PDF and CDF manipulation

  • Thread starter EngWiPy
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  • #1
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Main Question or Discussion Point

Hello,
I have this equation:

[tex]\int_{-\infty}^{\gamma}f_X(x)\,dx+\int_{\gamma}^{\infty}F_Y(x-\gamma)\,f_X(x)\,dx[/tex]

where [tex]f_X(x)[/tex] and [tex]F_Y(y)[/tex] are the PDF and CDF of the randome variables X and Y, respectively.

Now the question is: can I write the above equation in the form:

[tex]1-\int_{0}^{\infty}(...)[/tex]

Regards
 

Answers and Replies

  • #2
statdad
Homework Helper
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Try starting with

[tex]
\int_{-\infty}^\gamma f_X(x) \, dx = \int_{-\infy}^\infty f_X (x) \, dx - \infty_\gamma^\infty f_X(x) \, dx = 1 - \infty_\gamma^\infty f_X(x) \, dx
[/tex]
 
  • #3
D H
Staff Emeritus
Science Advisor
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What statdad meant to say was: Try starting with

[tex]\int_{-\infty}^{\gamma} f_X(x)\,dx =
\int_{-\infty}^{\infty} f_X(x)\,dx \;- \,\int_{\gamma}^{\infty}f_X(x\,)dx =
1 \,- \int_{\gamma}^{\infty}f_X(x)dx[/tex]
 
  • #4
statdad
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What statdad meant to say was: Try starting with

[tex]\int_{-\infty}^{\gamma} f_X(x)\,dx =
\int_{-\infty}^{\infty} f_X(x)\,dx \;- \,\int_{\gamma}^{\infty}f_X(x\,)dx =
1 \,- \int_{\gamma}^{\infty}f_X(x)dx[/tex]
Yes indeed, but statdad, in his advanced age, was interrupted by some annoying folks at the door and neglected to fix his post. Thanks.
 
  • #5
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61
What statdad meant to say was: Try starting with

[tex]\int_{-\infty}^{\gamma} f_X(x)\,dx =
\int_{-\infty}^{\infty} f_X(x)\,dx \;- \,\int_{\gamma}^{\infty}f_X(x\,)dx =
1 \,- \int_{\gamma}^{\infty}f_X(x)dx[/tex]
Yes, but I want the whole right side be one minus single integral. Is this still doable in some how?
 
  • #6
statdad
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Yes: you can write

[tex]
1 - \int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx
[/tex]

You should be able to take this and write it as

[tex]
1 - \int_0^\infty ( \cdots ) \, dx
[/tex]

just play around with the integrand.
 
  • #7
1,367
61
Yes: you can write

[tex]
1 - \int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx
[/tex]

You should be able to take this and write it as

[tex]
1 - \int_0^\infty ( \cdots ) \, dx
[/tex]

just play around with the integrand.
Thank you, but this form is not the one in my mind. I need, if possible, in some how, to eliminate the first term, so that the equation looks like:

[tex]1-\int_0^{\infty}F_Y(a)\,f_X(a+\gamma)\,da[/tex]​

Regards
 
  • #8
statdad
Homework Helper
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Look at the integrand in

[tex]
\int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx
[/tex]

You should see a very simple way to factor it and then rewrite it in a form more suitable to your desires for this problem.

Try it - do some work - then post again.
 
  • #9
1,367
61
Look at the integrand in

[tex]
\int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx
[/tex]

You should see a very simple way to factor it and then rewrite it in a form more suitable to your desires for this problem.

Try it - do some work - then post again.
I can't see anything that I can do. :shy:
 
  • #10
statdad
Homework Helper
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Look a little harder. I won't give away the answer.
 
  • #11
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Look a little harder. I won't give away the answer.
Just give me a hint, I am not strong in probability.
 
  • #12
statdad
Homework Helper
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Just give me a hint, I am not strong in probability.
Work with the integrand.
 
  • #13
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Work with the integrand.
Are you sure that, we can write [tex]\int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx[/tex] as [tex]\int_0^{\infty}F_Y(a)\,f_X(a+\gamma)\,da[/tex]?
 

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