for your example. If you would like to use the Kalman filter there is a great example at the SciPy cookbook
I have modified the code to include a step change so you can see the convergence.
# Kalman filter example demo in Python
# A Python implementation of the example given in pages 11-15 of "An
# Introduction to the Kalman Filter" by Greg Welch and Gary Bishop,
# University of North Carolina at Chapel Hill, Department of Computer
# Science, TR 95-041,
# http://www.cs.unc.edu/~welch/kalman/kalmanIntro.html
# by Andrew D. Straw
import numpy as np
import matplotlib.pyplot as plt
plt.rcParams['figure.figsize'] = (10, 8)
# intial parameters
n_iter = 400
sz = (n_iter,) # size of array
x1 = -0.37727*np.ones(n_iter/2) # truth value 1
x2 = -0.57727*np.ones(n_iter/2) # truth value 2
x = np.concatenate((x1,x2),axis=0)
z = x+np.random.normal(0,0.1,size=sz) # observations (normal about x, sigma=0.1)
Q = 1e-5 # process variance
# allocate space for arrays
xhat=np.zeros(sz) # a posteri estimate of x
P=np.zeros(sz) # a posteri error estimate
xhatminus=np.zeros(sz) # a priori estimate of x
Pminus=np.zeros(sz) # a priori error estimate
K=np.zeros(sz) # gain or blending factor
R = 0.1**2 # estimate of measurement variance, change to see effect
# intial guesses
xhat[0] = 0.0
P[0] = 1.0
for k in range(1,n_iter):
# time update
xhatminus[k] = xhat[k-1]
Pminus[k] = P[k-1]+Q
# measurement update
K[k] = Pminus[k]/( Pminus[k]+R )
xhat[k] = xhatminus[k]+K[k]*(z[k]-xhatminus[k])
P[k] = (1-K[k])*Pminus[k]
plt.figure()
plt.plot(z,'k+',label='noisy measurements')
plt.plot(xhat,'b-',label='a posteri estimate')
plt.plot(x,color='g',label='truth value')
plt.legend()
plt.title('Estimate vs. iteration step', fontweight='bold')
plt.xlabel('Iteration')
plt.ylabel('Voltage')
And the output is:
I want to compare (sorty by) growth rates and disadvantage high rates with very low starting values.
Example:
1.
Start: 1.000.000
End: 1.100.000
Growth: +10%
Start: 100.000
End: 120.000
Growth: +20%
3.
Start: 1
End: 10
Growth: +900%
Start: 10
End: 15
Growth: +50%
Sorting just by growth, descending would result in: 900% (3.), 50% (4.), 20% (2.), 10% (1.)
But I want to have: 20% (2.), 10% (1.), 900% (3.), 50% (4.), because in my case the chance is high, that 3. and 4. are statistical outliers.
What's the best way to solve this problem and do I've to define a threshold for the start values?
Thanks!
Based on the description you have provided, the problem can be split into 2:
Finding and excluding Statistical Outliers from the data set
Sorting the resulting values in descending (or just in any) order
The general solution to the first problem and example using Microsoft Excel is described at : Statistical Outliers detection in Microsoft Excel worksheet (http://www.codeproject.com/Tips/214330/Statistical-Outliers-detection). Following is a bit of theory and a sample pertinent to your case.
Finding "Outliers" in a data set could be done by calculating the deviation for each number, expressed as either a "Z-score" or "modified Z-score" and testing it against certain predefined threshold. Z-score typically refers to number of standard deviation relative to the statistical average (in other words, it's measured in "Sigmas"). Modified Z-score applies the median computation technique to measure the deviation and in many cases provides more robust statistical detection of outliers. Mathematically the Modified Z-score could be written (as suggested by Iglewicz and Hoaglin - see the referenced article) as:
Mi = 0.6745 * (Xi - Median(Xi)) / MAD,
where MAD stands for Median Absolute Deviation. Any number in a data set with the absolute value of modified Z-score exceeding 3.5 is considered an "Outlier". Modified Z-score could be used to detect outliers in Microsoft Excel worksheet pertinent to your case as described below.
Step 1. Open a Microsoft Excel worksheet and in Cells A1, A2, A3 and A4 enter the values: 900%, 50% 20% and 10%, correspondingly.
Step 2. In C1 enter the formula: =MEDIAN(A1:A4) . The value in this cell corresponds to the median calculated on a data set entered at step 1.
Step 3. In C2 enter the array formula: {=MEDIAN(ABS(MEDIAN(A1:A4)-A1:A4))} . As a reminder, in order to enter the array formula, select the cell, type the formula in Excel Formula Bar and then click on the combination: CTRL-SHIFT-ENTER (notice the curly brackets surrounding the expression, which indicates the array formula). The value in this cell (C2) corresponds to MAD.
Step 4. Enter the formula: =IF((0.6745*ABS(C$1-A1)>3.5*C$2), "OUTLIER", "NORMAL") in the first row of column B and extend it down to the 4th row. Final result of “Outlier’s detection” should appear in column B.
