Calculating Randomly Truncated PDF for X given T1 < X < T2

[benjaminmar8]

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.

 

[John Creighto]

f(x|t_1&lt;x&lt;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

where rect(x,t_1,t_2) is defined to be 1 if t_1&lt;x&lt;t_2 and 0 otherwise.

 

[benjaminmar8]

f(x|t_1&lt;x&lt;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

where rect(x,t_1,t_2) is defined to be 1 if t_1&lt;x&lt;t_2 and 0 otherwise.

But the question is, how do I know when t1<X<t2 since t1 and t2 are random?

 

[John Creighto]

But the question is, how do I know when t1<X<t2 since t1 and t2 are random?

You don't. You consider all possibles for t1, and t2 and the probability of each possibility.

 

[benjaminmar8]

f(x|t_1&lt;x&lt;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

where rect(x,t_1,t_2) is defined to be 1 if t_1&lt;x&lt;t_2 and 0 otherwise.

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.

 

[John Creighto]

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.

I'm sorry. What I gave you wasn't really f(x|t1<x<t2). To get the conventional probability, simply divide 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.

 

[benjaminmar8]

I'm sorry. What I gave you wasn't really f(x|t1<x<t2). To get the conventional probability, simply divide 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.

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.

 

[bpet]

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.