# Thread: Difference of expectations of RV's given arbtrary pdfs

1. ## Difference of expectations of RV's given arbtrary pdfs

I've been beating my head on this for a while...

Let X and Y have pdf's f & g respectively such that . Show that .

As the expectation is a probability weighted average, it stands to reason that if there is more "weight" given to larger values (i.e. larger values are more likely to occur), then the expectation is larger. I just don't know how to show mathematically. So far, I've tried looking at the difference in the pdfs:
. I can show that h(t) integrates to zero (as the difference of the integrals of two pdf's) so , but then I don't know where to go from there.

Any thoughts are appreciated!

2. ## Re: Difference of expectations of RV's given arbtrary pdfs

There maybe a simpler and more direct method, but you may need to make use of the following formula:

Then you may obtain the following:

where are the CDF of respectively.

Next you may show that the integrand tends to zero when it goes to infinity (property of CDF) and has a unique global and local maximum at by usual calculus check with the given property. Thus showing that the integrand, and therefore the integral is non-negative which completes the proof.

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