Let $\eta$ be non-negative (a.c.) continuous random variable. How to prove that $\mathbf{E}\eta = \int_{0}^{\infty}\mathbf{P}(\{\eta > z\})dz$ for arbitrary non-negative random variable?

If $\eta$ is discrete random variable it's quite simple. I have an idea that we can prove it easily for random variable $$\xi = \sum_{i=1}^{n}\xi_i\mathbf{1}_{\Delta_i}$$ that can take only finite set of values , where $\mathbf{1}_{\mathbf{X}}$ is an indicator function of a set $\mathbf{X}$. Since for every measurable function $\xi$ there is an approximating sequence $\{\xi_i\}_{i\ge1}$ of the functions of such kind that tends to $\xi$ pointwise, I would like to make a passage to the limit to prove this statement for an arbitrary continuous non-negative random variable? How can one do that rigorously?

  • Have you considered Fubini theorem ? – Gabriel Romon Apr 26 '17 at 19:06
  • @LeGrandDODOM I can't clearly understand what use it could be here? We don't have any measure products. Could you explain? – Illia Yurtsiv Apr 26 '17 at 19:09
  • 1
    Note that $P(\eta >z)=P_\eta((z,\infty))=\int 1_{(z,\infty)}(u)dP_\eta (u)$. Hence $\int_{0}^{\infty}\int 1_{(z,\infty)}(u)dP_\eta (u)dz=\int \int1_{(0,u)}(z)dz dP_\eta(u)=\int u dP_\eta(u) = E(\nu) $ – Gabriel Romon Apr 26 '17 at 19:17
  • @LeGrandDODOM, Thanks a lot! – Illia Yurtsiv Apr 26 '17 at 19:19
up vote 2 down vote accepted

Do the following. Let $\mathbb{P}$ be the underlying probability measure. Then, \begin{align} \int_0^{\infty}\mathbb{P}(X(\omega)\geq x)dx & = \int_0^{\infty}\mathbb{E}\left[\mathcal{X}_{X(\omega)\geq x}\right]dx \\ &= \int_0^{\infty}\int_{\Omega}\mathcal{X}_{X(\omega)\geq x}d\mathbb{P}(w)dx\\ & =\int_{\Omega}\int_{0}^{X(\omega)}dxd\mathbb{P}(\omega) \\ & = \int_{\Omega}X(\omega)d\mathbb{P}(\omega) = \mathbb{E}[X]. \end{align}

Above, the first line uses the fact that expected value of an indicator of an event is simply the probability of the event itself, second line is valid due to expressing expected value as an integral with respect to probability measure, and third line follows from Fubini.

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