# Equivalent way of rewriting a two-component mixture

I'm confused on the following equivalent way of rewriting a two-component mixture.

Consider the two-component conditional mixture $$F(z|x)=\lambda F_1(z|x)+(1-\lambda)F_2(z|x)$$ where all the $$F$$'s are conditional CDFs and $$\lambda\in (0,1)$$.

A1: Assume that $$F(z|x)=0$$ $$\forall z\leq 0$$.

A2: Assume that $$\int_{\mathbb{R}}zdF_j(z|x)$$ exists $$\forall j\in \{1,2\}$$.

Consider a random variable $$Z_j$$ with CDF conditional on $$x$$ equal to $$F_j(\cdot|x)$$ $$\forall j \in \{1,2\}$$.

Define the random variable $$\epsilon_j\equiv Z_j-\int_{\mathbb{R}}zdF_j(z|x)$$ $$\forall j \in \{1,2\}$$. Let $$G_j(\cdot|x)$$ denote the CDF of $$\epsilon_j$$ conditional on $$x$$ $$\forall j \in \{1,2\}$$. Then, $$F_j(z|x)=G_j\Big(z-\int_{\mathbb{R}}zdF_j(z|x)\Big|x\Big)$$ $$\forall j \in \{1,2\}$$. Hence we can rewrite $$(\star) \hspace{1cm} F(z|x)=\lambda\times G_1\Big(z-\int_{\mathbb{R}}zdF_1(z|x)\Big|x\Big)+(1-\lambda)\times G_2\Big(z-\int_{\mathbb{R}}zdF_2(z|x)\Big|x\Big)$$

Question 1: I believe that A1 implies that $$F_j(z|x)=0$$ $$\forall z\leq 0$$ and $$\forall j\in \{1,2\}$$. Is this correct? If this is true, then can we say that, for example, A1 implies that $$F_1(\cdot|x)$$ and $$F_2(\cdot|x)$$ cannot be normal CDF?

If I'm correct about Question 1:

Question 2: A1 remains valid also under the new representation of the mixture $$(\star)$$. However, I can't see how $$G_1\Big(z-\int_{\mathbb{R}}zdF_1(z|x)\Big|x\Big)$$ and $$G_2\Big(z-\int_{\mathbb{R}}zdF_2(z|x)\Big|x\Big)$$ can be zero for negative values of their arguments, given that the definition of $$\epsilon_1,\epsilon_2$$ does not prevent these two random variables from having negative realisations.

You are correct about Question 1. Since both $$\lambda$$ and $$1-\lambda$$ are positive, while $$F_j(z\mid x)\ge0$$, the only way for $$\lambda F_1(z\mid x)+(1-\lambda)F_2(z\mid x)$$ to be zero is if both $$F_1(z\mid x)$$ and $$F_2(z\mid x)$$ are zero. So it follows that $$F_1(\cdot\mid x)$$ and $$F_2(\cdot\mid x)$$ can never be the cdf of a Normal distribution.
For question 2, you are misreading the conclusion. The same argument as above shows that $$G_j(z-\int zdF_j(z\mid x)\mid x)$$ equals zero for all $$z\le 0$$. However, this does not imply that $$G_j(t\mid x)$$ equals zero for all $$t\le0$$, which seems to be your conclusion.