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Homework Help: 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/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


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    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


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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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