Showing equilibrium probability density function is a density function

Question

Given the equilibrium probability density function: $$ g_Y(x)=\frac{1}{\mu}\int_x^\infty\!f_Y(y)\,\mathrm{d}y $$ check that $g_Y(x)$ is indeed a probability density function.

Solution attempt: Integrating $g_Y(x)$ with respect to $x$, I was able to get $$ \int_0^\infty\!g_Y(x)\mathrm{d}x=\int_0^\infty\!\frac{1}{\mu}\int_x^\infty\!f_Y(y)\,\mathrm{d}y\mathrm{d}x=\int_0^\infty\!\frac{1}{\mu}(F_Y(\infty)-F_Y(x))dx=\int_0^\infty\frac{1}{\mu}(1-F_Y(x))\mathrm{d}x $$ I am unsure how to proceed from here. I know that I have to prove that the integral is equal to 1, but can't find a next step.

Edit: Changed $g_y(x)$ to $g_Y(x)$ and $f_y(y)$ to $f_Y(y)$

Answer 1

If $Y$ is a non-negative random variable with mean $\mu$, you can use the result Explain why $E(X) = \int_0^\infty (1-F_X (t)) \, dt$ for every nonnegative random variable $X$.