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.

  • $\begingroup$ An identity I completely forgot! Thanks! $\endgroup$ – user661541 Oct 22 at 9:42

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