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Randomly truncated pdf

  1. Jul 19, 2009 #1
    Hi, all,

    I am having a problem in calculating a randomly truncated pdf. Let x be a random variable, it's pdf f(x) is known. Let t1 and t2 be anther two random variables, their pdf f(t1) and f(t2) are known as well. Now, how do I compute the pdf f(x|t1<x<t2)?

    Thks a lot.
     
  2. jcsd
  3. Jul 19, 2009 #2
    [tex]f(x|t_1<x<t_2)=\int_{-\infty}^{-\infty}\int_{-\infty}^{-\infty}f(x)rect(x,t_1,t_2)f(t_1)f(t_2)dt_1dt_2[/tex]

    where [tex]rect(x,t_1,t_2)[/tex] is defined to be [tex]1[/tex] if [tex]t_1<x<t_2[/tex] and [tex]0[/tex] otherwise.
     
  4. Jul 19, 2009 #3
    But the question is, how do I know when t1<X<t2 since t1 and t2 are random?
     
  5. Jul 19, 2009 #4
    You don't. You consider all possibles for t1, and t2 and the probability of each possibility.
     
  6. Jul 20, 2009 #5
    I did a couple of simulations and found that the pdf f(x|t1<x<t2) seems need to be scaled. Maybe I have miss out some conditions, say the support of x, t1 and t2 are all [0,R]. In this case, how do I compute the truncated pdf? Thanks a lot.
     
  7. Jul 20, 2009 #6
    I'm sorry. What I gave you wasn't really f(x|t1<x<t2). To get the conventional probability, simply devide f(x) by the integral of f(x) from t1 to t2. However, the contional probability is not the same thing as a randomly truncated PDF. What I gave you is the distribution of f(x) given some random truncation. I'm not sure which you want because I don't know much about the problem you are trying to solve.
     
  8. Jul 22, 2009 #7
    what I am trying to solve is the desnity function of f(x|t1<x<t2), therefore, its intergral over the support should be 1. What you gave me seems should be devided by 1/(F(t2)-F(t1)) (and you mentioned that), however, since t2 and t1 are random, I use its expectation instead. That's to say, the scaling is 1/(F(E[t2])-F(E[t1])). I know this is an approximation, how do I compute it in an exact manner? Thank u very much.
     
  9. Aug 4, 2009 #8
    I'd start with the CDF and differentiate.

    F[x|t1<x<t2] = P[t1<X<t2 & X<=x] / P[t1<X<t2]

    both those probabilities can be written as integrals of functions of the pdf's.
     
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