Why Averages Are (Almost) Always Wrong: Jensen's Inequality and the Flaw of Averages

[@jonmacdonald1413]

This analysis is very insightful. The concept is on full display in financial markets with daily leveraged ETFs where the average return of the underlying instrument(s) over time is meaningless. It is the distribution of daily returns of the underlying that determines the ETF return.

[@RiskByNumbers]

Great example. The original plan for this video was providing the example of the value of a financial option, so it is great to hear that this concept resonated with your past experience!

[@Mike-oz3ox]

@@RiskByNumbers In fact, ignoring transaction costs and fees, etc, the excess return on a 2x leveraged ETF expressed as a factor on the original investment is roughly the SQUARE of the excess return (also expressed as a factor) on the underlying reduced by multiplying by function of the total-variance (

[@Koroistro]

The intuition that I came up with is that from 10 to 20 you're going twice as fast, but from 20 to 30 you're only going 1.5 times faster. Therefore the impact on time spent by going slower is higher than the impact of time saved by going faster. Thanks for the content!

[@RiskByNumbers]

Thanks, @Koroistro for the feedback and for providing this nice explainer to others!

[@whataboutthis10]

Yep, also the computed 'average' of velocity "20" is per unit length. Lower speeds yield longer time at that speed, greater speeds yield shorter time. Travel average is however an average over _time_ so such average speed is always lower - because time spent with different speeds is inversely proportional to the speed.. When probabilities are given per 'equal length', the _harmonic mean_ has to be used to determine travel average, not arithmetic mean which gives the expected value, let's see this example: as the probabilities stand, the quarter of _path_ is done with 10, half the path with 20 and quarter of the path with 30mph. Since 20 is exactly twice as likely, we simply use it twice. Average speed of "10,20,20,30" is then 4 / ( 1/10 + 2/20 + 1/30) which is 4*30/7 = 20*6/7 mph. For 20 miles the travel then takes 7/6 h Do check about the harmonic mean, it gives the average speed when sectional speeds are known for equal lengths. Eg going up the hill 10 and down the hill 30, the average isn't 20, it's 2 / ( 1/10 + 1/30) = 15 enjoy and good luck!

[@Lolwutdesu9000]

That intuition doesn't help, actually.

[@jadolphson]

Many (typically discrete) systems don’t even accept average in, nor can they ever output the average out. For example, I could never roll a 3.5, and you would never need to pay me $12.25 or $15. This is true even for linear systems.

[@RiskByNumbers]

This is a wonderful comment, @jadolphson! Thanks for pointing that out. When I re-watched the video prior to posting it, my last thought to myself around a possible edit was: "maybe I should point out that a mean of 3.5 is not actually feasible, highlighting an important point...". I didn't make that change, but perhaps that was a good thing; it allowed for great comments like this one. I always find it much more straightforward to explain discrete rather than continuous cases. If a nice, simple continuous situation comes to mind to you, don't hesitate to let me know, and I'll see if there is a way to briefly touch upon it in a future video. Thanks again!

[@jadolphson]

@@RiskByNumbers Dropping a ball from a random height and measuring the time to impact?

[@百合仙子]

yes you can't have that value in, but you can have nearby values in. it's no problems to use averages to understand a discrete system, as long as it is used correctly.

[@m.fazlurrahman5854]

Better stick with below average right!! Always under the umbrella… because above average will never need a middle-man!!

[@RiskByNumbers]

@@jadolphson I like it! I still struggle with trying to make densities as clear as possible, but I'll keep this suggestion in mind for the future. Thanks again.

[@ClearerThanMud]

This came up for me at work in the context of agile software development. The agile instructor explained that the estimated difficulty of fixing a problem would be coded as follows: 0 = minutes, 1 = hours, 2 = days, 3 = weeks, 4 = months, 5 = years. Then he went on to say that we could use the average of the codes for all of our bugs to estimate how long it would take to fix them all. Say what??? You can''t do that; the encoding is not linear! In general, avg(f(x)) = f(avg(x)) only if f is linear. He didn't understand what that meant, unfortunately, and thought I was just being a PITA. OK, imagine you have 1000 bugs, all 0s except for one 5. The average difficulty estimate is going to be very close to 0, but it will take you years to fix them all!

[@RiskByNumbers]

@ClearerThanMud: this is such an excellent example. Thanks for sharing!

