- #1
jimbobian
- 52
- 0
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) = [itex]\frac{4}{3x^{3}}[/itex] 1≤x<2
f(x) = [itex]\frac{x}{12}[/itex] 2≤x≤4
f(x) = 0
So the CDF for this variable comes out as:
F(x) = [itex]\frac{-2}{3x^{2}}[/itex] 1≤x<2
F(x) = [itex]\frac{x^{2}}{24}[/itex] 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?
Someone enlighten me... please!
Let's take as an example the probability density function for a random variable X so that:
f(x) = [itex]\frac{4}{3x^{3}}[/itex] 1≤x<2
f(x) = [itex]\frac{x}{12}[/itex] 2≤x≤4
f(x) = 0
So the CDF for this variable comes out as:
F(x) = [itex]\frac{-2}{3x^{2}}[/itex] 1≤x<2
F(x) = [itex]\frac{x^{2}}{24}[/itex] 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?
Someone enlighten me... please!