8
$\begingroup$

I am having a huge problem with a conceptual problem that I came up with.

Say a company has a distribution that is highly skewed. Something similar to an exponential or lognormal only more extreme. Now pretend the distribution is so skewed that the mean of the distribution is higher than the 99% Percentile of the distribution. (Aka 1-2 extreme higher values caused the mean to be extremely high compared to the rest of distribution).

By definition, if this distribution was used to forecast a future value (aka a random sample from the distribution) would it be true that mean would not be in the 95% Prediction interval?

In my brain, a 95% predition interval is a range that 95% of all future values will fall between. For any distribution this should exactly equal the .025 Percentile on the lower bound, and the .975 percentile on the upper bound... If the mean is higher than the .975 Percentile, then the mean would not be within the '95% prediction interval'.

Am I thinking of this incorrectly? It seems strange to report a forecast as

  • Mean Forecasted Value: 6,000,0000
  • 95% Prediction Interval: [400,5000].
$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – whuber Mar 16 '15 at 23:37
  • $\begingroup$ 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. $\endgroup$ – Anotherdream Mar 17 '15 at 13:23
  • $\begingroup$ 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'? $\endgroup$ – Anotherdream Mar 17 '15 at 13:25
  • $\begingroup$ 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? $\endgroup$ – whuber Mar 17 '15 at 15:00
  • $\begingroup$ 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? $\endgroup$ – Anotherdream Mar 18 '15 at 14:51
7
$\begingroup$

No, a prediction interval need not contain the mean. I think some of your confusion might be mixing prediction intervals and confidence intervals. While the goal of a prediction interval is to contain with some certainty future values of the random variable, the goal of a confidence interval is to contain the true mean of distribution.

As you mentioned in highly skewed distributions these ideas seem to be at odds with each other. The important thing is to recognize the value in each of the statistics provided.

The predictive value of the mean is:

1) Cumulative: As more samples come in, their average will tend toward the true mean. So if the cumulative value is of interest (for instance, if you're gambling and dealing with winnings or losses you're interested in cumulative effects) then the mean is very useful.

2) Minimizes Squared Residuals: While squared residuals are a somewhat arbitrary quantity of interest it is worthwhile to know what your prediction is minimizing.

If however your goal is to minimize the absolute error in your predictions, the mean forecasted value of 6,000,000 is not what I would go with.

$\endgroup$
  • $\begingroup$ Thanks for the time jlimahaverfold. So if I understand you correctly is the following a true statement (I think I do, it just 'feels wrong' haha). If I had a variable where I was given a "point" estimate (using the mean), but the residuals were extremely non-normal (exponential for example) I could get the 'forecast distribution' by basically randomly sampling from the residual distribution 10k times (monte carlo) and then the newly created distribution would the forecast interval? I think this is how this should be done, but wwant to confirm i'm understanding correctly $\endgroup$ – Anotherdream Mar 16 '15 at 21:39
  • $\begingroup$ To clarify my question a bit further. If someone took a 6 month moving average forecast, but had non-normal residuals in this estimate... Is it correct to create the forecast distribution by sampling from the residual distribution and adding the value to the Mean forecast point estimate, and then calculating the 95% prediction interval from the percentiles of this resulting distribution? Also, can you specify what else you might go with besides the "mean" if I wanted to minimize the absolute error in a given prediction for highly skewed data? Again I truly appreciate your help! $\endgroup$ – Anotherdream Mar 16 '15 at 21:48
  • $\begingroup$ I'm still having trouble interpreting the question. Let me be clear about what I'm looking for. I have a random variable X, and data {x1, x2,... xN}. I assume this 6 month rolling average is something along the lines of \sum_{j=i}^{i+180} x_i / 180. Something along these lines. As for what I meant about minimizing the absolute residuals, it's simply another objective function. While the mean minimizes the sum of the squared residuals, this does not necessarily minimize the absolute residuals, but some value (not necessarily unique) does. $\endgroup$ – jlimahaverford Mar 17 '15 at 2:06
  • $\begingroup$ +1, very nice point about a possible confusion about prediction intervals and confidence intervals. Incidentally, if you want to minimize the expected absolute error, you use the median of the predictive distribution as your point forecast (see here). This will of course always be included in a (central) prediction interval. $\endgroup$ – Stephan Kolassa Mar 17 '15 at 7:28
  • $\begingroup$ Stephan. Your comment helps a BUNCH. I think this is what needs to happen in the future of these estimates. Really I think the problem is the mean was the wrong place to start with using such skewed distributions... But since they DID start here, I was confused with what I could do... Is it commonly 'acceptable' to use a median as a 'forecast point estimate' and give it bounds? I am very new to forecasting and am not sure if that is commonly done with skewed distributions.. $\endgroup$ – Anotherdream Mar 17 '15 at 13:27
1
$\begingroup$

Consider the distribution of possible returns in the St Petersburg paradox:

Prob(1)=1/2

Prob(2)=1/4

Prob(4)=1/8 ... Prob(2^n)=1/2^(n+1)

The mean diverges and is outside of any reasonable prediction interval. (The median is 1 in this case, but I don't know what I'd use for my point forecast. Maybe Stephan Kolassa, see above, has a suggestion.)

There's another complication: Let's say you want a 95% prediction interval for some distribution (other than the one I just mentioned). Do you go from the 2.5%tile to the 97.5%tile or the 0 to the 95th or the 5th to the 100th or....? The answer probably depends on why you are asking the question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.