# How to transform an exponential density mixture model from independent to dependent terms?

Keep in mind that I do not recommend univariate mixture models for pharmacokinetic usage, except as a default comparison to rather better models, e.g., see this, i.e., convolution models. I am attempting to ask if such univariate models can be improved upon by some other procedure that somehow incorporates covariance, apart from just replacement with better motivated convolution models. For example, see this answer, to this question. That is, some refinement of parameter estimation may be available from population estimates of parameters. Before we can adequately pose the question, we need to relate what is typically done.

The exponential distribution (ED) is given by

$$\text{ED}(t; \lambda)= \begin{cases} \lambda e^{-\lambda t} & \lambda,t\geq 0 \\ 0 & \text{True} \\ \end{cases}\;. \\$$

The simplest independent term mixture model of exponential densities is the biexponential case,

\begin{aligned} \text{pdf}(t)&=p\text{ ED}(\lambda_1)+q\, \text{ED}(\lambda_2)\\&=p\lambda_1 e^{-\lambda_1 t}+q\lambda_2 e^{-\lambda_2 t}, \end{aligned} \

where $0<p<1$ and $q=1-p$ are the fractional (dimensionless) contributions of each ED to the pdf. The triexponential and higher models are defined analogously with $\sum{p_i}=1$, where all $p_i>0$. Such distributions are in most common use for pharmacokinetic modelling, almost to the exclusion of other models. There are a number of problems with this, not the least of which, I suspect, is that the mixed distribution terms are independent, which obviates any interaction between terms, making for an inflexible model. Now this is not seen directly to be a mixture model, the biexponential model (E2) is used in practice to model concentration where $\kappa$ pdf$(t)$ is usually scaled to a concentration curve, where $\kappa$ is the $t=0,\infty$ area under the concentration curve, where the concentration is drug in serum, plasma, or whole blood, depending on which drug is being modeled. A direct stochastic model of this has not been explicitly specified in the literature, and such models are usually spoken of as indirectly implying body compartments of fixed volume(s). That is an unnecessary assumption. However, consider that the drug mass fraction $\frac{M(t)}{M(0)}$ remaining in any concentration model system is nominally $$\frac{M(t)}{M(0)} = \frac{\int_t^{\infty } C(\tau ) \, d\tau}{\int_0^{\infty } C(\tau ) \, d\tau}= \int_t^{\infty } \text{pdf}(\tau ) \, d\tau := S(t) ,$$

where $S(t)$ is, by definition, the survival function of relative drug mass. Next, considering this for the biexponential model yields $$S(t)=p e^{-\lambda_1 t}+q e^{-\lambda_2 t}\;,$$

which is then a stochastic model because a mass unit, (i.e., a drug molecule) has a probability of leaving the circulation through the "portal" corresponding to the first term of $p$, and the second term "portal" of $q=1-p$. No molecule can physically be eliminated via both "portals," thus this is a mixture model of independent processes. The existence of these "portals" is an assumption inherent to the compartmental assumption, e.g., see this, such that that is indeed what is being assumed. However, an assumption is not a proof of anything beyond the tautology implied by that assumption, and the question here is asking for a broader assumption, while still maintaining a mixture model format.

Although the answer is not likely unique, there should be a most logical method of introducing non-independence between terms.

Q: What would a logical method of introducing non-independence of terms for the biexponential mixture density function look like?

I do not know the answer to this question. At face value, this question may be self-contradictory. A possible lead is here. My motive for asking is because the multiexponential mixture model is problematic. That is, even for the bi-exponential case, the parameters have a nasty tendency of failing partial probability ANOVA testing, i.e., overfitting is common, also, the derivative of concentration is mismatched to concentration slope, etc., i.e., lots and lots of problems, and this question is also important as the vast plurality of drugs are quantified by this methodology.

Let $X_1$ be the random variable from component 1, and $X_2$ be the random variable from component 2. The mixture model paradigm dictates that we always only see one of $\{X_1, X_2\}$ as the observed value $Y$.

Now, if you wanted to create a dependence structure between $X_1$ and $X_2$, this could be done with something like a copula model.

However, if given that you will never observe the pair $\{X_1, X_2\}$, the data will never provide you will any insight into the dependency between these two values. Since presumably you will never observe $X_1$ with the intent of estimating $X_2$, this relation will not be helpful to you so it's simplest to just treat them as independent.

If, for some reason, you do have the ability to observe both $\{X_1, X_2\}$, then you should not be modeling them as coming from a mixture model.

• Well, see above for details. There is no contemporaneity. However, an event occurrence is physically associated with one term at a time.
– Carl
Oct 18, 2017 at 0:27
• @Cliff_AB Can you give me an answer for the biexponential mixture case, please? I am finding the cupola link a bit too general for me to make use of.
– Carl
Oct 18, 2017 at 1:11
• @Cliff_AB I do not have as a goal using mixture models for anything. I would not go that route. I am exploring if there is enough wiggle room to improve upon them as applied to pharmacokinetics, and if my characterization of them is statistically correct.
– Carl
Oct 18, 2017 at 3:35
• +1 Because you at least gave me a hint as to what to explore further.
– Carl
Oct 18, 2017 at 3:44
• I know that using mixture models is a dead end for pharmacokinetics, my problem is explaining why. Almost all of the pharmacokinetic literature uses univariate multi-exponential mixture models, usually with two or three terms. The best workaround I have found is to use convolution and dump the mixture approach entirely. Question is, can the the mixture approach be extended to include covariance, and most assuredly it can, for example, E1*E2, where E1 is a monoexponential, E2 is biexponential, and * is convolution. Question is, can this be done without convolution?
– Carl
Oct 18, 2017 at 5:16