Understanding probability density function

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Discussion Overview

The discussion revolves around the concept of the probability density function (pdf) for continuous random variables, particularly addressing the nature of f(x) and its interpretation in relation to probabilities. Participants explore theoretical implications and analogies to better understand the function's role in probability theory.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant questions the nature of f(x), noting that while it produces a value for every x, it cannot represent a probability since the probability of an exact value for a continuous random variable is zero.
  • Another participant explains that the probability for a continuous random variable is given by the distribution function, with f(x) being its derivative, assuming the distribution function is well-behaved.
  • A different perspective is introduced using an analogy of a rod of variable mass, where f(X) is interpreted as the instantaneous rate of change of mass, suggesting that while a specific point does not have mass, the mass in an interval can be approximated by f(X) dX.
  • One participant mentions that thinking of f(X) as "the probability that x = X" can be a useful mnemonic, despite being an incorrect interpretation.

Areas of Agreement / Disagreement

Participants express differing interpretations of f(x) and its implications, indicating that multiple competing views remain regarding its meaning and role in probability theory. The discussion does not reach a consensus.

Contextual Notes

There are unresolved assumptions regarding the behavior of the distribution function and the implications of interpreting f(x) in various ways. The discussion highlights the complexity of relating continuous probability functions to discrete probability concepts.

cdot
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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)?
 
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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).
 
cdot said:
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

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.
 
Thank you Stephen.
 

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