I found the equation mentioned in the title while reading through a proof, and was wondering if there is a way to prove it without using measure theoretic induction, I think I remember having seeing it in a stochastic 1 script before, but don't know where exactly. Any help would be appreciated.
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1Write the definition if $E[X^2]$. Integration by parts might work after that. Could you try it? – Shreyas Pimpalgaonkar Jul 05 '18 at 11:31
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I think integration by parts would only work if we have a suitable density function. Then we could of course simply use integration by parts on $\int ^{\infty} _{- \infty} x^2f(x) dx$. But equality should also hold without a density. – El Duderino Jul 05 '18 at 11:39
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The equality is not true for $X$ negative. It must be nonnegative. Moreover, how you can even define the expectation without measure theory if $X$ does not have a density? – Shashi Jul 05 '18 at 11:45
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If i'm not mistaken the expectation is simply defined as the limit of the expectation of a series of simple functions that almost surely converge towards X. – El Duderino Jul 05 '18 at 11:52
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And almost sure convergence is defined without measure theory? I don't really get it, but it's okay. – Shashi Jul 05 '18 at 11:54
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1If you use the Riemann-Stieltjes integral, then you can use Shreyas Pimpalgaonkar's first suggestion. – robjohn Jul 05 '18 at 12:26
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This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post. – Lee David Chung Lin Nov 13 '18 at 12:19
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For non-negative $X$ you can prove it easily using Fubini's Theorem: $2\int_0^{\infty} xP\{X>x\}\, dx=2\int_0^{\infty} x \int_{\{X>x\}} \, dP\, dx=2 \int \int_0^{X}x\, dx dP=\int X^{2} \, dP$.
Kavi Rama Murthy
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Assuming $X$ is nonnegative, one has:
$$\int^\infty_0 2x\mathbb{I}_{X>x} \,dx = X^2$$ where $\mathbb I$ is the indicator function. Taking expectation on both sides using Fubini gets you pretty much fast to the result.
Shashi
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Ah I think I see where our misunderstanding is, no you're right, of course you need measure theory. But one can show the result by first proving it for indicator functions, then simple functions, and finally measurable functions. This is referred to as measure theoretic induction but I find it doesn't provide very good intuition as to why a result is true which is why I asked for a different method of proof. Thanks for the answer! – El Duderino Jul 05 '18 at 12:07
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If $Y$ is non-negative then $$\mathbb{E}[Y]= \int\limits_{y=0} ^{\infty} P(Y>y)\, dy$$
Now suppose $X$ is also non-negative and let $y=x^2$ (strictly increasing) so $\frac{dy}{dx}=2x$ and $P(Y > x^2) = P(X^2 > x^2) = P(X>x)$. Then by substitution $$\mathbb{E}[X^2]= \int\limits_{x=0} ^{\infty} P(X>x) \,2x \,dx$$
This will therefore not be true if $P(X <0) > 0$ but you could say more generally $$\mathbb{E}[X^2]= 2\int\limits_{x=0} ^{\infty} x\,P(|X|>x) \,dx$$
Henry
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