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When you multiplied all the x's in the e^ßx^(Delta)) portion, why is it not e^-nß∑x^(delta) but just e^ß∑x^(delta)? Wouldn't we have to multiply ß n times too?
you would have to verify it is a maximum by taking the second partial derivative of the likelihood function and if its negative it is a maximum
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It is a pdf of weibull distribution
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where is come from beta^n and delta^n ??
what about BETA
Why dont you take the summation of the other x in the problem?
When you multiplied all the x's in the e^ßx^(Delta)) portion, why is it not e^-nß∑x^(delta) but just e^ß∑x^(delta)? Wouldn't we have to multiply ß n times too?
What you're multiplying is (e^Bx)*(e^Bx)*.... n times. So for example (e^Bx)*(e^Bx) = e^(Bx+Bx) = e^(B sigma x)
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