# Restricting the median of marginals has implications for the joint distribution?

Consider a bivariate cumulative distribution function (cdf) $F: \bar{\mathbb{R}}^2\rightarrow [0,1]$ where $\bar{\mathbb{R}}^2\equiv \mathbb{R}\cup\{-\infty, \infty\}\times \mathbb{R}\cup\{-\infty, \infty\}$

Let $F_1, F_2$ denote the marginal cdf from $F$ and suppose they both have median zero, i.e., $$F_1(0)=\frac{1}{2} \text{, } F_2(0)=\frac{1}{2}$$

Question:

1) Can we conclude that $F(0,0)=\frac{1}{2}$? Or any other value?

2) Can we say something about $F(0,a)$ or $F(b,0)$ for any $(a,b)\in \bar{\mathbb{R}}^2$?

• If anything, we might be able to conclude $F(0,0)=\frac14$, not $\frac12$? – joriki Jul 25 '18 at 20:20
• Thank you, I see. – TEX Jul 25 '18 at 20:31

Fixing the median of the margins certainly has implications, but you need to consider the joint distribution as well. Denote by $(X,Y)$ the two random variables making up your bivariate distribution.

With regard to 1)

The answer is No! If $X$ and $Y$ are independent, $P(X\leq 0, Y\leq 0) = P(X\leq 0)P(Y\leq 0) = \frac{1}{4}$ as joriki pointed out. But this is not the only possibility. Take $X$ standard normal and $Y=-X$. Then $$P(X\leq 0, Y\leq 0) = P(X\leq 0, X\geq 0) = 0.$$

With regard to 2):

You certainly can say "something". In the example above you can explicitly calculate $F$. Hence $F(0,a) = P(X\leq 0, -X\leq a)$ which is zero if $a\leq 0$ and $\frac{1}{2} - P(X\leq -a)$ if $a\geq 0$.

In general it is true that $0\leq F(0,0) \leq \frac{1}{2}$. My example above and joriki's observation in the comment (Y = X) show both extremes. Due to the properties of probabilities you always know that $0 \leq F(0,0)\leq F_1(0) = \frac{1}{2}$.

Furthermore, every intermediate value in the interval $[0, \frac{1}{2}]$ is possible for $F(0,0).$

• Also, $F(0,\infty)=F(\infty,0)=\frac12$. Also, if $X=Y$ (with any distribution), $F(0,0)=\frac12$. – joriki Jul 25 '18 at 20:44