[@mutanttipossu]

I wouldn't be so quick to dismiss the idea! The proposed metric lies between the median and the actual average, which seems useful! The actual average might be of little interest when you know the total time. The median might be bad since it goes to 0 faster. It's a tradeoff between two location measures.

[@ritwikism]

"Average speed" is universally understood as the true average though, and not a mean of multiple speeds over time so the beginning of the video is confusing to me

[@ryanlohbrunner7760]

What is “true average”? Wouldn’t that literally be a mean of multiple speeds over time?

[@acidnik00]

@@ryanlohbrunner7760 if you were driving for 1 hour with a speed of 20 km/h and last 2 seconds with a speed of 100 km/h, would you say that you average speed is (20 + 100)/2?

[@TheRealBroodax]

You're thinking of the average speed of a trip, which by definition would be the actual average speed of that specific trip. The video would make no sense at all if that's what he was talking about. He's talking about a trip with an unknown true average speed, but whose average speed is some distribution based on traffic that day, etc.

[@OMGclueless]

The speed given at the start of the video is not "average speed" it is "expected average speed". In fact, it was just written as "expected speed" omitting the words average entirely in the video graphics, because it's not the important bit. The important word is "expected". "Average speed" is just total average speed of a trip computed the natural way as you say. "Expected average speed" is the mean of those calculated average speeds over many trips. This also gives a useful intuition as to why they are different: "average speed" of a trip is a time-weighted average of your travel-speeds. "Expected average speed" is not a time-weighted average, it's just one data point per trip regardless of how long that trip took.

[@RiskByNumbers]

It is wonderful to see all of these great comments. Apologies as well for just chiming in -- we just got back from the hospital with our newborn, so a bit sleep deprived. @OMGclueless: spot on! I subtly mentioned 'expected' value at the start of the video to try and highlight a really important point. The motivation for this video (outside of trying to distill Jensen's inequality in an understandable manner) stemmed from some of my past consulting experiences I've done outside of my university day job. I have found that there is a tendency to frequently only discuss and use 'expected values' in making decisions without recognizing that, for non-linear systems, it may be quite important to know the underlying distribution for your random variable of interest. My hope is that this video helps clarify that point. The original motivation for this video was going to be determining the value of a financial option (e.g., call or put), but I realized pretty early on that it would make more sense to use an example familiar to most everyone. I might, though, come back to that in the future (ideally when I put together a couple of videos around reinforcement learning).

[@hdthor]

I think a better lesson would be that the weights matter: equal weighted, time weighted, and distance weighted mean speeds will all be different. And only the distance weighted mean speed has the property that its reciprocal multiplied by distance equals the time taken.

[@RiskByNumbers]

@hdthor: this is a wonderful comment. A couple of others have also mentioned that the harmonic mean is something worth bringing up (I've discussed the geometric in the past), so I'll think about how to do so in future videos. Thanks again for the great comment and feedback -- definitely helpful food for thought!

[@33gbm]

The beginning of the video is not fair. 1.167 hours is not the answer if you only have the information about the expected value for the velocity, that will follow an unknown distribution. Indeed if you are driving at about 20 mph, under certain assumptions, 1 hour is an approximation for the time taken to cover 20 miles.

[@RiskByNumbers]

@33gbm: absolutely spot on and correct! One of the motivators for this video was that I have worked quite a bit with industry over the years and found that there is a tendency to avoid modeling uncertainties and to operate in a deterministic world. The consequences of doing so are not necessarily intuitive or apparent, particularly for more complex problems. Therefore, the goal was to show the possible consequences in a more straightforward example. Great job and catch! - Omar

[@33gbm]

​@@RiskByNumbersthanks for the reply. Nice to hear from you. By the way, the discussion about the problem is very well presented and I hope to see a lot more from you! 😊

[@RandomBurfness]

"Think about subscribing to the channel." You already earned my subscription when you said we were rolling a singular die, and not """singular dice""".

[@RiskByNumbers]

Haha, @RandomBurfness! Perhaps next video I’ll try to throw in the word ‘datum’. Appreciate the note and subscription!

