# Suppose $X, X'$ are variables on different probability spaces with equal distributions. Do they have the same expectation?

Suppose $$X: \Omega \to \mathbb{R}$$ is a random variable on $$(\Omega, \mathcal{F}, \mathbb{P})$$ and $$X': \Omega' \to \mathbb{R}$$ is a random variable on $$(\Omega', \mathcal{F}', \mathbb{P}')$$. Assume that $$\mathbb{P}_X = \mathbb{P'}_{X'}$$ (the distribution of $$X$$ is equal to the distribution of $$X'$$). Is it true that

$$\int_\Omega X d \mathbb{P} = \int_{\Omega'} X' d \mathbb{P'}$$

I.e. is the $$\mathbb{P}$$-expectation of $$X$$ equal to the $$\mathbb{P'}$$-expectation pf $$X'$$?

Intuitively, this ought to be true but how can I formally show this?

I tried the approach where you first show this for indicatorfunctions, then for positive functions etc but this doesn't work because we work on different probability spaces.

Maybe I can argue in the following way, if $$X\geq 0$$:

$$\int_\Omega X d \mathbb{P} = \int_0^\infty \mathbb{P}(X \geq t)dt = \int_0^\infty \mathbb{P}'(X'\geq t)dt = \int_{\Omega'}X'd\mathbb{P'}$$

and in the general case, the result then follows if we can prove that $$X^+=XI_{\{X \geq 0\}}$$ and $$(X')^+ = X' I_{\{X' \geq 0\}}$$ have equal distribution (and similarly for $$X^-$$ and $$(X')^-$$.

Any ideas?

$$EX=\int_{\mathbb R} x dP_X(x)$$ and $$EX'=\int_{\mathbb R} x dP_{X'}(x)$$, so the answer is YES. $$EX$$ exists iff $$EX'$$ exist and they are equal when they exist.