# Probability Density Functions

Just like the mass
density function, the
probability density function is defined by a limiting process. But there
is an easier way to calculate the density function: You find the
cumulative distribution function F(x)=P(X £ x) then take the derivative
of F(x). For two dimensional case, you find F(x,y)=P(X £ x, Y £ y) and
take partial derivative w.r.t. x then w.r.t. y.
It is important to know that if f(x) is the (probability) density function of a random variable X,
then f(x) times dx (a small positive quantity) is approximately the probability that X falls in the
interval [x, x+dx]. So the integal of f(x) over an interval [a,b] is the probability that a < X
< b (This
just follows from the Riemann sum).

Similarly, for pair of continuous random variables (X,Y), you have the density function f(x,y)
defined through a limiting process. And again, it is important to know that f(x,y) times dx dy (both
dx and dy are small positive quantities) is approximately the probability that (X,Y) falls in the
rectangle [x, x+dx]×[y,y+dy]. So by definition of double integral over a set R using Riemann sum, the
integral of f(x,y) over a set R is the probability that (X,Y) falls in the set R.

This definition of probability density function generalizes to more than two random variables in
the same manner.

## Example

Suppose f(x,y)=2x 0 < x < 1, 0 < y < 1 (f(x,y)=0 elsewhere) is the probability density
function of
(X,Y).Find an approximate probability that (X,Y) falls in the rectangle [0.5, 0.6]x[0.3,0.34], i.e.
0.5 < X < 0.6 and 0.3 < Y < 0.34.
Answer: It is f(0.5, 0.3) × 0.1 × 0.04 -- 2×0.5×0.1×0.04= 0.004.

To get the exact answer you will have to integrate f(x,y) over the rectangle. Try it. You answer
should be close to the approximate answer 0.004.

Send comments and questions to tlim@gmu.edu