# I A conditional probability

1. Apr 25, 2016

### rabbed

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))

2. Apr 25, 2016

### mathman

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

3. Apr 26, 2016

### rabbed

As a step in using the CDF method for a random variable as a function of X where i have X_PDF, I came from:

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. Apr 26, 2016

### rabbed

By the way..

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?