Joint Distribution – Calculating Quantiles of a Joint Distribution

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Give iid random vectors $(x_1,y_1),\dots, (x_n,y_n)$ from the two dimensional cumulative distribution function $J(x,y)$. The marginal CDF of $x$ (rep. $y$) is $F(x)$ (reps. same $F(y)$). They have same marginal CDF $F$. Assume that the density $f$ is continuous and positive for all $x$. For $p\in (0,1)$, define the quantile $\xi_p=\inf\{x: F(x)\ge p\}$.

Q1: Do we have $E[I(x_i\le \xi_p)\cdot I(y_i\le \xi_p)]=J(\xi_p, \xi_p)?$

Q2: Is this $p-J(\xi_p, \xi_p)>0$ true?

Best Answer

Let's generalize and simplify a little.

Question (1)

Suppose $(X,Y)$ is a bivariate random variable with distribution function $J$. By definition, for any numbers $(x,y),$

$$J(x,y) = \Pr(X\le x\text{ and }Y \le y).$$

This event can also be expressed in terms of indicator functions $\mathcal I$ as

$$X\le x\text{ and }Y \le y\quad =\quad \mathcal{I}(X\le x)=1\text{ and }\mathcal{I}(Y\le y)=1.$$

The laws of arithmetic permit us to write the right hand expression in an equivalent way as $\mathcal{I}(X\le x)\mathcal{I}(Y\le y)=1.$ Consequently,

$$J(x,y) = \Pr(\mathcal{I}(X\le x)\mathcal{I}(Y\le y)=1) = E\left[\mathcal{I}(X\le x)\mathcal{I}(Y\le y)\right].$$

Suppose now that the marginal distributions of $J,$ which I will write $F_X$ and $F_Y,$ are continuous. Let $p$ and $q$ be numbers between $0$ and $1.$ For $\xi_p = \inf \{ x\mid F_X(x) \ge p\},$ this implies

$$F_X(\xi_p)=p.$$

A similar relationship holds for the $q^\text{th}$ quantile of $F_Y,$ which I will call $\eta_q.$

Notice that

$$X\le \xi_p \equiv F(X)\le p \equiv \Pr(X\le \xi_p)=p.$$

It follows immediately that

$$J(\xi_p, \eta_q) = E\left[\mathcal{I} (X\le \xi_p)\, \mathcal{I} (Y\le \eta_q)\right].$$

When $q=p$ this answers question (1) in the affirmative (and generalizes the result).

Question (2)

For the second question, notice that

$$J(\xi_p, \eta_q) = \Pr(X \le \xi_p\text{ and } Y \le \eta_q) \le \Pr(X\le \xi_p) = p.$$

The same argument applies to $Y.$ Thus

$$J(\xi_p, \eta_q) \le \min(p,q).$$

(This is known as the upper Fréchet–Hoeffding bound.)

This inequality resolves question (2) when $p=q,$ because it shows

$$p - J(\xi_p, \eta_p) \ge 0.$$

Comments

Equality can hold here. For instance, when $J$ is the uniform distribution on the union of squares $[0,1/2]\times[0,1/2]\cup [1/2,1]\times[1/2,1]$ and $p=1/2,$ both marginals are the uniform distribution on $[0,1]$ (which is continuous) yet we find $\xi_p=\eta_p=1/2$ and $J(1/2, 1/2) = 1/2 = p.$

Notice that this relationship requires continuity. For instance, let $X$ and $Y$ both have Bernoulli$(3/4)$ distributions and let both $p$ and $q$ exceed $1/4$ but be less than $1.$ Then $\xi_p=\eta_q=1$ and we have

$$J(\xi_p, \eta_q) = J(1,1) = 1 \gt \max(p,q),$$

offering a counterexample to (2).

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