Page 441 - Contributed Paper Session (CPS) - Volume 2

CPS1917 Trijya S.
estimation of (1) is essential. Let = and = be the estimates which
̂
̂
2
1
1
2
satisfy (2). Then after extensive algebra, it can be shown that,
+ + = 0 (8)
3
1
1
where = ( 4 −4 1 3 ) and = (2 2 3 − 1 4 )
2
2
12(2 − 2 ) 12(2 − 2 )
1
1
It is well known that the cubic equation (8) will have three real roots if < 0
and 4 + 27 < 0. Therefore, if these conditions are satisfied and we get a
3
2
real root of (8), the second root can be obtained using the following
expression from Rider (1961),
2 −
̂
̂
= 2 1 1
2
̂
−
1
1
Using the estimates and thus obtained, the estimate of mixing
̂
̂
2
1
proportion can be obtained as well. It is important to note that the exact
distributional properties of the estimators are difficult to obtain in general.
However, Rider (1961) showed that the estimators are consistent and obtained
the expressions for the asymptotic variances of the estimators assuming to
be known. For estimating the asymptotic standard errors, we use our estimates
of moments in Rider’s formulas.
A major advantage of the proposed method lies in its application for
estimating parameters of the widely used two-compartment model in
(9)
pharmacokinetics. Pharmacokinetic analysis reveals the rates (amount per unit
time) of drug movement into, within and exiting from the body. A commonly
used model in such analyses is the two compartment model given by () =
− + − , ≥ 0, where () ≥ 0 is concentration of drug, that is, the
1
amount of drug per milliliter in the blood stream at time . Here, = and
1,
1
= . The first term of the model, − , corresponds to the drug
2,
absorption stage and dominates, so > . The second term corresponds to
the drug elimination phase. Integrating () over , we obtain the area under
∞

the curve, = ∫ () = + . Dividing () by , we have
0

1 − 1 − (9)
(| ) = ∙ 1 + (1 − ) ∙ ∙ 2 ,
1, 2,
1 2

−1

where = ( + ) . If both A and B are positive, then () could be

treated as a mixture of two exponential densities of random observation time
. If A is negative, () shall still be a probability density function as () ≥ 0
∞
for all and ∫ () = 1. Therefore, the methods described could be
0
applied without any technical problem.

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