- #1

- 234

- 2

## Main Question or Discussion Point

Is it possible to solve something like this generally or does it depend on the pdf's of the variables?

P(x < f(y) | x > -f(y))

P(x < f(y) | x > -f(y))

- I
- Thread starter rabbed
- Start date

- #1

- 234

- 2

Is it possible to solve something like this generally or does it depend on the pdf's of the variables?

P(x < f(y) | x > -f(y))

P(x < f(y) | x > -f(y))

- #2

mathman

Science Advisor

- 7,766

- 417

Your expression can be given as [itex]\frac{P(-f(y)<x<f(y))}{P(-f(y)<x<\infty)}[/itex].

- #3

- 234

- 2

P(x > -f(y) AND x < f(y)) =

P(x > -f(y)) * P(x < f(y) | x > -f(y))

The aim is to convert the P()'s to X_CDF()'s.

Your answer led me back a step, which made me think that maybe P(x > -f(y) AND x < f(y)) might be expressed like (1-X_CDF(-f(y))) - X_CDF(f(y))

It seems to be correct for my case, so thank you :)

- #4

- 234

- 2

In the CDF method, I understand that I need to reformulate expressions to get something like P(X < y) which equals X_CDF(y) or P(X > y) which equals (1-X_CDF(y)), since I know the expression of X_PDF(x) = X_CDF'(x).

What if I have P(A + B < y), knowing A_PDF(a) and B_PDF(b)?

Would that require that I know AplusB_PDF(a,b) and some transformation from y to a and y to b?

- Replies
- 4

- Views
- 352

- Last Post

- Replies
- 2

- Views
- 663

- Last Post

- Replies
- 3

- Views
- 142

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 980

- Replies
- 3

- Views
- 516