Expected Shortfall for asset returns

Hi everybody!!!

I'm trying to compute the Conditional VaR (Expected Shortfall) for some funds' returns. I have a large matrix (time series) of returns (132x1773) where 132 indicates the number of months and 1773 the number of funds. For those values has been quite easy to compute the Value at Risk for each fund but now I need to compute the conditional one.

As theory suggets, CVaR is the mean of the values above the VaR itself but I'm having lots of trouble ciomputing it.

I already tried 1) storing the value in a matrix with a for loop but since each fund has different number of values below tha VaR, the storage matrix is no more a matrix and 2) using the find function

I'll try to write down some codes to make it easier to understand:

Ret %is my 132x1773 matrix of returns
VaR %is my 1x1773 row vector

[rownum, columnum] = size(Ret);
store = zeros(1,columnum)

for i = 1:columnum
    ind(i) = find(Ret(:,i)>VaR(i))
    Ret(:,ind(i)) = [];
    CVaR(i) = mean(Ret(:,i));
end

Unable to perform assignment because the indices on the left side are not compatible with the size of the
right side.

Thanks a lot in advance!!!

Answer 1

[Star Strider]

If I understand correctly what you want to do, it would likely be easier to create ‘ind’ as a logical vector. There is also no specific need to save it to a matrix (although you can do that if it is absolutely necessary), since it is created and used once in each iteration.

% Ret %is my 132x1773 matrix of returns
% VaR %is my 1x1773 row vector

Ret = randn(5,12);
VaR = randn(1,12);

[rownum, columnum] = size(Ret);
store = zeros(1,columnum);

for i = 1:columnum
    ind = Ret(:,i)>VaR(i);                              % Logical Vector
    CVaR(i) = mean(Ret(~ind,i));                        % Use Only Those Values That Do Not Meet The 'ind' Criteria
end

.

[Marco Piazza]

Thanks a lot for the fast answer. It seems to work, at least i do not get any error. Since I have some NaNs in the funds' return, I did the following

for i = 1:columnum
    ind = Ret(:,i)<=-VaR95(i);
    CVaR(i) = nanmean(Ret(ind,i));
end

and I also tried with

for i = 1:columnum
    ind = Ret(:,i)<=-VaR95(i);
    CVaR(i) = mean(Ret(ind,i), 'omitnan');
end

But still got 42 NaNs. Can you guess why?

Thank you very much. Really appreciate.

[Marco Piazza]

Maybe the problem is that for that specific vector (where the NaN appears) the for loop did not find any values smaller than the VaR. I used a standardized VaR, maybe I should look for the historical one (based on realized returns) so that the for loop will always find smaller values.

[Star Strider]

As always, my pleasure!

That is certainly possible.

‘‘Can you guess why?’

I am not certain, however there are at least three possibilities that I can think of.

It could be that none of the values met the criteria and ‘ind’ returned a vector of logical 0. This would result in ‘Ret(ind,i)’ being empty.

Comparing any numeric value to NaN results in a logical value of 0 (false). In the context of ‘ind’ that would mean that none of the values matched, creating an empty vector, and the mean of an empty vector would be NaN.

If there are NaN values in the data, then it could be that all the selected rows in any particular column are all NaN.

For example:

v = NaN(5,1);
ind = v <= 2                                % In This Instance 'Ret(ind,i)' Would Have No Elements

ind = 5×1 logical array
   0
   0
   0
   0
   0

v = NaN(5,1);                               % If All The Elements Are 'NaN'
meanv = mean(v, 'omitnan')

meanv = NaN

So, if all the values were NaN, then mean with 'omitnan' ends up calculating the mean of an empty vector, resulting in (0/0) that is of course NaN.

The mean documentation does not specifically discuss this particular issue (at least that I could find) so this is my best guess as to what is causing the NaN values in ‘CVaR’.

.

Answer 2

[KSSV]

[rownum, columnum] = size(Ret);
store = zeros(1,columnum)
CVaR = cell(columnum,1) ; 
for i = 1:columnum
    ind(i) = find(Ret(:,i)>VaR(i))
    Ret(:,ind(i)) = [];
    CVaR{i} = mean(Ret(:,i));
end

[Marco Piazza]

Thanks KSSV, but also with your guess I still get the same error from Matlab.