Changing order of expectation and integration

Question

If I know that

$$E\biggl(\int_{0}^{\infty} e^{-\lambda t}X^{2}_{t}dt\biggr)<\infty$$

Can I say

$$\int_{0}^{\infty}E (e^{-\lambda t}X^{2}_{t})dt<\infty$$

Or do I need that $E(X_{t}^{2})<\infty$

Answer 1

Let $(\Omega, \mathcal F, P)$ denote the probability space $X_t$ is defined.

You can do this without any additional assumptions on $X_t$ besides that the map $X_t : \Omega \times [0,\infty) \to \mathbb R$ defined by $(\omega, t) \mapsto X_t(\omega)$ is measurable on the product $\sigma$-algebra $\mathcal F \otimes \mathcal B([0,\infty))$, where $\mathcal B([0,\infty))$ is the Borel $\sigma$-algebra on $[0,\infty)$ (which seems to be true in your case.)

This is because Fubini's theorem for positive functions (noting that $(\omega, t) \mapsto e^{-\lambda t}X_t(\omega)^2$ is a positive function): $$\mathbb E\left[\int_0^\infty e^{-\lambda t}X_t^2\,dt\right] = \int_{\Omega}\int_0^\infty e^{-\lambda t}X_t(\omega)^2\,dt\,dP(\omega) = \int_0^\infty \int_{\Omega}e^{-\lambda t}X_t(\omega)^2\,dP(\omega)\,dt = \int_0^\infty \mathbb E[e^{-\lambda t}X_t^2]\,dt.$$