tag:blogger.com,1999:blog-593563533834706486.post6343958284044487764..comments2024-03-28T10:35:10.453-04:00Comments on Yet Another Math Programming Consultant: Generating random variables having a GAMMA distributionErwin Kalvelagenhttp://www.blogger.com/profile/09496091402502236997noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-593563533834706486.post-43952918322540861432010-06-07T11:00:20.230-04:002010-06-07T11:00:20.230-04:00The test was just to make sure I did not make a mi...The test was just to make sure I did not make a mistake in the expression to calculate the scale parameter from a known mean. Nothing to get excited about.<br /><br />Point 2 does not make sense to me.Erwin Kalvelagenhttps://www.blogger.com/profile/09496091402502236997noreply@blogger.comtag:blogger.com,1999:blog-593563533834706486.post-50500818460102929232010-06-07T06:19:22.185-04:002010-06-07T06:19:22.185-04:00Erwin, 2 questions:
this post seems to be similar ...Erwin, 2 questions:<br />this post seems to be similar to the Random Number Generator code you built back to 2005, but you added up a statement of the failure of the condition <br />abort$(abs(m0 - b0*c0) > 0.001) "calculation of c0 was not correct"; <br /><br />But since b0=m0/c0, m0-b0*c0 should always be 0, isn't it? how is it possible abs(m0 - b0*c0) > 0.001? I think i've missed some point... could you please help me with that?<br /><br />2. it comes up to my mind another question that what if the parameter needed to generate the random number, say, mean or lambda, is a variable rather than a known parameter? can we still generat a random number through this? (in the sense that the random number generated is a function of the unknown parameter, rather than a specific number).Does the code above work?Unknownhttps://www.blogger.com/profile/01557317190815419692noreply@blogger.com