# Mixture of two components

Consider a univariate probability density function $$p(x)$$ that is a mixture of $$2$$ probability density functions with weights $$\eta, 1-\eta$$ and $$\eta\in (0,1)$$: $$p(x)=(1-\eta)g(x)+\eta f(x) \hspace{1cm} \forall x \in \mathbb{R}$$

Suppose we know $$p(x)$$ and $$g(x)$$ at every $$x\in \mathbb{R}$$. Our professor made the following point: without further restrictions we cannot back out $$\eta, f(x)$$. To see that, it is sufficient to observe that $$(1-\eta)g(x)+\eta f(x)=(1-\eta/2)g(x)+\eta/2 (-g(x)+2f(x))$$ Hence, $$p(x)$$ can be generated by $$\{\eta, f(x)\}$$ and by $$\{\eta/2, -g(x)+2f(x)\}$$.

Question: I see the point. However, I don't understand how we are sure that $$(g(x)+2f(x))$$ is a probability density function. Specifically, it is not necessarily positive. Can't that requirement allow to get rid of the second solution?

• I think you have a mistake in the second formula. It should be $\left(1 - \frac{\eta}{2}\right) g\left(x\right) + \frac{\eta}{2} \left(2f\left(x\right) - g\left(x\right)\right)$. Then, it integrates to 1. But there is another problem - you are not promised that density is positive, thus it is not a distribution. – tmrlvi Feb 5 at 16:17
• Thanks. I've corrected the typo. Am I right to say then that the point of my professor does not actually show what he wanted to say? – user3285148 Feb 5 at 16:29
• First, their point stands because they showed that exist case where the pair $\left(f\left(x\right), \eta\right)$ is not uniquely recoverable. Note that in fact, this example can be strengthened to show that you can always find other decompositions. – tmrlvi Feb 5 at 16:59
• Thanks. Is this correct: they have shown that $p(x)$ can be generated by $\{\eta, f(x)\}$ with $2f(x)\geq g(x)$ and by $\{\eta/2, -g(x)+2f(x)\}$ – user3285148 Feb 5 at 17:20
• Yes, though I would phrase it as "they have shown that, in the case where $2f\left(x\right) \ge g\left(x\right)$, $p\left(x\right)$ can be generated by ..." – tmrlvi Feb 5 at 17:23

The following presents a general way to recover another decomposition of a mixture model. In essence, we can always move some of the mass covered by $$g\left(x\right)$$ to the second component.
Let $$\beta>\frac{\eta}{1-\eta}$$ and $$\alpha=\frac{1}{1+\frac{1}{\beta}}>\eta>0$$ (so that $$\frac{1}{\alpha}-\frac{1}{\beta}=1$$). Note that $$p\left(x\right)=\left(1-\frac{\eta}{\alpha}\right)g\left(x\right)+\frac{\eta}{\alpha}\left(\alpha f\left(x\right)+\alpha\frac{g\left(x\right)}{\beta}\right)$$
Note that $$\frac{\eta}{\alpha} \in \left(0, 1\right)$$ and that $$\int_{-\infty}^{\infty}\alpha f\left(x\right)+\alpha\frac{g\left(x\right)}{\beta}dx=\alpha+\alpha\frac{1}{\beta}=\alpha+\alpha\left(\frac{1}{\alpha}-1\right)=1$$
Thus the pair $$\left(\frac{\eta}{\alpha}, \alpha f\left(x\right) + \frac{\alpha}{\beta}g\left(x\right)\right)$$ generates $$p\left(x\right)$$ from $$g\left(x\right)$$.
We can set $$\beta = 2 \frac{\eta}{1 - \eta}$$, and we get
$$p\left(x\right) = \left(1-\frac{1+\eta}{2}\right)g\left(x\right)+\frac{1+\eta}{2}\left(\frac{2\eta}{1+\eta}f\left(x\right)+\frac{1-\eta}{1+\eta}g\left(x\right)\right)$$