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.


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.

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