# Thread: Need help with this distribution

1. ## Need help with this distribution

Below is the distribution function:

$F(x)=\frac{1}{4}1_{[0,\infty)}(x)+\frac{1}{2}1_{[1,\infty)}(x)+\frac{1}{4}1_{[2,\infty)(x)$

What does $1_{[1,\infty)}(x)...$etc mean? Are they like some sort of indicator function? For example in this case, when x= 1/2, it's 0 and when x=3/2, it's 1.

2. P(X=0)=.25

P(X=1)=.5

P(X=2)=.25

You can see the jumps at those point, which are the probabilities.
That is a sum of indicator functions that achieves one after 2.
Both of you comments are wrong.
At 1/2, only the first indicator is 1, hence F(.5)=1/4
At 3/2, the first two indicators are 1, hence F(1.5)=1/4+1/2=3/4.
Finally at 2 or afterwards F(x)=1/4+1/2+1/4=1

3. Originally Posted by matheagle
P(X=0)=.25

P(X=1)=.5

P(X=2)=.25

You can see the jumps at those point, which are the probabilities.
That is a sum of indicator functions that achieves one after 2.
Both of you comments are wrong.
At 1/2, only the first indicator is 1, hence F(.5)=1/4
At 3/2, the first two indicators are 1, hence F(1.5)=1/4+1/2=3/4.
Finally at 2 or afterwards F(x)=1/4+1/2+1/4=1
Just Want to make sure I know what u mean.

Suppose $A=(-\frac{1}{2},\frac{1}{2})$ The probability is
$F(\frac{1}{2})-F(-\frac{1}{2})=\frac{1}{4}$. Is that true?

4. $1_{[0,\infty)}(x)$ is ONE if x is greater than or equal to 0.