Timeline for Simulating continuous distribution using discrete distribution
Current License: CC BY-SA 3.0
10 events
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| Oct 29, 2011 at 16:58 | history | edited | JohnRos | CC BY-SA 3.0 |
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| Oct 29, 2011 at 16:56 | comment | added | JohnRos | @whuber: The "only of" relates to the distributions for which the mid-points are the solution. You are right that there is indeed freedom in choosing the representative points, and there might be points which are not the conditional expectancy which will return the right answer. Anyhow, since the "only if" is immaterial to the answer, I will remove it. | |
| Oct 29, 2011 at 16:02 | comment | added | whuber♦ | Good point John. Focusing on the expectation, though, it appears there are many choices of representative of each interval that will reproduce the first moment. As an extreme example, consider a distribution known only to be symmetric. You can discretize it into two bins with cutpoint at 0. Choosing "representative points" at $\pm a$ for any real $a$ will give the correct expectation. This provides plenty of counterexamples to the "only if" assertion in your reply. | |
| Oct 28, 2011 at 17:11 | comment | added | JohnRos | @whuber: but the question specifically deals with the first moment. | |
| Oct 28, 2011 at 14:19 | comment | added | whuber♦ | It's hard to say what the midpoints are the "solution" to, given that we haven't any objective function to optimize. For instance, if the purpose is to reproduce the second moment of the distribution, then it's usually not the case that midpoints are best. | |
| Oct 28, 2011 at 13:14 | comment | added | JohnRos | @cardinal: true. Changed answer accordingly. | |
| Oct 28, 2011 at 13:13 | history | edited | JohnRos | CC BY-SA 3.0 |
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| Oct 28, 2011 at 13:05 | history | edited | JohnRos | CC BY-SA 3.0 |
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| Oct 28, 2011 at 12:18 | comment | added | cardinal | You should really reconsider your last statement. particularly the only if part. | |
| Oct 28, 2011 at 11:52 | history | answered | JohnRos | CC BY-SA 3.0 |