Equivalent way of rewriting a two-component mixture

Question

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

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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 $\forall j\in \{1,2\}$.

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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 $$ (\star) \hspace{1cm} F(z|x)=\lambda\times G_1\Big(z-\int_{\mathbb{R}}zdF_1(z|x)\Big|x\Big)+(1-\lambda)\times G_2\Big(z-\int_{\mathbb{R}}zdF_2(z|x)\Big|x\Big) $$

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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?

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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.

Answer 1

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.