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3Â¥: Originally posted by »Ã÷Ê at 2015-11-03 21:12:27
¶àл»Ø¸´£¬²»¹ýÎÒÖ¸µÄÊÇwgt = 1 ./ sqrt(pEmp.*(1-pEmp))Õâ¸öÈ¨ÖØ£¬ÎªÊ²Ã´Õâôȡ£¿...

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4Â¥2015-11-03 23:08:53
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To fit a Weibull distribution to these data, notice that the CDF for the Weibull is p = Pr{X <= x} = 1 - exp(-(x/a)^b). Transforming that to log(a) + log(-log(1-p))*(1/b) = log(x) again gives a linear relationship, this time between log(-log(1-p)) and log(x). We can use least squares to fit a straight line on the transformed scale using p and x from the ECDF, and the slope and intercept of that line lead to estimates of a and b.
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2Â¥2015-11-03 20:26:57
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2Â¥: Originally posted by feixiaolin at 2015-11-03 20:26:57
To fit a Weibull distribution to these data, notice that the CDF for the Weibull is p = Pr{X <= x} = 1 - exp(-(x/a)^b). Transforming that to log(a) + log(-log(1-p))*(1/b) = log(x) again gives a li ...

¶àл»Ø¸´£¬²»¹ýÎÒÖ¸µÄÊÇwgt = 1 ./ sqrt(pEmp.*(1-pEmp))Õâ¸öÈ¨ÖØ£¬ÎªÊ²Ã´Õâôȡ£¿
3Â¥2015-11-03 21:12:27
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ÒýÓûØÌû:
4Â¥: Originally posted by feixiaolin at 2015-11-03 23:08:53
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