Definition of conditional probability density

Click For Summary
SUMMARY

The discussion centers on the definition of conditional probability density, specifically the expression f(X = x | Y = y) = f(X=x)/f(Y=y). The confusion arises when considering the scenario where the random variable Y is the same as X, leading to the conclusion that f(X = a | X = b) should equal zero if a is not equal to b. The correct interpretation is clarified with the definition P(A|B) = P(A and B)/P(B). Additionally, the notation X_{(n)} is identified as the n-th order statistic, representing the maximum of a set of random variables.

PREREQUISITES
  • Understanding of conditional probability and its definitions
  • Familiarity with random variables and their distributions
  • Knowledge of order statistics in probability theory
  • Basic proficiency in mathematical notation and expressions
NEXT STEPS
  • Study the derivation of conditional probability density functions
  • Learn about order statistics and their applications in probability
  • Explore the properties of joint and marginal distributions
  • Investigate the implications of independence in probability theory
USEFUL FOR

Students and professionals in statistics, data science, and mathematics who seek to deepen their understanding of conditional probability and its applications in statistical analysis.

nonequilibrium
Messages
1,412
Reaction score
2
Hello, I'm somewhat confused by the expression f(X = x | Y = y) = \frac{f(X=x)}{f(Y=y)} (which, if I'm right, is the definition of a conditional probability density? My course seems to state it as a theorem, without proof, but then again my course is a little bit vague; although I welcome replies on this part, this is not the essential of this topic)

Anyway, the confusion is the following: let the s.v. Y be the s.v. X, then of course f(X = a | X=b) should be zero if a is not equal to b (if the expression means what it is meant to mean), however it is equal to \frac{f(X = a)}{f(X=b)} and there's no real reason why this should be zero.
 
Physics news on Phys.org
Your definition is incorrect. Let A and B be events, then P(A|B) = P(A and B)/P(B).
 
Oh chucks that was stupid of me :) thank you...
 
While we're at it, could you suggest me how to prove the following? (NOT a homework question)

f(X_1 = x_1, \dots, X_n = x_n, X_{(n)} = a) =\left\{<br /> \begin{matrix}<br /> f(X_1 = x_1, \dots, X_n = x_n) &amp; \quad \text{if max($x_1, \dots, x_n$)$=a$}\\<br /> 0 &amp; \quad \text{otherwise}\\<br /> \end{matrix} \right.
 
Could you explain the notation X(n)?
 
My apologies, I should have:

it is the n-th order statistic,
i.e. X_{(n)} := \textrm{max}(X_1,\dots,X_n)
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 12 ·
Replies
12
Views
5K