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

  1. Aug 17, 2008 #1
    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 text book, my lecture notes and even wikipedia dont refrer to the 't', apart from in the definition. I wouldnt 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.

    1. The problem statement, all variables and given/known data
    'X is a continuous random quantity with probability density function f(x) = x for 0<x<1, f(x)=2-x for 1 [tex]\leq[/tex] 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 [tex]\leq[/tex] t [tex]\leq[/tex] 2 )'


    2. Relevant 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?)

    3. 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[tex]^{2}[/tex]/2 and for 1<x<2 F(x)= 2x - x[tex]^{2}[/tex]/2 , but its not...
     
  2. jcsd
  3. Aug 17, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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:
    [tex]\sum_{n=1}^5 3n- 1= (3(1)-1)+ (3(2)-1)+ (3(3)-1)+ (3(4)-1)+ (3(5)-1)[/tex]
    [tex]= 2+ 5+ 8+ 11+ 14= 40[/tex]
    the "n" is a "dummy index" which exists only in the sum.
     
  4. Aug 18, 2008 #3
    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.
     
  5. Aug 18, 2008 #4

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

    and for 0<t<1 F(t)= t[tex]^{2}[/tex]/2 and for 1[tex]\leq[/tex]t<2 F(x)= 2t - t[tex]^{2}[/tex]/2 , but im not being asked for that, im being asked for 0[tex]\leq[/tex]t[tex]\leq[/tex]2, what is the answer to this?
     
  6. Aug 18, 2008 #5

    statdad

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    Homework Helper

    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: [tex] 1 \le t < 2 [/tex]. Here is where you miss something. The correct value for [tex] F(t) [/tex] is found by integrating the density from 0 to t. Since the density is in two pieces, and [tex] t [/tex] 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.
     
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