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    Note: Instead of "a^2" it should read "sigma^2" for the variance.

    Anyway, I'm stuck on Eq. 3.51. I don't know where the 1/y comes from. I believe its to do with implicit differentiation maybe but not sure.

    Any help?

    I tried a basic equation, f(x) = x^3 + x^2. I then said what is f'(x^2). So I subbed in x^2 into the equation and got my answer. I then tried D{f(x^2)} = 2x * D{f(x)} and this gives a different answer which is why I don't understand the derivation in the book.
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    (Original post by djpailo)




    Note: Instead of "a^2" it should read "sigma^2" for the variance.

    Anyway, I'm stuck on Eq. 3.51. I don't know where the 1/y comes from. I believe its to do with implicit differentiation maybe but not sure.

    Any help?

    I tried a basic equation, f(x) = x^3 + x^2. I then said what is f'(x^2). So I subbed in x^2 into the equation and got my answer. I then tried D{f(x^2)} = 2x * D{f(x)} and this gives a different answer which is why I don't understand the derivation in the book.
    Using (3.50), fy(y) = .... their second thing in 3.51 = d/dy F(lny) = dlny/dy f(lny) = 1/y * f(lny) (the last bit in the first line of 3.51).

    The bit in the middle is where you're getting stuck. d'F(g(x))/dx where F is the CDF and g(x) is some function of x, is g'(x) * f(g(x)).
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    (Original post by SeanFM)
    Using (3.50), fy(y) = .... their second thing in 3.51 = d/dy F(lny) = dlny/dy f(lny) = 1/y * f(lny) (the last bit in the first line of 3.51).

    The bit in the middle is where you're getting stuck. d'F(g(x))/dx where F is the CDF and g(x) is some function of x, is g'(x) * f(g(x)).
    Do you know where I can find more examples of this?

    Also, if F was not the CDF, would that still apply?

    Also, should it be: d(F(g(x)))/dx (I'm a bit confused what the apostrophe is doing in the first bold term)
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    (Original post by djpailo)
    Do you know where I can find more examples of this?

    Also, if F was not the CDF, would that still apply?
    I'm not sure where you'd be able to find more examples, sorry! :getmecoat:

    It's the chain rule in disguise. If F were the PDF then differentiating the PDF would be meaningless (as far as I am aware...) but what's applied there is just the chain rule.
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    So if I had:
    F(x) = x^3 and g(x) = x^2.

    d/dx ( F(g(x)) ) = g'(x) * f(g(x)) = 2x . 3(x^2)^2 = 2x . x^4 = 6x^5 ?

    And when I substitute g(x) into F, you get x^6 and obviously the derivative of that is the same, 6x^5.

    Okay, I understand the workings of the formula, but I am still a bit confused. Why isn't the answer just 1/y then?
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    I see my mistake now. I kept thinking ln(y) was a variable but its a function and so the chain rule is required. I've also found some helpful understanding of the chain rule here for anyone interested:

    http://betterexplained.com/articles/...t-power-chain/
 
 
 
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