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If $f(t)$ is a Probability density function of a positive RV.

$\int_0^\infty\int_x^{\infty}f(t)dtdx$ Using fubini theorem should become $\int_0^\infty\int_0^{t}f(t)dxdt$

But why? Surely the answer is very simple but I don't see it. Fubini theorem allows the change of the order of integration. So it should become $\int_0^\infty\int_x^{\infty}f(t)dxdt$

but then I don't understand what meaning could have $\int_x^{\infty}dx$

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  • $\begingroup$ Note that $\int_a^b \int_c^d f(x,y) \,dx \, dy = \int_c^d \int_a^b f(x,y) \,dy \, dx$. So you probably forgot to change the order of $\int_0^\infty$ and $\int_x^\infty$ in "So it should become...". (Nevertheless, $\int_x^\infty \int_0^\infty f(t) \,dx \,dt$ is not sensible either, so your question remains.) $\endgroup$ – JiK Jun 30 '14 at 11:45
  • $\begingroup$ Did you follow the first rule of word problems and draw a picture? $\endgroup$ – Eric Towers Jun 30 '14 at 14:53
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Note that the integral can be written as $$ \int_0^\infty \int_x^\infty f(t)\,\mathrm dt\,\mathrm dx=\int_0^\infty \int_0^\infty f(t)\mathbf{1}_{t\geq x}\,\mathrm dt\,\mathrm dx $$ and using Fubini we get $$ \int_0^\infty \int_0^\infty f(t)\mathbf{1}_{t\geq x}\,\mathrm dx\,\mathrm dt=\int_0^\infty \int_0^t f(t)\,\mathrm dx\,\mathrm dt. $$

Here $\mathbf{1}_{t\geq x}$ is the indicator functions which is $1$ if $t\geq x$ and $0$ otherwise.

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  • $\begingroup$ As clear as water! $\endgroup$ – Benzio Jun 30 '14 at 11:45
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In the original integral, you first pick an $x$, and then let $t$ run from $x$ to infty. This means that you are integrating over the area $\{(x,t)| x>0\wedge t>x\}$

When you flip the order of operations, you must integrate over $t$ first, and as before, $t$ can go from $0$ to $\infty$. However, once picking the value of $t$, you can only integrate over $x$ for the values $x\in(0,t)$, otherwise you fall out of your integration area.

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Hint: when $g\ge 0$, $$ \int_0^\infty dx\int_x^\infty dt g(x,t) = \iint 1_{0\le x} 1_{x\le t} g(x,t) dxdt $$

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