Stats/prob: finding cumulative distribution function

[Phox]

Homework Statement

given pdf:

f(x) = 2/3x for 0<=x<=1
f(x) = 2/3 for 1<x<=2
f(x) = 0 elsewhere

Find the CDF.

Homework Equations

The Attempt at a Solution

I've found:

F(x) = 0 for x<= 0
F(x) = 1 for x>=2
F(x) = (1/3)x² for 0<=x<=1

and I found:

F(x) = (2/3)x for 1<x<=2

However, the last bit is incorrect. It should be F(x) = (2/3)x -(1/3)

I'm unclear as to why. I think it has something to do with solving for the constant of integration, but I'm not sure exactly.

 

[Mute]

Yes, it has to do with the integration constant. Your CDF has to be continuous, so you need to fix that constant value so that

$$\lim_{\epsilon \rightarrow 0} F(1 - \epsilon) = \lim_{\epsilon \rightarrow 0} F(1 + \epsilon).$$

That is, the limit of the CDF on either side of x = 1 have to be the same, which wouldn't be true if you didn't fix the constant of integration to be -1/3 just above x=1.

 

[Phox]

Yes, it has to do with the integration constant. Your CDF has to be continuous, so you need to fix that constant value so that

$$\lim_{\epsilon \rightarrow 0} F(1 - \epsilon) = \lim_{\epsilon \rightarrow 0} F(1 + \epsilon).$$

That is, the limit of the CDF on either side of x = 1 have to be the same, which wouldn't be true if you didn't fix the constant of integration to be -1/3 just above x=1.

Could you clarify.. what is epsilon?

 

[Ray Vickson]

Homework Statement

given pdf:

f(x) = 2/3x for 0<=x<=1
f(x) = 2/3 for 1<x<=2
f(x) = 0 elsewhere

Find the CDF.

Homework Equations

The Attempt at a Solution

I've found:

F(x) = 0 for x<= 0
F(x) = 1 for x>=2
F(x) = (1/3)x² for 0<=x<=1

and I found:

F(x) = (2/3)x for 1<x<=2

However, the last bit is incorrect. It should be F(x) = (2/3)x -(1/3)

I'm unclear as to why. I think it has something to do with solving for the constant of integration, but I'm not sure exactly.

You have F correct for x ≤ 1 and for x ≥ 2. Since the random variable has a finite density function, its F(x) must be a continuous function, and since F'(x) = 2/3 on [1,2], F must increase linearly with slope 2/3, starting from F(1) = 1/3 and ending at F(2) = 1. You can figure out what the formula must be for F(x) in the region 1 ≤ x ≤ 2.

Basically, you need to use
$$F(x) = F(1) + \int_{1}^{x} f(t) \, dt, \: 1 \leq x \leq 2.$$
Note that in this calculation there is NO constant of integration!