If X and Y are continuous random variables, then the random vector (X,Y) is a continuous random vector? Just need help understanding this question, as I don't really understand what a continuous random vector is. My limited understanding is that a continuous random vector must be completely continuous (so for continuous X and Y this is satisfied) and that to get the probability of the random vector occurring, we double integrate over the supports (of X and Y obviously). My guess is then, that the c.r.v (X,Y) IS a continuous random vector.
Also, what of the converse? If (X,Y) is a continuous random vector, then are X and Y continuous random variables? I also think this is correct, as the continuity of the vector MUST imply the continuity of X and Y, although I'm honestly not sure.
Thanks
 A: A random vector is called continuous iff its CDF is a continuous function.
• If $F_{(X, Y)}$ is continuous in the first coordinate (for every fixed second coordinate), then $X$ is a continuous random variable. To see this, let $x_0 \in \mathbb{R}$ and $\varepsilon > 0$ be arbitrary. By the continuity of measures from below we can find a $y_0$ so that $P(X \le x_0) - P(X \le x_0, Y \le y_0) < \varepsilon$ holds. From our assumption we can find a $\delta > 0$ so that for all $x \in (x_0 - \delta, x_0]$ the inequality $P(X \le x_0, Y \le y_0) - P(X \le x, Y \le y_0) < \varepsilon$ holds. Now this means that for any $x \in (x_0 - \delta, x_0]$ the following holds:
$$\begin{align*}
&P(X \le x_0) - P(X \le x)\\
\le{}& \{P(X \le x_0) - P(X \le x_0, Y \le y_0)\}\\
&+ \{P(X \le x_0, Y \le y_0) - P(X \le x, Y \le y_0)\} \\
&+ \{P(X \le x, Y \le y_0) - P(X \le x)\} \\
\le{}& 2 \varepsilon
\end{align*}$$
But this means that $F_X$ is left-continuous and since every CDF is right-continuous, this implies that $X$ is a continuous random variable.
• If $X$ and $Y$ are both continuous, then $(X, Y)$ is also continuous. To see this, let $x_0, y_0 \in \mathbb{R}$ and $\varepsilon > 0$ be arbitrary. Choose $\delta > 0$ so that for all $x \in U_\delta(x_0)$ and $y \in U_\delta(y_0)$ the inequalities $|F_X(x) - F_X(x_0)| \le \varepsilon$ and $|F_Y(y) - F_Y(y_0)| \le \varepsilon$ hold. Then for all $(x, y) \in U_\delta(x_0) \times U_\delta(y_0)$:
$$\begin{align*}
&|P(X \le x, Y \le y) - P(X \le x_0, Y \le y_0)|\\
\le{}& |P(X \le x, Y \le y) - P(X \le x_0, Y \le y)| + |P(X \le x_0, Y \le y) - P(X \le x_0, Y \le y_0)| \\
={}&P(\min\{x, x_0\} < X \le \max\{x, x_0\}, Y \le y) + P(X \le x_0, \min\{y, y_0\} < Y \le \max\{y, y_0\}) \\
\le{}&P(\min\{x, x_0\} < X \le \max\{x, x_0\}) + P(\min\{y, y_0\} < Y \le \max\{y, y_0\}) \\
={}& |P(X \le x) - P(X \le x_0)| + |P(Y \le y) - P(Y \le y_0)| \le 2\varepsilon
\end{align*}$$
• A similar calculation shows that if $X$ is continuous, then $F_{(X, Y)}$ is continuous in the first coordinate for every fixed value of the second coordinate.
A: A random variable is continuous iff it's CDF is continuous.
So the following is easy to see:
i) Assume $(X,Y)$ is continuous, so is its CDF $F_{(X,Y)}$ but then  $$F_X(x) = P(X \le x) = P(X \le x, Y \le \infty) = F_{(X,Y)}(x,\infty)$$ is also continuous and so is X (the same for Y)
ii) Assume $(X,Y)$ is NOT continous, then there exists $(x,y) \in [-\infty,\infty]\times [-\infty,\infty]$ that the CDF $F_{X,Y}$ is not continous on $(x,y)$. It's an easy calculation then, that either $P(X \le x)$ or $P(Y\le y)$ is not continuous.
So it follows: $(X,Y)$ continuous $\iff$ $X$ and $Y$ continuous
