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Math Help - Help with a derivation!

  1. #1
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    Unhappy Help with a derivation!

    Help with a derivation!-ask.jpg


    Can someone help me with this derive. I have attached my solution in the picture link here, and in attachment to. Thank you in advance!
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  2. #2
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    Re: Help with a derivation!

    It would help greatly if we had the context for this problem. It looks to me like some sort of least-squares fit (maybe log-likelihood?). Please define every single variable I see. That is, what are each of the following:

    \alpha

    \beta

    n

    LL

    \mu_{i}?

    Do any of these variables depend on any of the others, aside from LL depending on \alpha and \beta?
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  3. #3
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    Re: Help with a derivation!

    Thanks for asking!
    beta - is a modeling parameter
    alfa - is chosen as the maximum likelihood estimate
    LL - is the first equation (in square shape)
    n - is the index of the numbers in the samle (from i - n)
    mu(i) is the value of the element ( for example i = 3; mu(i)=254.2...etc)
    This is a sample with some positive numbers. I'm asked to do this: Suppose you are free to choose both parameters alfa as well as beta . Can you derive
    the MLE (maximum likelihood estimates) for both parameters simultaneously? I've got the derive fore alfa, but I'm stuck for beta. Is it more clear now? It's a long problem, this was the reason didn't want to post much info, just the equations. Thanks a lot!
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  4. #4
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    Re: Help with a derivation!

    P.s.: LL is the log of this:
    Attached Thumbnails Attached Thumbnails Help with a derivation!-untitled.jpg  
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  5. #5
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    Re: Help with a derivation!

    What you need to know, then, is this:

    \frac{d}{dx}\,a^{x}=\ln(a)\,a^{x}.

    That will get you what you need.
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  6. #6
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    Re: Help with a derivation!

    Come to think of it, I'm not sure I buy the result you've got there. I'd have thought it would be this:

    \frac{\partial LL}{\partial\beta}=\frac{n}{\beta}+\sum_{i=1}^{n} \ln (\mu_{i})-\alpha\sum_{i=1}^{n}\ln (\mu_{i})\mu_{i}^{\beta}.
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  7. #7
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    Re: Help with a derivation!

    I was just typing the same thing that you just wrote in the last post Reading you comments, it makes sense now! I wasn't sure abt the formula.
    Many thanks!!
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  8. #8
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    Re: Help with a derivation!

    Can I ask you a final question? Pls
    Attached Thumbnails Attached Thumbnails Help with a derivation!-final.jpg  
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  9. #9
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    Re: Help with a derivation!

    So, more generally, you're asking if

    \sum_{i=1}^{n}a_{i}b_{i}=\left(\sum_{i=1}^{n}a_{i}  \right)\left(\sum_{i=1}^{n}b_{i}\right).

    I do not think it very difficult to convince yourself that this is false in general. Let a_{i}=2, a constant, and let b_{i}=3. Then the equation would have us believe that

    \sum_{i=1}^{n}2\cdot 3=\left(\sum_{i=1}^{n}2\right)\left(\sum_{i=1}^{n}  3\right), or

    6n=\left(2n\right)\left(3n\right)=6n^{2},

    which is certainly not true if n=2, say. The result is only true if n=1, in which case your summation really isn't doing anything, as there is only one term!

    Does that clear things up?
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  10. #10
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    Re: Help with a derivation!

    I tried it with doing some test after I asked, and I concluded that it was a stupid question. I wanted to simplify some things through my equations, but when you are not able to do so, it's better not to invent formulas :PP

    Thanks a lot!
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