Cumulative distribution function (what is this 't'?)

[Mattofix]

my question is regarding 'continuous' cumulative distribution functions.

i kind of get it apart from that darn 't' in the definition (see http://upload.wikimedia.org/math/f/2/4/f24252ffb5e5e747b246189b7e1cfcce.png). My textbook, my lecture notes and even wikipedia don't refrer to the 't', apart from in the definition. I wouldn't mind apart from that i am working my way through some questions and have come across quite a few
asking me for the 'c.d.f of X for all t' (for example t<0 and t>2), not asking for the c.d.f of x values (like all of the worked examples i have come across) so what are the questions after?

Here is an example so you understand what my problem is.

Homework Statement

'X is a continuous random quantity with probability density function f(x) = x for 0<x<1, f(x)=2-x for 1 \leq x < 2 with f(x)=0
for all other x. Find the value of Fx(t), the cumulative distribution function of X, for all t (that is t<0 and t>2 as well as 0 \leq t \leq 2 )'

Homework Equations

http://upload.wikimedia.org/math/f/2/4/f24252ffb5e5e747b246189b7e1cfcce.png
http://upload.wikimedia.org/math/f/d/e/fdec25ee8674e78b0bad557daa923a41.png (maybe for when i find the new boundaries they are asking for?)

The Attempt at a Solution

If it was like all of the examples iv seen id say for x<0 F(x)=0, for x>2 F(x)=1, for 0<x<1 F(x)= x^{2}/2 and for 1<x<2 F(x)= 2x - x^{2}/2 , but its not...

 

[HallsofIvy]

If that definition makes any sense to you at all, you must know how to integrate. And if that is true, then you should understand that the t is a "dummy variable"- it does not exist and has no meaning "outside" the integral.

Perhaps it would be easier to understand by looking at the analogous situation in a sum:
\sum_{n=1}^5 3n- 1= (3(1)-1)+ (3(2)-1)+ (3(3)-1)+ (3(4)-1)+ (3(5)-1)
= 2+ 5+ 8+ 11+ 14= 40
the "n" is a "dummy index" which exists only in the sum.

 

[Mattofix]

ok, but what about the example? i have integrated.

if you can help me get the correct answer for this then hopefully i will be able to make the links and get to grips with it.

 

[Mattofix]

The Attempt at a Solution

for x<0 F(x)=0, for x>2 F(x)=1, for 0<x<1 F(x)= x^{2}/2 and for 1<x<2 F(x)= 2x - x^{2}/2 , but its not...

ok so for t<0 F(t)=0, for t>2 F(t)=1,

and for 0<t<1 F(t)= t^{2}/2 and for 1\leqt<2 F(x)= 2t - t^{2}/2 , but I am not being asked for that, I am being asked for 0\leqt\leq2, what is the answer to this?

 

[statdad]

You are close, but still miss one point:

Since the density is in two "pieces", you need to calculate the distribution function F(t) for two cases.

case 1: 0 < t < 1. Integrate the density from 0 to t to obtain F(t) for this case.

case 2: 1 \le t &lt; 2. Here is where you miss something. The correct value for F(t) is found by integrating the density from 0 to t. Since the density is in two pieces, and t is in the second interval, the integral here is broken into two pieces.
Write it out - it is MUCH easier to see in symbols than in my cryptic explanation.