Say we obtain a random number from an exponential distribution with $\lambda = \frac{1}{4}$. The numbers are passed into $f(x) = (2x + 1)(2x - 1)$. How do i find the expected value of the result?
My first assumption was to do $$ E(X) = E(g(\frac{1}{4}e^{-\frac{1}{4}x})) = \dfrac{1}{4}[-2e^{-\frac{1}{4}x}]_{-\infty}^{\infty} - [x]_{-\infty}^{\infty} $$ However that would be incorrect as we would get an infinity.
I came across this, which would imply the right way to solve this problem is $$ E(g(f(x))) = \int_{-\infty}^{\infty} (\frac{1}{4}e^{-\frac{1}{4}x})(2x + 1)(2x - 1) $$ Though I am unsure if that is correct, or why it would be right. Could someone help explain to me if this is the right way to solve it and the intuition behind it?
The correct expression is $\int_0^{\infty} \frac 1 4 e^{-\frac 1 4 x} (2x+1)(2x-1) \, dx$. This can be evaluated by repeatedly integrating by parts.