# Understanding probability density function

1. Mar 21, 2013

### cdot

So I understand how for a continuous random variable the probability of an exact value of X is zero, but then what is the value of f(x) if it's not a probability? I thought it was a probability similar to how the pmf for a discrete random variable was a piece-wise function that gave the probability for various values of X. But it can't be a probability because the function f(x) DOES take on a value for every single value of x. You plug in an x and out pops an f(x). If f(x) is indeed a probability then doesn't this contradict the idea that the probability for any single value of x is zero. So what is f(x)?

2. Mar 21, 2013

### mathman

The probability for a continuous random variable (X) is given by the distribution function. Specifically P(a<X<b) = F(b)-F(a), where F(x) is the distribution function. The density function f(x) is simply the derivative, f(x) = F'(x)

Note: this assumes F is well behaved (absolutely continuous).

3. Mar 22, 2013

### Stephen Tashi

Visualize a rod of variable mass lying along the x-axis with it's left end at zero. Let F(X) be the total mass of the rod between x= 0 and x = X. The interpretation of F(X) is straightforward, but what is the meaning of f(X) = F'(X)? You have to accept the idea of an "instantaeous rate of change of mass" at the point x = X. The point x = X does not have a mass, but the mass in a small interval of length dX around it can be approximated by f(X) dX.

The probability density function f(X) of a continuous random variable has an analogous interpretation. It is the instantaneous rate of change of the cumulative probability function.

Often when you are trying to remember or derrive formulas in probability, you can cheat and think about f(X) as being "the probability that x = X" to remember the correct answer.

This way of incorrect thinking seems to work out more often in probability theory than in physics. In physics, if you have a instantaeous rate, you often have to keep the dX's in picture and your answer may have powers of the dX's and ratio's of them, some of which vanish and some of which produce the answer.

4. Mar 22, 2013

### cdot

Thank you Stephen.