How to write $\Pr(z<x)$ in the form of integral.

Question

If I have two random variables $z$ and $x$ with P.D.F. of $f_z(z)$ and $f_x(x)$, respectively. It is obvious that $\Pr(z \geq x)$ expression is evaluated as follows, $$\Pr(z \geq x) = \int_z \int_x f_z(z) f_x(x) \,\Bbb dz \,\Bbb dx.$$

Right?

But, how can I write $\Pr(z<x)$ in the form of a double integral? Like, $$\Pr(z < x) = \ldots\ ?$$

Answer 1

$$\Pr(Z<X)=\int_{-\infty}^{\infty} \int_z^{\infty} f_{Z,X} (z,x) \,\Bbb dx \,\Bbb dz$$ or $$\Pr(Z<X)=\int_{-\infty}^{\infty}\int_{-\infty}^{x} f_{Z,X} (z,x) \,\Bbb dz \,\Bbb dx.$$ You can write $f_{Z,X} (z,x)$ as $f_Z(z)f_X(x)$ only when you know that $Z$ and $X$ are independent.

Answer 2

Let $[z<x]$ denote the function $\mathbb R^2\to\mathbb R$ that takes value $1$ if $z<x$ and take value $0$ otherwise.

Then:

$$P\left(Z<X\right)=\mathbb{E}\left[Z<X\right]=\int\int\left[z<x\right]f_{Z,X}\left(z,x\right)dzdx=$$$$\int_{-\infty}^{\infty}\int_{-\infty}^{x}f_{Z,X}\left(z,x\right)dzdx\tag1$$

and with different order of integration:

$$P\left(Z<X\right)=\mathbb{E}\left[Z<X\right]=\int\int\left[z<x\right]f_{Z,X}\left(z,x\right)dxdz=$$$$\int_{-\infty}^{\infty}\int_{z}^{\infty}f_{Z,X}\left(z,x\right)dxdz\tag2$$

If moreover $Z$ and $X$ are independent and with PDF's $f_{Z}$ and $f_{X}$ respectively then: $$f_{Z,X}\left(z,x\right)=f_{Z}\left(z\right)f_{X}\left(x\right)$$ leading to:

$$\cdots\stackrel{\left(1\right)}{=}\int_{-\infty}^{\infty}f_{X}\left(x\right)\int_{-\infty}^{x}f_{Z}\left(z\right)dzdx=\int_{-\infty}^{\infty}f_{X}\left(x\right)P\left(Z<x\right)dx$$

and:

$$\cdots\stackrel{\left(2\right)}{=}\int_{-\infty}^{\infty}f_{Z}\left(z\right)\int_{z}^{\infty}f_{X}\left(x\right)dxdz=\int_{-\infty}^{\infty}f_{Z}\left(z\right)P\left(X>z\right)dz$$