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I was trying to solve the following simple integration involving indicator function $I_{(a,b]}$ in a journal article. Here are the equations (in LaTeX notation): $$ f(u) = \int_{0}^{1} (I_{(0,s]}(u) - s)\; ds\tag{1} $$ $$ g(u,v) = \int_{0}^{1} (I_{(0,s]}(u) - s)(I_{(0,s]}(v) - s)\; ds\tag{2} $$ where $0 < u, v < 1$. I was thinking that the integration will be simply just $$ f(u) = \int_{0}^{1} (1 - s)\; ds\tag{1} $$ $$ g(u,v) = \int_{0}^{1} (1 - s)(1 - s)\; ds \tag{2} $$ But, I'm not so sure about this. The constraint on both $u$ and $v$ confused me. Any pointer to this solution?

Thanks

Wayan

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    The integral gives $f$ as a function of $u$, so your formula which just gives $f$ as a number would seem unlikely.2012-04-02

1 Answers 1

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The integrals are functions of parameters, so I advise you to interchange the set and the variable in the indicator function: for a fixed $u$ we have $$ I_{(0,s]}(u) = \begin{cases} 1,&\text{ if }00$. Hence $$ f(u) = \int\limits_{0}^1(I_{(0,s]}(u) - s)ds = \int\limits_0^1(I_{[u,\infty)}(s) - s)ds = (1-u)-\int\limits_0^1 sds = \frac12-u $$ and similar for the function $g$: when you open the brackets you get 4 terms - of which 3 you should know how to compute (they involve at most 1 indicator function) and for the last term you have: $$ \int\limits_0^1I_{(0,s]}(u)\cdot I_{(0,s]}(v)ds = \int\limits_0^1I_{[u,\infty)}(s)I_{[v,\infty)}(s)ds = \int\limits_0^1I_{[\max(u,v),\infty)}(s)ds = 1-\max(u,v). $$

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    Where did $t$ come from?2012-04-02
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    Thanks Ilya. But, I am still a bit confused how do you get $\int_0^1 I_{[u,\infty)}(s) = (1 - u)$? Is it $\int_0^1 1 ds = s|_{s = u}^{s = 1}$ since $0 < s \leq 1$?2012-04-02
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    @GerryMyerson: I don't know how to say it in English. Apparently, from nowhere - it was a typo.2012-04-02
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    @Wayan: sure, $\int\limits_0^1 I_{[u,\infty)}(s)ds = \int\limits_u^1 ds = 1-u$ - and you had a typo in your comment, in the lower limit of the latter integral2012-04-02
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    @Ilya: Thank you for pointing the typo. What about in the second equation where $$\int_{0}^{1} I_{(0,s]}(u)I_{(0,s]}(v)?$$ If I do it similar like $f$ then $$\int_{u}^{1} I_{u,\infty]}(s)I_{(v,\infty]}(s)?$$ But what about the condition $0< u, v< 1$. Is it safe to assume $u < v$?2012-04-02
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    @Wayan No, it's not safe. Think of taking $\max(u,v)$. And again, you have a typo in the latter integral - although it does not matter formally, educationally it should be from 0 to 1.2012-04-02
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    @Ilya: Thank you for pointing the typo again. After checking this website (http://turing.une.edu.au/~stat354/notes/node16.html) is it similar to property (e)? Thus it becomes $$\int_{0}^{1} \max\{I_{(u,\infty]}(s),I_{(v,\infty]}(s)\} ds = \int_{0}^{1} ds ?$$ Sorry, I become a bit lost here.2012-04-02
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    @Wayan: no, I rather meant that $I_A\cdot I_B = I_{A\cap B}$ (property (c)) and $$ [u,\infty)\cap [v,\infty) = [\max(u,v),\infty). $$ Take a look on my answer - I've edited it.2012-04-02
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    @Ilya: Thank you so much. I really appreciate it!2012-04-02
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    @Ilya: One more thing: After checking (http://en.wikipedia.org/wiki/Indicator_function) and (http://planetmath.org/encyclopedia/CharacteristicFunction.html). Do you mean $I_{A}\cdot I_{B} = I_{A}\cap I_{B} = \min\{I_A,I_B\}$?2012-04-02
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    @Wayan: no, I didn't mean it - instead of reading wiki and planetmath, would you just read the last part of my answer and tell if there anything unclear to you?2012-04-03