A B C
900% OUTLIER 35%
50% NORMAL 0.35
20% NORMAL
10% NORMAL
thus the value 900% is found an "Outlier" while other values are OK. Sorting the result set will be just a trivial task.
Excel Worksheet example is included for the clarity of explanation. The algorithm itself could be implemented in any programming languages (VBA, C#, Java, etc). Hope this will help.
my solition
private static List<double> StatisticalOutLierAnalysis(List<double> allNumbers)
{
List<double> normalNumbers = new List<double>();
List<double> outLierNumbers = new List<double>();
double avg = allNumbers.Average();
double standardDeviation = Math.Sqrt(allNumbers.Average(v => Math.Pow(v - avg, 2)));
foreach (double number in allNumbers)
{
if ((Math.Abs(number - avg)) > (2 * standardDeviation))
outLierNumbers.Add(number);
else
normalNumbers.Add(number);
}
return normalNumbers;
}
I have 5 non-parametric models all with 5 to 8 parameters. This models are used to fit longitudinal data y(t) with t being time. Every datafile is fitted by all 5 models for comparison. The model itself cannot be altered.
For fitting starting values are used and these are fitted into a lsqcurvefit model using a levenberg-marquardt algortihm. So I've written a script for several models and one function for curvefitting
if i perform the curve fitting a lot of the starting values are wandering off to extreme values. This is the thing I want to avoid since these parameters should stay in the proximity off it's starting values and should only change between a well defined range or so that only curve fits within a standard deviation are included.Important to note here is that this restrictions should be imposed during the curve fitting (iterative numerization techique) and not afterwards.
The function I've written to fit models into height:
% Fit a specific model for all valid persons
try
opts = optimoptions(#lsqcurvefit, 'Algorithm', 'levenberg-marquardt');
[personalParams,personalRes,personalResidual] = lsqcurvefit(heightModel,initialValues,personalData(:,1),personalData(:,2),[],[],opts);
catch
x=1;
end
The function I've written for one of my models
elseif strcmpi(model,'jpss')
% y = h_1(1-(1/(1+((t+0.75)^c_1/d_1)+((t+0.75)^c_2/d_2)+((t+0.75)^c_3/d_3)))
% heightModel = #(params,ages) params(1).*(1-1./(1+((ages+0.75).^params(2))./params(3) + ((ages+0.75).^params(4))./params(5) + ((ages+0.75).^params(6))./params(7)));
heightModel = #(params,ages) params(1).*(1-1./(1+(((ages+0.75)./params(3)).^params(2)) + (((ages+0.75)./params(5)).^params(4)) + ((ages+0.75)./params(7)).^params(6))); % Adapted 25/07
modelStrings = {'h1','c1','d1','c2','d2','c3','d3'};
% Define initial values
if strcmpi('male',gender)
initialValues = [174.8 0.6109 2.9743 3.614 9.88 22.393 13.59];
else
initialValues = [162.7 0.6546 2.43 4.011 8.579 18.394 11.846];
end
What I would like to do:
Is it possible to place restrictions on every startingvalue #initial values? Putting restrictions on lsqcurvefit wouldn't be a good idea I think since there are different models with different starting values and different ranges that are allowed.
I had 2 things in my mind:
1. using range and place this between the initial values
initialValues = [162.7 0.6546 2.43 4.011 8.579 18.394 11.846]`
if range a1=[150,180]; range a2=[0.3,0.8] and so one
place lb and ub restrictions seperatly on all my initialvalues between lsqcurvefit
if Heightmodel='name model'
initial value* 1.2 and lb = initial value* 0.8
Can someone give me some hints or pointers because I can't make it work.
Thanks in advance
Lucy
Could somebody help me out
You state: there are different models with different starting values and different ranges that are allowed. This is where you can use ub and lb. How to do this is outlined in the lsqcurvefit documentation:
X=LSQCURVEFIT(FUN,X0,XDATA,YDATA,LB,UB) defines a set of lower and
upper bounds on the design variables, X, so that the solution is in the
range LB <= X <= UB. Use empty matrices for LB and UB if no bounds
exist. Set LB(i) = -Inf if X(i) is unbounded below; set UB(i) = Inf if
X(i) is unbounded above.
For instance in the following example the parameters are constrained within limits during the fit. The lower bound (lb) and upper bound (ub) are set to 20% below and above the starting values, respectively.
heightModel = #(params,ages) abs(params(1).*(1-1./(1+(params(2).* (ages+params(8) )).^params(5) +(params(3).* (ages+params(8) )).^params(6) +(params(4) .*(ages+params(8) )).^params(7) )));
initialValues = [161.92 0.4173 0.1354 0.090 0.540 2.87 14.281 0.3701];
lb = 0.8*initialValues; % <-- lower bound is 20% smaller than initial par values
ub = 1.2*initialValues;
[parsout,resnorm,residual] = lsqcurvefit(heightModel,initialValues,t,ht,lb,ub);