[@DenisZlokazov]

In the example provided at the end of the video let's see what happens with travelled distance. So, 15 minutes at 10 mph give us 2,5 miles. 30 mins at 20 mph - 10 miles. 15 mins at 30 mph - 7,5 miles. And that gives us 20 miles travelled in 1 hour with an average speed of 20 mph. I understand the concept, but either you assume that distance is linear function of time with speed as a coefficient or you can't just say "the probabilities of finding 10, 20 and 30 mph are distributed the way shown". And average speed in physics is total distance to total time by definition.

[@RiskByNumbers]

Thanks, @DenisZlokazov, for the comment! As I mentioned in a couple of responses, I was a bit hesitant around this example given the understanding of 'average' in a physics sense. Here, I do mean average speed as the total distance between 2 points, d2-d1, versus the total travel time, t2-t1. By its expected value, we can imagine three scenarios (given a deterministic distance): the travel time will be 120 minutes (10 mph), 60 minutes (20 mph), or 40 minutes (30 mph). With the travel times listed above, it becomes quite obvious what the expected travel time should be. Thanks for the comment and feedback.

[@elunedssong8909]

The videos not bad, but its mostly just a phrasing thing. If you had initially phrased the question: Given that a destination is 20 miles away, and given the average speed across many different journeys to that same destination is 20 miles per hour, what is the average journey's time to the destination, of all possible journeys. Then, i doubt everyone would jump to answer 1 hour. Plenty would say, who knows, or phrase "1 hour?" as a guess. Here is a much simpler explanation: If every 100 journeys you end up going, a single journey goes at a rate of 1/(20/100)th a mile per hour, then that one journey would take 100 hours. But we know, that the sum of all 100 journeys needs to be = 100. Therefore, adding that single case eliminates our conclusion from being possible. It is also the case that we can validly add a 1/5th mile per hour average into the distrbution of average travel speeds, as the average of all miles per hour needs to be 20, and we can simply add 20+(20-1/5) and have the distribution of average travel speeds already back exactly on expectation with a single next journey. I'm also not at all convinced we should be talking about "most" things following or not following this type of rules. I would instead say, when does it make sense for a distribution to be equally weighted(the actual mechanism behind when this works or not If there were a situation where someone did not drive past any red lights to a specific location, then there are only a few other red-light ish factors, like when you come to a turn is someone already occupying the road in the same direction. It's realistically possible for some drivers, going to some regular destinations, the mean of average speeds matches the mean of travel times. I would personally have delved into how to identify, without direct observation, what might a distribution look like, and then from that, how we might be able to make judgements about the relationship between the two averages. For instance, the squared distribution, obviously has its "weight" distributed to the right. So if we take a equally weighted distribution as an input, we should expect that to line up to the left of the true mean. No math needed!

[@NicolasChanCSY]

0:24 In the caption, it should be the "flaw" of averages, instead of "law". The explanation is simple and intuitive. Thank you!

[@RiskByNumbers]

Thanks, @NicolasChanCSY! Just updated the subtitle -- appreciate the comment and positive feedback!

[@DannyTobin-b2g]

Really nice job! Clear, concise, and simple enough while not shying away from the key terminology or notation. Would love to see more videos like this from you

[@RiskByNumbers]

Thanks, @DannyTobin-b2g! Great to hear from you, and glad to hear that you found that this struck the right balance. I'll keep those thoughts in mind as I put together the next video.

[@mgostIH]

I liked this, it helped me understand jensen's ineq geometrically which will help me remember which way the direction of it goes and also makes for a good example to show people new to the topic! On this note, I recently discovered that a similar problem appears on common statistical estimators, for example we have an unbiased estimator of the variance (n-1 in the denominator) but there's no general unbiased estimator for the standard deviation! Taking the sqrt of your estimated variance will lead to an underestimate for this same reason due the concavity of sqrt 👀

[@RiskByNumbers]

@mgostIH: thanks for this comment! This is a fantastic example of Jensen's inequality. Thank you for pointing it out. The more you delve into probability, statistics, and machine learning, the more you realize how much Jensen's inequality shows up in many important proofs and theorems.

[@andrashorvath2411]

Amazingly clear explanation, you didn't make too big jumps without explaining it which is not a common thing (of course hard to do as well). Thanks.

[@RiskByNumbers]

@andrashorvath2411 really appreciate the kind note and message! It means a lot. Cheers! -Omar

[@florafeldner]

this is a brilliant video, with a good example and reasonable length of only 10min that explains the issue at hand well. subscribed! :)

[@RiskByNumbers]

Thanks, @florafeldner! Really appreciate the kind note and message. Thanks for subscribing, too!

