# Essay on Sample Data And The Matrix

Consider the specific case where i=1 (like), so substitute the value of i in equation (11)

At the time t_0 the value of X is considered zero since at the beginning there won’t be any trust value. Therefore the equation (8) is expanded according to the data provided in the table2 are:

X_1 (t)=0+[X_10 (t_0 ) 〖(t〗_1-t_(0 ))+X_11 (t_1 ) 〖(t〗_2-t_(1 ))+X_12 (t_2 ) 〖(t〗_3-t_(2 ))+X_13 (t_3 ) 〖(t〗_4-t_(3 ))+ X_14 (t_4 ) 〖(t〗_5-t_(4 ))+X_15 (t_5 ) 〖(t〗_6-t_(5 ))] + 4/20 [X_10 (t_0 )[w(t_1 )-w(t_0 ) ]+X_11 (t_1 )[w(t_2 )-w(t_1 ) ] +X_12 (t_2 )[w(t_3 )-w(t_2 ) ]+X_13 (t_3 )[w(t_4 )-w(t_3 ) ]+X_14 (t_4 )[w(t_5 )-w(4) ]+X_15 (t_5 )[w(t_6 )-w(t_5 ) ] ] (14)

According to the equation(8) substitute the values from the table 2 as corresponding cells from the second column for X_ij (t_j ) and an eighth column for (t_(j+1)-t_j ). The wiener process W(t_(j+1) )-W(t_j) follows a cumulative normal distribution with X, mean and standard deviation as N(X,μ,σ). Where X take the value from the second column, μ=0 and σ take the value from the ninth column. Substituting these values in equation (14):

X_1 (t)=0+[0+1×16+1×102+1×126+1×293+0]+4/20 [0+1×N(434,0,16)+1×N(536,0,102)+1×N(662,0,126)+1×N(955,0,293)+0] (15)

X_1…