PDF and CDF Integration Simplification

AI Thread Summary
The discussion revolves around simplifying an integral equation involving the probability density function (PDF) and cumulative distribution function (CDF) of random variables X and Y. Participants suggest starting with the integral of the PDF and manipulating it to express it in a desired form. The goal is to rewrite the equation as one minus a single integral, specifically aiming for a format that eliminates the first term. Suggestions include factoring the integrand and experimenting with different forms to achieve the desired expression. The conversation emphasizes the importance of working through the integrand to find a suitable solution.
EngWiPy
Messages
1,361
Reaction score
61
Hello,
I have this equation:

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

where f_X(x) and F_Y(y) 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:

1-\int_{0}^{\infty}(...)

Regards
 
Physics news on Phys.org
Try starting with

<br /> \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 <br />
 
What statdad meant to say was: Try starting with

\int_{-\infty}^{\gamma} f_X(x)\,dx =<br /> \int_{-\infty}^{\infty} f_X(x)\,dx \;- \,\int_{\gamma}^{\infty}f_X(x\,)dx =<br /> 1 \,- \int_{\gamma}^{\infty}f_X(x)dx
 
D H said:
What statdad meant to say was: Try starting with

\int_{-\infty}^{\gamma} f_X(x)\,dx =<br /> \int_{-\infty}^{\infty} f_X(x)\,dx \;- \,\int_{\gamma}^{\infty}f_X(x\,)dx =<br /> 1 \,- \int_{\gamma}^{\infty}f_X(x)dx

Yes indeed, but statdad, in his advanced age, was interrupted by some annoying folks at the door and neglected to fix his post. Thanks.
 
D H said:
What statdad meant to say was: Try starting with

\int_{-\infty}^{\gamma} f_X(x)\,dx =<br /> \int_{-\infty}^{\infty} f_X(x)\,dx \;- \,\int_{\gamma}^{\infty}f_X(x\,)dx =<br /> 1 \,- \int_{\gamma}^{\infty}f_X(x)dx

Yes, but I want the whole right side be one minus single integral. Is this still doable in some how?
 
Yes: you can write

<br /> 1 - \int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx<br />

You should be able to take this and write it as

<br /> 1 - \int_0^\infty ( \cdots ) \, dx<br />

just play around with the integrand.
 
statdad said:
Yes: you can write

<br /> 1 - \int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx<br />

You should be able to take this and write it as

<br /> 1 - \int_0^\infty ( \cdots ) \, dx<br />

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:

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

Regards
 
Look at the integrand in

<br /> \int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx<br />

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.
 
statdad said:
Look at the integrand in

<br /> \int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx<br />

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
Look a little harder. I won't give away the answer.
 
  • #11
statdad said:
Look a little harder. I won't give away the answer.

Just give me a hint, I am not strong in probability.
 
  • #12
saeddawoud said:
Just give me a hint, I am not strong in probability.

Work with the integrand.
 
  • #13
statdad said:
Work with the integrand.

Are you sure that, we can write \int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx as \int_0^{\infty}F_Y(a)\,f_X(a+\gamma)\,da?
 
Back
Top