# CDF of a variable with a negative exponent in its PDF

1. May 3, 2012

### jimbobian

Ok, this one's got me stumped!

Let's take as an example the probability density function for a random variable X so that:

f(x) = $\frac{4}{3x^{3}}$ 1≤x<2
f(x) = $\frac{x}{12}$ 2≤x≤4
f(x) = 0

So the CDF for this variable comes out as:

F(x) = $\frac{-2}{3x^{2}}$ 1≤x<2
F(x) = $\frac{x^{2}}{24}$ 2≤x≤4

So how can the CDF be negative for 1≤x<2, the CDF is P(X≤x), so to my mind that makes no sense. And secondly I have never seen a CDF with a discontinuity like in this one. At x=2, it jumps from -1/6 to 1/6

This made me think that I should ignore the negative sign in the CDF for 1≤x<2, but then for 1≤x<2 F(x) is a decreasing function, how can that make sense?

2. May 3, 2012

### jimbobian

Ok I've figured out why the minus sign shouldn't have been there, I forgot that what you actually do to find the CDF is:

$\int_{-∞}^{x} p(t)dt$

Which sorts out the minus sign, other problem still remains?

EDIT: I'm talking rubbish this hasn't fixed a thing!

EDIT2: Turns out it fixed both problems. CDF is now a lovely increasing function that starts at 0 and ends at 1. Turns out I'd been doing an incorrect short cut on CDFs and never noticed because it has never not worked until tonight!

Last edited: May 3, 2012