Difference between two vector CDFs
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I am struggling to work out how to derive the vector for Prob(A - B > 0) where A and B are CDFs of independent variables in vector form.
I thought going through each point in the CDF vectors and multiplying 1-CDF_B by CDF_A would give the correct result, but the resulting vector doesn't sum to 1.
6 件のコメント
Torsten
2017 年 5 月 5 日
編集済み: Torsten
2017 年 5 月 5 日
The CDF of C=A-B at a point z can be obtained as follows:
Multiply PDF-vector of A at points x_i with CDF-vector of B at points x_i-z, sum the products and multiply the result by the distance of the points x_i (deltax). Then take the negative of this value and add 1.
The exact formula can be derived as follows :
F_C(z)
= Prob(A-B<=z)
= integral_{x=-oo}^(x=+oo) Prob(A=x)*Prob(B>=x-z) dx
= integral_{x=-oo)^(x=+oo) f_A(x)*(1-F_B(x-z)) dx
= 1 - integral_{x=-oo}^{x=+oo} f_A(x)*F_B(x-z) dx
Maybe MATLAB's "conv" for the vectors f_A and F_B can automatically perform the task you are looking for, but I don't have the time to go into detail.
Best wishes
Torsten.
回答 (1 件)
Torsten
2017 年 5 月 5 日
編集済み: Torsten
2017 年 5 月 5 日
Prob(A>B)
= integral_{a=-oo}^{a=+oo} Prob(A=a)*Prob(B<a) da
= integral_{a=-oo}^{a=+oo} f_A(a)*F_B(a) da
= integral_{a=-oo}^{a=+oo} (dF_A(a)/da)*F_B(a) da
where f_A, f_B denote PDFs of A and B and F_A, F_B denote CDFs of A and B.
Best wishes
Torsten.
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