I am having difficulty in solving the following practice problem:
Assume that $X_n \leq Y_n$ for every $n$, $X_n \overset{D}\longrightarrow X$, and $Y_n \overset{D}\longrightarrow Y$. Let $F$ and $G$ be the cumulative distribution functions of $X$ and $Y$, respectively. Show that $F(z) \geq G(z)$ for every $z$.
Here is my current solution:
Let $X_n \leq Y_n$, or $W_n = X_n - Y_n \leq 0$. Set a cumulative distribution function for $W_n$ as $H_n (z) \leq 0$. Then $H_n \left(z \right) \to H(z) = 0$. Thus,
$$H_n(z) \leq F_n (z) - G_n(z) \\ 0 \leq F_n (z) - G_n(z) \\ G_n(z) \leq F_n (z) $$
Since $X_n \overset{D}\longrightarrow X$ and $Y_n \overset{D}\longrightarrow Y$, we have $F_n (z) \to F(z)$ and $G_n (z) \to G(z)$. Therefore, $G_n (z) \leq F_n (z)$ approaches to $G(z) \leq F(z)$ for every $z$.
I have a feeling that my solution is incorrect already, unless you disagree. If my solution is incorrect, then I would appreciate it if you can provide me with the correct solution, or a hint at least. Thank you.
Without independence you cannot say anything about the distribution of $X_n-Y_n$ so your argument fails. [$H_n (z) \to 0$ is wrong].
$G(z)=P(Y \leq z)=\lim P(Y_n \leq z)\leq \lim P(X_n \leq z)=\lim F_n(z)=F(z)$ provided $F$ is continuous at $z$. [The inequality here follows from the fact that $(Y_n \leq z) \subseteq (X_n\leq z)$]. But the set of points of discontinuity of $F$ is at most countable. Thsi implies that $G(z) \leq F(z)$ at all points of a dense set. Since $F$ and $G$ are right continuous we can take limits through continuity points of $F$ to now see that $G(z) \leq F(z)$ holds for all real numbers $z$.