Timeline for Does a prediction interval have to contain the mean?
Current License: CC BY-SA 3.0
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Mar 18, 2015 at 14:52 | comment | added | Anotherdream | And that is what caused a lot of my confusion. It seems like when the assumption of normal residuals is broken, then a 'forecast point estimate' becomes a meaningless statistic, and really what you need is the original distribution.... Which can be made from the point estimate + residual distribution.... That 'realization' is what is confusing me the most I believe. | |
Mar 18, 2015 at 14:51 | comment | added | Anotherdream | Whuber. I agree completely with what you are saying.... It seems like the mean is not relevant at all in this example... But does that imply that if you ever run a simulation, and use "a varaible" (in this case the mean) as a point estimate, and your residuals are horribly skewed, you can simply re-make the original distribution by taking the skewed point estimate and randomly sampling from the residuals and add the results together. I have just re-made the original dist from the "biased" estimate and the residual dist... So what use is the original estimate at all? | |
Mar 17, 2015 at 21:18 | answer | added | Emil Friedman | timeline score: 1 | |
Mar 17, 2015 at 15:00 | comment | added | whuber♦ | Your first comment is interesting in how it seems to introduce the mean unnecessarily. Once you have a good simulation of the distribution of the variable itself, why is that not enough information to make a good prediction? Wouldn't it be likely that a future value would lie within the main body of that distribution? Why would the mean be relevant in that case? | |
Mar 17, 2015 at 13:25 | comment | added | Anotherdream | I guess part of the confusion is this. I was told to provide a prediction interval for a variable that behaves VERY similar to this. The "prediction point estimate" was the 6 month moving average. However the 6 month moving average was higher than the upper percentile... As such my "prediction interval" did not include my "prediction estimate". It sounds like everyone is saying the mean was a bad value to use to begin with (which I can see... I didn't build this thing haha). Am I following that correctly? Perhaps a different value should be used as the 'prediction point estimate'? | |
Mar 17, 2015 at 13:23 | comment | added | Anotherdream | Actuall Whuber... what would you do when predicting a value from a distribution with no mean... You can't do monte carlo because it would have no mean... You could show the distribution of the variable itself... Would you maybe use the median? I actually don't know the answer to that question, and maybe that's part of the confusion. | |
Mar 17, 2015 at 3:32 | history | tweeted | twitter.com/#!/StackStats/status/577673912483463168 | ||
Mar 16, 2015 at 23:37 | comment | added | whuber♦ | What would you do when predicting a value from a distribution that has no mean at all? Why do you think it would be strange to make a prediction for such a distribution? | |
S Mar 16, 2015 at 21:05 | history | suggested | user56674 | CC BY-SA 3.0 |
fixed typo in title, converted all-caps-for-emphasis into bold
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Mar 16, 2015 at 20:57 | review | Suggested edits | |||
S Mar 16, 2015 at 21:05 | |||||
Mar 16, 2015 at 20:49 | answer | added | jlimahaverford | timeline score: 7 | |
Mar 16, 2015 at 19:48 | history | asked | Anotherdream | CC BY-SA 3.0 |