Transitionmatrix needs to be converted from day to one tenth of a day.

Hello,

i have a transitionmatrix that describes a population from one day to the other. I need to convert the transitionsmatrix so that one multiplication of my population with the matrix equals one tenth of a day instead of a full day. I already got a suggestion to use the squareroot, that only works with a single number but not within a transitionmatrix (atleast it did not work for me).

The folowing code is used to calculate:

%development rates of each stage

dpreg = 0.172;

dprL1 = 0.217;

dprL2 = 0.313;

dprL3 = 0.222;

dprL4 = 0.135;

dprPu = 0.099;

%relative mortality rates of each stage

rmreg = 0.017;

rmrL1 = 0.060;

rmrL2 = 0.032;

rmrL3 = 0.022;

rmrL4 = 0.020;

rmrPu = 0.020;

rmrAd = 0.027;

B1 = [1-(dpreg+rmreg),0,0,0,0,0,0;

dpreg,1-(dprL1+rmrL1),0,0,0,0,0;

0,dprL1,1-(dprL2+rmrL2),0,0,0,0;

0,0,dprL2,1-(dprL3+rmrL3),0,0,0;

0,0,0,dprL3,1-(dprL4+rmrL4),0,0;

0,0,0,0,dprL4,1-(dprPu+rmrPu),0;

0,0,0,0,0,dprPu,1-rmrAd];

A = [10;10;10;10;10;10;10];

%%this loop can be used to calculate the result for 10 days to check whether it is the same %result, to do so change the 1:1 to 1:10

for i = 1:1

C = B1*A;

A=C;

end

Thanks a lot

Best regards Dominik

Answer 1

[Rohit]

Hi Dominik,

I understand that you are trying to convert the transition matrix so that one multiplication of the population with the matrix equals one tenth of a day instead of a full day.

To achieve this, you can simply take the transition matrix to the power of 1/10.

Taking the matrix to the power of 1/10 is equivalent to multiplying the matrix with itself 1/10 times, which is what we want to achieve to scale the transition matrix for one tenth of a day instead of a full day.

You can see modified code below and it shows both matrix are equivalent: -

%development rates of each stage
dpreg = 0.172;
dprL1 = 0.217;
dprL2 = 0.313;
dprL3 = 0.222;
dprL4 = 0.135;
dprPu = 0.099;
%relative mortality rates of each stage
rmreg = 0.017;
rmrL1 = 0.060;
rmrL2 = 0.032;
rmrL3 = 0.022;
rmrL4 = 0.020;
rmrPu = 0.020;
rmrAd = 0.027;
B1 =           [1-(dpreg+rmreg),0,0,0,0,0,0;
                dpreg,1-(dprL1+rmrL1),0,0,0,0,0;
                0,dprL1,1-(dprL2+rmrL2),0,0,0,0;
                0,0,dprL2,1-(dprL3+rmrL3),0,0,0;
                0,0,0,dprL3,1-(dprL4+rmrL4),0,0;
                0,0,0,0,dprL4,1-(dprPu+rmrPu),0;
                0,0,0,0,0,dprPu,1-rmrAd];
A = [10;10;10;10;10;10;10];

%%this loop can be used to calculate the result for 10 days to check whether it is the same %result, to do so change the 1:1 to 1:10
for i = 1:1
C = B1*A;
A=C;
end
disp(A);

    8.1100
    8.9500
    8.7200
   10.6900
   10.6700
   10.1600
   10.7200

%Converting transition matrix
A = [10;10;10;10;10;10;10];
B2=B1^(0.1);
for i = 1:10
C = B2*A;
A=C;
end
disp(A);

    8.1100
    8.9500
    8.7200
   10.6900
   10.6700
   10.1600
   10.7200