# PDF and CDF manipulation

## Main Question or Discussion Point

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

## Answers and Replies

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Homework Helper
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
Staff Emeritus
What statdad meant to say was: Try starting with

$$\int_{-\infty}^{\gamma} f_X(x)\,dx = \int_{-\infty}^{\infty} f_X(x)\,dx \;- \,\int_{\gamma}^{\infty}f_X(x\,)dx = 1 \,- \int_{\gamma}^{\infty}f_X(x)dx$$

Homework Helper
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.

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?

Homework Helper
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.

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

Homework Helper
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.

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:

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

Look a little harder. I won't give away the answer.
Just give me a hint, I am not strong in probability.

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