Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, $\alpha > 0$ and $X: \Omega \to \mathbb{R}$ a non-negative real-valued random variable. I need to prove that

$$\int_{[0,\infty)}\alpha^{-2}\mathbb{E}(\min(X, \alpha^2))\mathrm{d}\lambda^1(\alpha) = 2\mathbb{E}(\sqrt{X}),$$

where $\lambda^1$ denotes the lebesgue measure. In a first step I proved that

$$\mathbb{E}(\min(X, \alpha^2)) = \int_{[0,\alpha^2]}\mathbb{P}(X \geq t)\mathrm{d}\lambda^1(t).$$

Using this, we get

$$\int_{[0,\infty)}\alpha^{-2}\mathbb{E}(\min(X, \alpha^2))\mathrm{d}\lambda^1(\alpha) = \int_{[0,\infty)}\int_{[0,\alpha^2]}\alpha^{-2}\mathbb{P}(X \geq t)\mathrm{d}\lambda^1(t)\mathrm{d}\lambda^1(\alpha)$$

and here is where I get stuck. This smells like Fubini but I don't see how to reach

$$2\mathbb{E}(\sqrt{X}) = 2\int_{\Omega}\sqrt{X}\mathrm{d}\mathbb{P}$$ from here. I would appreciate any help.

  • 2
    $\begingroup$ Did you try to swap the two integrals using Fubini? $\endgroup$
    – gerw
    Jun 2 at 14:32
  • $\begingroup$ Yes but then I would get $\int_{[0,\infty)}\int_{[0,\alpha^2]}\alpha^{-2}\mathbb{P}(X \geq t)\mathrm{d}\lambda^1(t)\mathrm{d}\lambda^1(\alpha) = \int_{[0,\alpha^2]}\int_{[0,\infty)}\alpha^{-2}\mathbb{P}(X \geq t)\mathrm{d}\lambda^1(\alpha)\mathrm{d}\lambda^1(t)$ and the first integral would evaluate to $\infty$ $\endgroup$ Jun 2 at 14:36
  • 3
    $\begingroup$ No when you swap you have to adjust the bounds $\endgroup$
    – Kroki
    Jun 2 at 14:37

1 Answer 1


Hint: \begin{align} \int_0^\infty \int_0^{\alpha^2}\ldots \mathrm d\lambda(t)\mathrm d\lambda(\alpha) &=\int_0^\infty \int_0^{\infty}\ldots \, \mathbf 1_{t\le \alpha^2}\mathrm d\lambda(t)\mathrm d\lambda(\alpha)\\ &= \int_0^\infty \int_0^{\infty}\ldots \, \mathbf 1_{t\le \alpha^2}\mathrm d\lambda(\alpha)\mathrm d\lambda(t)\\ &= \int_0^\infty \int_0^{\infty}\ldots \, \mathbf 1_{\sqrt t\le \alpha}\mathrm d\lambda(\alpha)\mathrm d\lambda(t)\\ &=\int_0^\infty \int_{\sqrt{t}}^\infty \ldots \mathrm d\lambda(\alpha)\mathrm d\lambda(t) \end{align}

  • $\begingroup$ I'm sorry but I don't understand why the bounds change. The version of Fubini's theorem in our course only states that $\int_{\Omega}\int_{\Sigma}f(\omega,\sigma)\mathrm{d}\nu\mathrm{d}\mu = \int_{\Sigma}\int_{\Omega}f(\omega,\sigma)\mathrm{d}\mu\mathrm{d}\nu$ for general measure spaces $(\Omega, \mathcal{F}, \mu), (\Sigma, \mathcal{G}, \nu)$ $\endgroup$ Jun 2 at 15:02
  • 1
    $\begingroup$ In the version that mentioned you can see that the bounds are independent from the variables of integration. It is not the case in your integral. Can you make the bounds independent of the variables and exchange the integrals? $\endgroup$
    – Kroki
    Jun 2 at 15:04
  • $\begingroup$ I updated my answer with details. $\endgroup$
    – Kroki
    Jun 2 at 15:09
  • $\begingroup$ Thank you for the details, I think I get it now! $\endgroup$ Jun 2 at 15:11

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