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B How to evaluate this CDF?

  1. Aug 23, 2016 #1
    Hello all,

    I have the following random variable ##X=\frac{a_1}{a_2+1}##, where ##a_i=b_i/c_i##, where ##b_i## and ##c_i## are exponential random variables with mean 1. I need to evaluate the CDF of ##X## as

    [tex]F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(a_2+1)\right]f_{a_2}(a_2)\,da_2[/tex]

    I found the CDF and PDF of ##a_i## as ##F_{a_i}(x)=1-\frac{1}{1+x}## and ##f_{a_i}(x)=\frac{1}{(1+x)^2}##. My fist question is: are the limits of the integral above correct?

    Thanks
     
  2. jcsd
  3. Aug 23, 2016 #2

    micromass

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    You can't solve this unless you know the joint distributions of the variables.
     
  4. Aug 23, 2016 #3
    All random variables are independent. I forgot to mention this.
     
  5. Aug 23, 2016 #4

    micromass

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    OK. I agree with the pdf and cfd of the ##a_i## then. But ##a_2## is the name of a function. So you can't use that as integration variable.
     
  6. Aug 23, 2016 #5
    Is this OK?

    $$
    F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(\beta+1)\right]f_{a_2}(\beta)\,d\beta
    $$
     
  7. Aug 23, 2016 #6

    micromass

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    I don't see how that follows. I would agree with

    $$
    F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(a_2+1)\right]f_{a_2}(\beta)\,d\beta
    $$

    But I don't see why the ##a_2## can be a ##\beta##.
     
  8. Aug 23, 2016 #7
    OK, thanks. So, I guess this means that the limits are correct. I would like to proceed to evaluate the integral, because I did on my papers, and got a final expression, but evaluating it using software gave me values that aren't correct.
     
  9. Aug 23, 2016 #8

    micromass

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    The integral you posted is not correct. I don't see where it comes from at all.
     
  10. Aug 23, 2016 #9
    $$Pr\left[a_1\leq x(a_2+1)\right]=\int_{a_2}Pr\left[\left. a_1\leq x(\beta+1)\right| a_2=\beta\right]f_{a_2}(\beta)\,d\beta$$

    To my undertsnading, since ##a_2## is a random variable too, we average over all the values of ##a_2##. For the sake of argument, how would you find the CDF above?
     
    Last edited: Aug 23, 2016
  11. Aug 23, 2016 #10

    micromass

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    I would solve it by evaluating the following integral:
    [tex]\mathbb{P}\{a_1\leq x(a_2 + 1)\} = \int_0^{+\infty} \int_0^{x(\beta_2 + 1)} f_{a_1}(\beta_1)f_{a_2}(\beta_2) d\beta_1 d\beta_2[/tex]
     
  12. Aug 23, 2016 #11
    I think our disagreement is only about using symbols. Your evaluation can be re-written as
    $$
    \mathbb{P}\{a_1\leq x(a_2 + 1)\} = \int_0^{+\infty} \underbrace{\left[\int_0^{x(\beta_2 + 1)} f_{a_1}(\beta_1)\,d\beta_1\right]}_{F_{a_1}\left(x[\beta_2+1]\right)=Pr\left[a_1\leq x(\beta_2+1)\right]} f_{a_2}(\beta_2) d\beta_2
    $$

    which is effectively the same as I did in the the previous posts.

    Are we on the same page now?
     
  13. Aug 24, 2016 #12
    Substituting the CDF of ##a_1## and the PDF of ##a_2## in the integral yields

    $$F_X(x)=\int_0^{\infty}\left[1-\frac{1}{1+x(\beta+1)}\right]\frac{1}{(1+\beta)^2}\,d\beta=1-\frac{1}{x}\int_0^{\infty}\frac{1}{\beta+\left(1+\frac{1}{x}\right)}\frac{1}{(1+\beta)^2}\,d\beta$$

    Then I used the integral formula attached from the table of integrals. Is there anything wrong in my work?
     

    Attached Files:

  14. Aug 29, 2016 #13
    Using the integration formula attached yields to:

    [tex]F_X(x)=1-\frac{1}{x}B(1,2)_2F_1(2,1;3;1-\left(1+\frac{1}{x}\right))[/tex]

    where ##B(.,.)## is the Beta function, and ##_2F_1## is the Gauss Hypergeometric function. The constants in the integral attached are: ##\nu=1##, ##\beta=1## which implies that ##\mu=2##, and ##\gamma=1+\frac{1}{x}## which implies ##\rho=1##. I checked the conditions, and they are satisfied. However, the result of ##F_X(x)## must be in ##[0,\,1]## which isn't the case when I evaluate it in Mathematica. Why? I appreciate any help.
     
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