[@ReneKnuvers74rk]

1:47 faces 3 and 4 are adjacent. This is not true on normal d6-dice as opposite faces always sum up to total 7.

[@RiskByNumbers]

Thanks for catching this, @ReneKnuvers74rk! You know, I kept looking at this die thinking 'something looked off...' but could not put my finger on it. This is a good point that, in the future, I should be willing to spend a few bucks for a stock image and spend my time solely on the animations. Thanks again!

[@user-wr4yl7tx3w]

The introduction to motivate was excellent

[@RiskByNumbers]

Why thank you! Appreciate the kind comment.

[@theondono]

I wonder if it’s just me, but I think this is easier to observe by simply going over operations on random variables. It’s clearer to me that both sum and integral will behave this way, so a moment defined with them will as well. Sums of averages work, but multiplication/division fails.

[@RiskByNumbers]

Thanks, @theondon, for the feedback! Great idea and food for thought.

[@vlad_objective]

Thank you! This is actually very helpful 👍👍

[@RiskByNumbers]

Thanks, @vlad_objective! Very much appreciated.

[@TimschneiderSchneider]

Which of the two function does represent the correct value, f(avg(x)) or avg(f(x)) ? Or are both wrong ? Because the distribution is unknown ? And if so, is there away to calculate the correct value without a simulation ?

[@RiskByNumbers]

Great question, @TimschneiderSchneider! As highlighted at the start of this video, we are oftentimes interested in E[f(x)]. We want to know the distribution around some response, say 'Z', which is uncertain due to uncertainty for 'X'. In this video, we can directly determine the distribution for 'Z' (i.e., travel speed) by simply mapping the probability of each possible outcome for 'X' to 'Z'. However, that is not always the case. Or, it may be possible, but the problem we are looking at is quite large. Simulation is what I've found to be the quick solution in these cases. However, analytical approaches do exist, such as the use of Taylor expansions. Again, very good question(s).

[@MrHaggyy]

Mhm the speed example is excelent but tricky at the same time. If you do the physics on paper you always have an absolute value, if you look at the dash of your car you always get the average speed that got you from A-B. Speed is usually not experienced as a mean. Also once you do statistics you get a very weird looking probability function. You can't get faster without going illegal, you get a view local maxima due to intersections or common jams. And you never drive at average. You almost always faster as it needs to average out stopping times.

[@RiskByNumbers]

@MrHaggyy: thanks for the note -- you brought up some wonderful points.

[@theupson]

for the many variations where f and f' are monotonic over the range of relevant values of x, the cookie cutter proof for the relative size of E(f) and f(E) revolves around a first order taylor series with remainder. lots of results (the harmonic mean being smaller than the algebraic, et al; the economic idea of risk aversion) can be spun out of applying this to a suitable f(x)

[@RiskByNumbers]

@theupson: love this point. I debated if it would be worthwhile to delve into that towards the end of the video, but the video was already feeling a bit long. Great point again, and I'll see if I can bring this up in a follow up video. Thanks!

[@Rdffuguihug]

Superb content. Keep it going, brother.

[@RiskByNumbers]

@Rdffuguihug: many thanks - cheers!

[@boltez6507]

Hey nice video,but before/after the formal proof you could have just shown why does this happen,its easy to visualise why the equality would hold for linear functions and why it would deviate for convex/concave functions.

[@RiskByNumbers]

Thanks for the positive comment, @boltez6507, and feedback. Really do appreciate it.

[@fibbooo1123]

Very well done! For travel speed, I'm interested in using harmonic mean as opposed to arithmetic mean. But I couldn't figure out a way that this generalizes to that

[@RiskByNumbers]

Thanks for the comment, @fibbooo1123! Here, you could compute the inverse of the expected value of 1/X. E[1/X] = 1/4 x 1/10 + 1/2 x 1/20 + 1/4 x 1/30 = 7/120. Therefore, 1/E[1/X] = 120/7. We can then plug in our distance and 20/(120/7) = 140/120 = 7/6. Hopefully I've made no mistakes (welcoming a newborn to the family this week, so I've been a bit out of it...). Thanks again for the comment, and great catch around the relationship and importance of the harmonic mean!

[@fibbooo1123]

@@RiskByNumbers looks right to me, thanks!

