Joint Distribution – Calculating Quantiles of a Joint Distribution

joint distributionmathematical-statisticsquantiles

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?

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

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

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