# Definition of conditional probability density

1. Oct 8, 2011

### nonequilibrium

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.

2. Oct 8, 2011

### mathman

Your definition is incorrect. Let A and B be events, then P(A|B) = P(A and B)/P(B).

3. Oct 8, 2011

### nonequilibrium

Oh chucks that was stupid of me :) thank you...

4. Oct 8, 2011

### nonequilibrium

While we're at it, could you suggest me how to prove the following? (NOT a homework question)

$f(X_1 = x_1, \dots, X_n = x_n, X_{(n)} = a) =\left\{ \begin{matrix} f(X_1 = x_1, \dots, X_n = x_n) & \quad \text{if max(x_1, \dots, x_n)=a}\\ 0 & \quad \text{otherwise}\\ \end{matrix} \right.$

5. Oct 8, 2011

### Useful nucleus

Could you explain the notation X(n)?

6. Oct 8, 2011

### nonequilibrium

My apologies, I should have:

it is the n-th order statistic,
i.e. $X_{(n)} := \textrm{max}(X_1,\dots,X_n)$