PDF and CDF manipulation

[S_David]

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

[statdad]

Try starting with


\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

[D H]

What statdad meant to say was: Try starting with

\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

[statdad]

What statdad meant to say was: Try starting with

\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

Yes indeed, but statdad, in his advanced age, was interrupted by some annoying folks at the door and neglected to fix his post. Thanks.

[S_David]

What statdad meant to say was: Try starting with

\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

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

[statdad]

Yes: you can write


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


You should be able to take this and write it as


1 - \int_0^\infty ( \cdots ) \, dx


just play around with the integrand.

[S_David]

Yes: you can write


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


You should be able to take this and write it as


1 - \int_0^\infty ( \cdots ) \, dx


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

[statdad]

Look at the integrand in


\int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx


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.

[S_David]

Look at the integrand in


\int_\gamma^\infty \left(f_X(x) - F_Y(x-\gamma)f_X(x)\right) \, dx


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:

[statdad]

Look a little harder. I won't give away the answer.

[S_David]

Look a little harder. I won't give away the answer.

Just give me a hint, I am not strong in probability.

[statdad]

Just give me a hint, I am not strong in probability.

Work with the integrand.

[S_David]

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?