Suppose we have statistics N(m1, m2), where m1 is the value of the first factor, m2 is the value of the second factor, N(m1, m2) is the number of observations corresponding to the values of factors m1 and m2. In this case, the probability P(m1, m2) = N(m1, m2) /K, where K is the total number of observations. In real situations, detailed statistics N(m1, m2) is often unavailable, and only the normalized marginal values S1(m1) and S2(m2) are known, where S1(m1) is the normalized total number of observations corresponding to the value m1 of the first factor and S2(m2) is the normalized total number of observations corresponding to the value m2 of the second factor. In this case P1(m1) = S1(m1)/K and P2(m2) = S2(m2)/K. It is clear that based on P1(m1) and P2(m2) it is impossible to calculate the exact value of P(m1, m2). But how to do this approximately with the best confidence? Thanks in advance for any advice.
For your normalising constant or marginal distribution of vector case, you may use saddle point approximation, see e.g., http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.123.1487&rep=rep1&type=pdf
If the factor m1 and m2 are independent, you can always use P(m1,m2) = P(m1)P(m2). I am assuming that is not the case. Can you elaborate a little more? Thanks.
For your normalising constant or marginal distribution of vector case, you may use saddle point approximation, see e.g., http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.123.1487&rep=rep1&type=pdf
Dear Shah Mahdi Hasan , thanks for your answer. Certainly, the factors are not independent. For, example, m1 is value of the one marker of diagnose and m2 is the value of the second marker, N(m1, m2) is amount of patients with values of m1 and m2 of markers.
Dear Muhammad Ali , thanks a lot for your link. I will carefully read this article.
Dear Ette Etuk , Hello! Unfortunately, it is incorrect to use this expression for this task. according this expression P(m1,m2)=P(m1nm2) < max( P(m1), P(m2) ). But it is possible to build two-dimensional probability P(m1,m2) s.t.
P(m1, m2) > max( P(m1), P(m2) )
Shah Mahdi Hasan For simplicity we can assume, that both P(m1, m2), P(m1) and P(m2) are monotonic increasing functions for variables m1 and m2.
Ette Etuk , I fully agree with you, that " The data given are not enough for the calculation of P(m1,m2)". I also have written, that "The data given are not enough for the calculation of P(m1,m2)". But I have asked about approximate solution !
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