Timeline for PDF of sum of ordered weighted exponential RVs
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2015 at 10:39 | answer | added | tristan | timeline score: 0 | |
| Mar 31, 2015 at 21:06 | comment | added | tristan | Will have to wait till the morning. Taking me longer to work out than I thought! | |
| Mar 31, 2015 at 19:50 | comment | added | whuber♦ | My ordering was a hint at a simpler way to solve the problem. | |
| Mar 31, 2015 at 19:48 | comment | added | tristan | @Noor you don't have to add self-study if it's not homework, it just looked homework-y and you asked for hints. Good edits, I'll hopefully take a look later this evening. | |
| Mar 31, 2015 at 19:22 | history | edited | Noor | CC BY-SA 3.0 |
clearly mention one of the assumptions
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| Mar 31, 2015 at 19:20 | comment | added | Noor | @whuber actually the assumption here is that $X_{(1)} \geq X_{(2)} \geq ... X_{(N)}$. The question is updated to be more clear | |
| Mar 31, 2015 at 19:17 | comment | added | whuber♦ | The expression for the joint PDF lacks the critical information that $0 \le x_1 \le x_2 \le \cdots \le x_M $; for all other possible values, the PDF is zero. | |
| Mar 31, 2015 at 18:46 | comment | added | Noor | @Xi'an Yes, I want to divide by $M$ not $i$. Also, I updated the question with the joint PDF of ${{X_{(1)}},...,{X_{(M)}}}$. | |
| Mar 31, 2015 at 18:45 | comment | added | Noor | @tristan It is not a homework, i.e. a given question that I seek answer for. Do I still need to add [self-study]? Also, the question is updated with the PDF of the $n$th order statistics and as you can see they are dependent. | |
| Mar 31, 2015 at 18:42 | history | edited | Noor | CC BY-SA 3.0 |
adding more details as per the comments
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| Mar 31, 2015 at 8:44 | comment | added | tristan | Can you post the PDF of the $n$th order statistic $X_{(n)}$? I normally approach addition problems using convolution integrals, and in some cases there is a closed form solution that allows you to construct an induction formula. | |
| Mar 31, 2015 at 8:42 | comment | added | Xi'an | another indication: you need the joint pdf of $X_{(1)},\ldots,X_{(M)}$ and not of $X_{(n)}$ by itself. | |
| Mar 31, 2015 at 8:21 | comment | added | tristan | Just checking whether this is self study (e.g., homework)? In which case please add [self-study] tag so answers can be tailored appropriately. | |
| Mar 31, 2015 at 6:25 | history | edited | Glen_b | CC BY-SA 3.0 |
edited body
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| Mar 31, 2015 at 6:00 | comment | added | Noor | @Alecos Sorry for the typo. The question reads correctly now. | |
| Mar 31, 2015 at 6:00 | comment | added | Noor | @Glen_b Sorry for the typo. The question reads correctly now. | |
| Mar 31, 2015 at 5:57 | history | edited | Noor | CC BY-SA 3.0 |
fixing typo in question
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| Mar 31, 2015 at 3:02 | comment | added | Glen_b | @Alecos I hold a similar suspicion, but we can't tell for sure. | |
| Mar 31, 2015 at 2:03 | comment | added | Alecos Papadopoulos | @Glen_b I suspect the OP meant to write $X_{(i)}$ instead of $X_{(M)}$, but let's see how he responds. | |
| Mar 31, 2015 at 1:10 | comment | added | Glen_b | Since neither numerator nor denominator in your sum changes with $i$, both terms can be taken out the front of the sum, and then the $M$ in the denominator will be cancelled by $\sum^M 1$. Please check your question says what you mean. | |
| Mar 30, 2015 at 23:21 | review | First posts | |||
| Mar 30, 2015 at 23:32 | |||||
| Mar 30, 2015 at 23:18 | history | asked | Noor | CC BY-SA 3.0 |