[@whataboutthis10]

@@fibbooo1123Harmonic mean works for average, when sectional speeds are known for equal lengths. This is actually the case here, where "1/4,1/2,1/4" probabilities determine the distribution along the *path* not over time. Because it's not "1/4 of time the speed is 10" etc, it's "1/4 of the path the speed is 10". Sure the speed 20 is twice as likely, but that's a nice multiple - so just use it twice, basically the overall average speed is the harmonic mean of "10, 20, 20, 30"

[@darrenlefcoe]

In summary, the expected value of non linear systems will be different to the expected value of linear systems.

[@denizersoz7012]

Isnt there a confusion here between 'average' and 'expected' speed ? Average speed by definition comes from total distance over total time. So the example to begin with is not correct I think.

[@RiskByNumbers]

@denizersoz7012: thanks for the clarifying question! Indeed, I found myself nervous using a speed example in a discussion around 'averages'. By 'average speed', I do mean total distance over time. By 'expected average speed', I mean the expected value across all possible average speeds. Here, we have three possible average speeds: 10, 20, and 30. As some have mentioned, this case is a great problem to highlight the harmonic mean. I'll think about how to fit that into future videos. Feel free to reach out to me directly to continue the conversation -- Omar

[@leo_tra]

Interesting video. Can you recommend any literature on this topic, please? Aside from the example in the video, where/how can this idea be applied in real life? Like business, engineering, social studies...

[@RiskByNumbers]

@leo_tra: great questions! First, in terms of applications, I'll note a couple of items. Jensen's inequality can show up quite a bit for certain proofs important in the areas of probability, statistics, and machine learning. A common example is Kullback-Leibler divergence (where we are measuring the difference between 2 distributions), where Jensen's inequality can be used to prove its non-negative property. For myself, the more interesting application of this idea is in the modeling of dynamic systems. A very intuitive example that I use to motivate one of my courses is the construction of a parking garage to maximize financial return. It costs you a certain amount of money to build each floor. Your revenue is based on demand for your parking facility, though there is a capacity constraint. You now want to balance your added revenue with each floor versus its cost to build. For an average demand, E[X], I can determine the design that maximizes my profit, f(E[X]). However, the expected profit, E[f(x)] is likely lower for that design when we introduce uncertainty. If demand is low, we are making less than we thought. If demand is high, the capacity constraint comes into play, and we can't take advantage of 'good times'. The solution then, in this uncertain system, may be to create a garage that is smaller (so that, if demand is low, we spent less upfront) but with beefed up columns and a foundation (so that, if demand is high, we can take advantage of that high demand). The reason that I like this example is that it does not require a background in optimization, reinforcement learning, etc. to appreciate that recognizing uncertainty allows you to identify adaptable policies that allow you to do well in the real world. As I worked out the script for the above example, I realized it was getting to be a bit much, which is why I just did this travel speed example. In terms of references, Warren Powell has provided excellent references on the topic, and he does a good job of trying to unify the perspectives of those working in the areas of optimization and reinforcement learning. Feel free to shoot me an email if you'd like to know more -- I'm planning to expand on the topic soon. Cheers -- Omar

[@leo_tra]

@@RiskByNumbers Thank you for the details. I`ll try to go through Powell's work and if I have further questions I`ll message you.

[@DJWESG1]

"imagine a river bed of pebbles, now imagine the agerage pebble weighs just 5 grams, whats the chance of you putting your hand in and grabbing a 5 gram pebble?" - jung (as best as i can remember)

[@RiskByNumbers]

I believe it comes from "The Undiscovered Self": "If, for instance, I determine the weight of each stone in a bed of pebbles and get an average weight of 145 grams, this tells me very little about the real nature of the pebbles. Anyone who thought, on the basis of these findings, that he could pick up a pebble of 145 grams at the first try would be in for a serious disappointment. Indeed, it might well happen that however long he searched he would not find a single pebble weighing exactly 145 grams."

[@DJWESG1]

@@RiskByNumbers was this written b4 or after the mathematics you referenced in your video??

[@PeterZaitcev]

The problem with that video is the definition of average speed which is exactly total distance divided by total time.

