Show joint cdf is continuous

Question

Let $X$ have a standard normal distribution and let $Y=2X$. Show the cdf $F(x, y)$ of $(X, Y)$ is continuous.

I have attempted this below, but I think I am pretty far off track.

$$F_{XY}(x, y) = P(X \le x, 2X \le y) = P(X \le (x \wedge \frac{y}{2}))$$

$$ \Phi(x) = F_{XY}(x, y) = \begin{cases} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^\frac{-z^2}{2}\text{d}z, \quad\text{ if } x \le \frac{y}{2}\\ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^\frac{-z^2}{8}\text{d}z, \quad\text{ if } x > \frac{y}{2} \end{cases} $$

$$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^\frac{-z^2}{2}\text{d}z = \frac{\operatorname{erf}(\frac{x}{\sqrt{2}})}{2}+1 $$

$$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^\frac{-z^2}{8}\,\text{d}z = \operatorname{erf} \left(\frac{x}{2^\frac{3}{2}}\right) $$

$$ \operatorname{erf}(x) \text{ continuous } \implies F_{XY}(x,y) \text{ continuous } $$

Answer 1

Your first line after the word "below" looks good. I'd extend it to say $$ F_{X,Y}(x, y) = P(X \le x\ \&\ 2X \le y) = P\left(X \le \left(x \wedge \frac y 2 \right) \right) = \Phi\left( x \wedge \frac y 2 \right). $$ Then the question is: How do you show that $\displaystyle (x,y) \mapsto \Phi\left( x \wedge \frac y 2 \right)$ is continuous?

If you can show $(x,y) \mapsto x \wedge \dfrac y 2$ is continuous and $\Phi$ is continuous, then citing a theorem that says a composition of continuous functions is continuous will do it.

Answer 2

Falrach’s comment is right and your approach is correct. But here is a briefer way to say it.

Let $m(x, y)=x \wedge \frac{y}{2}$. Then $m: {\mathbb{R}}^2 \to \mathbb{R}$ is continuous. As you have noted,

$$F_{X,Y}(x, y) = P(X \le m) = G(m(x,y))$$ where $G$ is the CDF of the standard normal distribution, a continuous function.

Being the composition of two continuous functions $G$ and $m$, $F : {\mathbb{R}}^2 \to \mathbb{R}$ is continuous.

How to show that $m$ is continuous? It is again the composition of $\wedge$ and the continuous halving function $y\mapsto \frac{y}{2}$. To show that $x \wedge z$ is continuous: easy at $x=z$. Anywhere else, there is a small $\delta$-ball within which it is simply the projection $x$ or $z$.