For actuarial mathematics there is an important identity to calculate the life expectancy for X defined on $[0,\infty)$: identity $$ E[X]=\int_0^\infty S_X[x]\,\mathrm{d}x $$ where $S_{x}(x)$(survival function) is equal to $1-F_{x}(x)$, where $F_{x}$ is the CDF, which theorem or lemma exists so i can understand this identity?
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It is a consequence of Fubini's Theorem and the fact that $1-F_x(x)=\int_{x}^{\infty}$ dF_X(x)$. – Kavi Rama Murthy Feb 01 '21 at 23:46
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2Note how $$E(X)=\int_{0}^{\infty}xf_X(x)dx=\int_{0}^{\infty}\int_{0}^xf_X(x)dydx=\int_0^{\infty} \int_y^{\infty}f_X(x)dxdy$$ The last integral equals $\int_{0}^{\infty}\big[1-F_X(y)\big]dy$ – Matthew H. Feb 01 '21 at 23:49
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1@KaviRamaMurthy the result comes from a calculus theorem, so there isnt a specific probability theorem?, thx – minibeto666 Feb 02 '21 at 00:16
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@MatthewPilling i never thought of doing it as double integrals, thank u – minibeto666 Feb 02 '21 at 00:17
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@minibeto666 The calculus theorem is Fubini's Theorem. – Graham Kemp Feb 02 '21 at 01:38
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I will elaborate a little on the general measure-theoretical context here, which does not rely on the existence of a density $f_X$ (which would simply be the Radon-Nikodym derivative of the pushforward $X_\ast \mathbb{P}$ with respect to the Lebesgue measure $\lambda$).
I will use the following notation: $(\Omega, \mathcal{B}, \mathbb{P})$ for the probability space on which $X \in L^0(\Omega)$ is defined. Then by definition we have that $\mathbb{E}[X] = \int_\Omega X d\mathbb{P} = \int_{[0, \infty)} x \, dX_\ast\mathbb{P},$ where I used the fact that your variable takes values in $[0, \infty)$ and applied the change of variables formula in the measure-theoretical context at play here. Note that the pushforward $X_\ast\mathbb{P} = \mathbb{P} \circ X^{-1}$ is just the cummulative distribution function of $X.$ But we also have that $x = \int_0^x d\lambda(y),$ so $\mathbb{E}[X] = \int_0^\infty \int_0^x d\lambda(y) \, d X_\ast \mathbb{P}(x).$ Note that we may indeed apply Fubini's Theorem here, finally yielding $\mathbb{E}[X] = \int_0^\infty \int_y^\infty dX_\ast\mathbb{P}(x) \, d\lambda(y) = \int_0^\infty (1 - F_X(y)) \, d \lambda(y) = \int_0^\infty S_X(y) d\lambda(y),$ which is the desired equality.
I hope this is clear, but if not do let me know!
Actually Fritz
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