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So I came across below symmetry in my probability course that I can't understand.

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I understand how the lower bound changes when removing the absolute value operator, but how does the 2 disappear?

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  • $\begingroup$ Draw the graphs of the integrands. Sketch in the areas represented by the integrals. The result will be obvious. $\endgroup$ – whuber Nov 20 '19 at 16:17
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In the left hand you have the integral of a function of $|v|$ (considering the quadratic term too), hence that function is symmetric on $v$'s domain. This comes simply from the fact that $f(|-v|) = f(|v|)$.

Integrating that function on positive values of $v$ yields the same result as integrating it on negative values of $v$, so, to compute the whole integral, it is sufficent to only perform the integral on positive $v$ and to double it (if you don't you will not account for negative $v$).

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  • $\begingroup$ Hi, thank you for answer. Just for clarification, when you say "to double it", do you mean that I have to multiply the integral by 2, when changing the domain and removing the absolute value operator? $\endgroup$ – FAHRB Nov 20 '19 at 22:26
  • $\begingroup$ yes.___________ $\endgroup$ – carlo Nov 21 '19 at 11:13

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