Let X and Y have pdf's f & g respectively such that [math]

\begin{cases}

f(x) >= g(x) & \mbox{if } x \leq a \\

f(x) <= g(x) & \mbox{if } x > a\\

\end{cases} [/math]. Show that [math] E[X] \leq E[Y][/math].

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:

[math] E[Y]-E[X] = \int_{-\infty}^{\infty} t g(t)dt - \int_{-\infty}^{\infty} t f(t)dt = \int_{-\infty}^{\infty} t \cdot [g(t)-f(t)]dt = \int_{-\infty}^{\infty} t \cdot h(t)dt [/math]. I can show that h(t) integrates to zero (as the difference of the integrals of two pdf's) so [math]\int_{-\infty}^a h(t)dt=-\int_a^{\infty} h(t)dt[/math], but then I don't know where to go from there.

Any thoughts are appreciated!