Using Expected Shortfall Estimation and Backtesting Code

조회 수: 19 (최근 30일)
jens pauwels
jens pauwels 2020년 4월 26일
댓글: jens pauwels 2020년 4월 30일
Hi All,
I use the following coding and would like to test it against the portfolio I have created. Unfortunately I get the same error every time:Array indices must be positive integers or logical values.
H = readtable('CAPM2.csv');
symbol = {'PBIT'};
nAsset = numel(symbol);
Convert prices to returns
Returns = (H{:,symbol});
Select assets into portfolio
symbol2 = {'Datum'};
Datum = (H{:,symbol2});
DateReturns = Datum(1:end);
SampleSize = length(Returns);
TestWindowStart = find(year(DateReturns)==2015,1);
TestWindowEnd = find(year(DateReturns)==2019,1,'last');
TestWindow = TestWindowStart:TestWindowEnd;
EstimationWindowSize = 250;
DatesTest = DateReturns(TestWindow);
ReturnsTest = Returns(TestWindow);
VaRLevel = 0.975;
VaR_Hist = zeros(length(TestWindow),1);
ES_Hist = zeros(length(TestWindow),1);
for t = TestWindow
i = t - TestWindowStart + 1;
EstimationWindow = t-EstimationWindowSize:t-1;
[VaR_Hist(i),ES_Hist(i)] = hHistoricalVaRES(Returns(EstimationWindow),VaRLevel);
end
Array indices must be positive integers or logical values.
The problem arises when creating the EstimationWindow. The result is always negative. Unfortunately I have already tried to convert it to positive, unfortunately without success.

답변 (1개)

Sai Sri Pathuri
Sai Sri Pathuri 2020년 4월 29일
I am assuming that you want to declare EstimationWindow as first 250 indices for t = 1 and shift it by 1 for every loop. The window you declared is,
EstimationWindow = t-EstimationWindowSize:t-1;
For t = 1, this EstimationWindow will take the values -249 to 0. But, in MATLAB, the indices start from 1. Hence, the error is being thrown. You may declare the EstimationWindow as
EstimationWindow = t:EstimationWindowSize-t+1;
  댓글 수: 2
jens pauwels
jens pauwels 2020년 4월 29일
thanks that took me a step further, unfortunately he gives me the next error in the next line
VaR_Hist = zeros(length(TestWindow),1);
ES_Hist = zeros(length(TestWindow),1);
for t = TestWindow
i = t - TestWindowStart + 1;
EstimationWindow = t:EstimationWindowSize-t+1;
[VaR_Hist(i),ES_Hist(i)] = hHistoricalVaRES(Returns(EstimationWindow),VaRLevel);
end
Operator '.*' is not supported for operands of type 'cell'.
Error in ExpectedShortfallBacktestingExample2>hHistoricalVaRES (line 249)
ES = ((k - N*VaRLevel).*z(k) + sum(z(k+1:N)))/(N.*(1 - VaRLevel));
Local Functions
function [VaR,ES] = hHistoricalVaRES(Sample,VaRLevel)
% Compute historical VaR and ES
% See [5] for technical details
% Convert to losses
Sample = Sample;
N = length(Sample);
k = ceil(N*VaRLevel);
z = sort(Sample);
VaR = z(k);
if k < N
ES = ((k - N*VaRLevel)*z(k) + sum(z(k+1:N)))/(N*(1 - VaRLevel));
else
ES = z(k);
end
end
function [VaR,ES] = hNormalVaRES(Mu,Sigma,VaRLevel)
% Compute VaR and ES for normal distribution
% See [4] for technical details
VaR = -1*(Mu-Sigma*norminv(VaRLevel));
ES = -1*(Mu-Sigma*normpdf(norminv(VaRLevel))./(1-VaRLevel));
end
function [VaR,ES] = hTVaRES(DoF,Mu,Sigma,VaRLevel)
% Compute VaR and ES for t location-scale distribution
% See [4] for technical details
VaR = -1*(Mu-Sigma*tinv(VaRLevel,DoF));
ES_StandardT = (tpdf(tinv(VaRLevel,DoF),DoF).*(DoF+tinv(VaRLevel,DoF).^2)./((1-VaRLevel).*(DoF-1)));
ES = -1*(Mu-Sigma*ES_StandardT);
end
jens pauwels
jens pauwels 2020년 4월 30일
I was able to solve this problem but instead i recieve the next error
Array indices must be positive integers or logical values.
Error in ExpectedShortfallBacktestingExample2>hHistoricalVaRES (line 246)VaR = z(k);
but k i positive
function [VaR,ES] = hHistoricalVaRES(Sample,VaRLevel)
% Compute historical VaR and ES
% See [5] for technical details
% Convert to losses
Sample = -Sample;
N = length(Sample);
k = ceil(N*VaRLevel);
z = sort(Sample);
VaR = z(k);
if k < N
ES = ((k - N*VaRLevel)*z(k) + sum(z(k+1:N)))/(N*(1 - VaRLevel));
else
ES = z(k);
end

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