[@RiskByNumbers]

@PeterZaitcev: thanks for the comment -- very much appreciated. I mentioned a couple of points in other comments that I'll mention here. Indeed, when I put together this video, I got a bit worried about using the term 'average' speed in the same video where I would be discussing expectations/means. Here, I meant by expected average speed as travel distance over time, as you noted, but considering that there are 3 possible situations. There is a 25% chance that total distance to total time is one ratio, etc. Now, the original problem was going to be a case of modeling a financial option, as I find them to be a great example of this idea. I actually will buy one each term in my class to motivate the topic and keep folks in the class engaged. They require knowing quite a few terms, though, hence I switched things up. Very much appreciate the feedback -- it is helpful as I think through ways to improve things in the future. Thanks again.

[@maths.visualization]

Can you share video code ?

[@RiskByNumbers]

@maths.visualization: great to hear from you. I'll work on cleaning up the code on my end and eventually share it on GitHub. We are welcoming a new member to the family this week, so apologies ahead of time for the delay!

[@百合仙子]

How can you define average speed as the average of numbers you get when you periodically check your speedometer?!

[@RiskByNumbers]

@百合仙子: thanks for the comment! I definitely see the confusion with using 'average speed' in the same video that I'm discussing 'expected values/means/averages'. Here, I meant that the average speed is the ratio between the distance travelled between 2 points relative to the time it takes to complete that trip (i.e., (d2-d1)/(t2-t1)). The 'expected average speed' is meant to be the expected value of the parameter 'v' knowing that the term 't2-t1' is uncertain. Hopefully that clarifies the confusion.

[@百合仙子]

@@RiskByNumbers thanks for the explanation, but if you define the average speed as the ratio like in textbooks, there will be no issue to calculate the time, no matter if it is actual values, or expected values. If the time is uncertain, then the expected average speed is uncertain too, just a different range of uncertainty. I expected to hear about the pitfalls of using averages in statistics when starting watching this video btw.

[@whataboutthis10]

@@百合仙子 it's still well defined, just a bit confusing because there's certain ambiguity to what 'average' can mean. Instead of checking the odometer periodically in *time* - average would be 20*6/7, the 'average' "20" is obtained by checking the odometer periodically according to *distance* or some road signs, eg every 100meters

[@winchesterdown]

Does this mean average speed cameras over or underestimate your speed?

[@RiskByNumbers]

@winchesterdown: this comment caused me to imagine myself successfully arguing against a future speeding ticket based on some obscure math/proof. On a more serious note: I would imagine that an average speed camera measures your 'average speed' by computing the ratio of distance over time for 2 different points. Therefore, if your instantaneous speed was not constant over that distance, then in theory your instantaneous speed at some point was higher than that average. I suppose that I wouldn't mention this fact in court...😂

[@winchesterdown]

@@RiskByNumbers haha. Yeah I thought about it a bit and came to the same conclusion as you.

[@wstaempfli]

Great work

[@RiskByNumbers]

Thanks, @wstaempfli!

[@vitalysarmaev]

Bravo! 👏

[@RiskByNumbers]

@vitalysarmaev many thanks!

[@lachlanperrier2851]

Wonderful video thankyou!

[@RiskByNumbers]

Thanks, @lachlanperrier2851! Your note is much appreciated.

[@BigDBrian]

If average in does not lead to average out... then you're using the wrong average.

[@berlinisvictorious]

Another banger to watch while I eat

[@RiskByNumbers]

Haha, cheers and thanks, @berlinisvictorious!

[@DanielHsHu]

The car example in the video is incorect. Average speed of 20, means that in the past many drivers rode this 20 km, and they did it in exactly in 1 hour. So your expectation is 1 hour of drive, and the video is wrong. However, if you define an average speed as (many drivers look instantly on their speedometer and report the number, and i averaged this and got 20) then the expected time travel would be more than 1 hour. But this is not how average speed is defined.

[@lonjil]

should've used the harmonic mean :p

[@RiskByNumbers]

@lonjil: I have gone over geometric means in a past video, but I really should also find a point to bring up the harmonic mean (and their application). Thanks for the comment!

[@orterves]

It's so obvious when you put it like that

[@RiskByNumbers]

@orterves: that's wonderful to hear -- much appreciated!

[@atreidesson]

Okay, so. Referring to functions not all being straight lines as "flaw of averages" is really messed up.

[@peterkiedron8949]

Who you trying to dazzle with your bs? Just show that when f(x) is nonlinear usually avg[f(x)] is not equal to f(avg(x)).