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Let $X$ and $Y$ be non-negative random variables with an arbitrary joint probability distribution function. Let $I(x, y) = 1$ if $X > x, Y > y$ and $0$ otherwise.

Show that $\int_0^\infty \int_0^\infty I(x, y) dxdy = XY$.

I'm stuck, but if I assumed what is to be proven, then can it be concluded that both $X$ and $Y$ are constant? Is part of the problem to first show that $X$ and $Y$ are constant?

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    $\begingroup$ The notation might be deceiving you. $I(x,y)$ is a random variable because it is a function of $(X,Y).$ If we were to make that explicit, by writing (say) $I(X,Y;x,y),$ then the left hand side would be $$\int_0^\infty\int_0^\infty I(X,Y;x,y)\,\mathrm{d}x\mathrm{d}y = \int_0^X\int_0^Y\mathrm{d}x\mathrm{d}y.$$ This equality doesn't reflect anything of substance: it's purely a change of notation, because finite integrals are defined in terms of the indicator function: $\int_a^b f(x)\,\mathrm{d}x = \int_{-\infty}^\infty I(a\le x\le b)f(x)\,\mathrm{d}x.$ Can you proceed from here? $\endgroup$
    – whuber
    Aug 6 at 13:09
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    $\begingroup$ @whuber Is there a $f(x)$ missing in that very last integral in your comment? $\endgroup$ Aug 6 at 13:39
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    $\begingroup$ @Dilip Yes, that's how I read it. I didn't pay attention to the inequalities because it was clear the intention was for the integration to be over a compact region. $\endgroup$
    – whuber
    Aug 6 at 17:52
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    $\begingroup$ This identity is useful in establishing the famous identity that $$\mathbb E[X]=\int_0^\infty\text{Pr}(X>x)\,\text dx$$since$$\int_0^\infty x\,\text dP(x)=\int_0^\infty\int_0^\infty \mathbb I_{x>z}\,\text{dz}\,\text dP(x)=\int_0^\infty\int_0^\infty \mathbb I_{x>z}\,\text dP(x)\,\text{dz}$$by Fubini. $\endgroup$
    – Xi'an
    Aug 7 at 10:59
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    $\begingroup$ @whuber: I finally appreciate the use of $X$, $Y$ as upper limits of integration. Thanks - I think I can proceed. $\endgroup$
    – johnsmith
    Aug 9 at 22:39

1 Answer 1

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Because notation is the crux of the matter, let's be rigorous and explicit about it.

To say that $(X,Y)$ has a joint distribution means $(X,Y)$ is a (measurable) function from a measure space $(\Omega, \mathfrak F)$ to $\mathbb R^2.$ The definition of the integrand $I$ involves four quantities: $x,$ $y,$ $X,$ and $Y,$ so let's make this explicit by writing the integral in the question as

$$\int_0^\infty\int_0^\infty I(x,y)\,\mathrm d x \mathrm d y = \int_0^\infty\int_0^\infty I(x,y,X,Y)\,\mathrm d x \mathrm d y.$$

What does this mean? The integrand $I$ is a (measurable) function of four real variables $z_1,z_2,z_3,z_4$ defined by the (usual) indicator function $\mathscr I$

$$I(z_1,z_2,z_3,z_4) = \mathscr{I}(z_3\gt z_1,z_4\gt z_2) = \left\{\begin{aligned} 1 & \text{ if } z_3 \gt z_1\text{ and } z_4 \gt z_2 \\ 0 & \text{ otherwise.}\end{aligned}\right.$$

Now, integration of a (measurable) function $g:\mathbb R^n\to\mathbb R$ over any (measurable) region $\mathscr R \subset \mathbb R^n$ is defined in terms of an integral over the entire space as

$$\iint_{\mathscr R} g(\mathbf x)\,\mathbf{\mathrm{d}}\mathbf x = \iint_{\mathbb R^n} \mathscr{I}(\mathbf x \in \mathscr R) g(\mathbf x)\,\mathbf{\mathrm{d}}\mathbf x.$$

Let's pause to note a basic property of the indicator function: multiplication corresponds to intersection in the sense that for any two sets $\mathscr R$ and $\mathscr S,$

$$\mathscr{I}(x \in \mathscr R)\mathscr{I}(x \in \mathscr S) = \mathscr{I}(x\in \mathscr R \cap \mathscr S).$$

Use this to define a function $f:\mathbb R^2\to \mathbb R$ as follows:

$$\begin{aligned} f(u,v) &= \int_0^\infty\int_0^\infty I(x,y, u,v)\,\mathrm d x \mathrm d y\\ &=\iint_{\mathbf R^2}\mathscr{I}(0 \le x, 0 \le y)\mathscr{I}(x \lt u, y \lt v) \,\mathrm d x \mathrm d y \\ &=\iint_{\mathbf R^2}\mathscr{I}(0 \le x \lt u, 0 \le y \lt v) \,\mathrm d x \mathrm d y \\ &= \int_0^{u}\int_0^{v}\mathrm d x \mathrm d y = uv \end{aligned}$$

(assuming $u$ and $v$ are both non-negative; the result is $0$ otherwise).

That's the area of a rectangle of width $v$ and height $u,$ equal to $uv.$ (Don't read too much into these calculations. This is all elementary set theory and geometry at this point, having nothing to do with random variables or even with any theory of integration. The only conception of integration you need is that the integral of the constant $c$ over a rectangle is $c$ times the area of the rectangle: this is the starting point of all theories of integration.)

Composing $f$ with $(X,Y)$ when both $X$ and $Y$ always have non-negative values gives the (necessarily measurable) function

$$f\circ (X,Y):(\Omega,\mathfrak F) \to \mathbb R$$

where, by definition, the value of the composition at any $\omega\in\Omega$ is

$$(f\circ (X,Y))(\omega) = f(X(\omega), Y(\omega)) = X(\omega)Y(\omega).$$

That's what "$XY$" means, QED.

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