How to represent the value of player production

[TooLiveBrew]

I apologize for the relatively vague thread title. I think my question is fairly simple, but I don't trust my own mathematical 'abilities'.

 

Is the best way to think of the value of player production an exponential curve, or is it more linear? I ask, because obviously an OPS of, say, .750, is far more common than an OPS of 1.000. I hope this makes enough sense for others to understand what I'm asking. I swear it's a valid question inside my head.

[Baldkin]

It depends what exactly you're trying to represent...

 

Showing all the data for the entire league, you'd end up with some sort of a bell curve... the terrible and the great being few, and the masses being between .700 and .850 OPS.

 

As for player value, using something like WAR, it would actually be linear. Every WAR is worth approximately the same so a 6 WAR player is worth 6x as much as a 1 WAR player.

[sbrylski]

Here's one study that kind of gets at the question: http://baseballanalysts.com/archives/2009/05/findings_from_t.php

 

I only read the first half, and not the part about pitchers velocity, but it seems to conclude that in terms of how much they are paid, player value is linear. Of course, that might just be because it is an inefficient market, and there's an advantage to be gained here through proper analysis.

 

(And here's a quote from Bill James used in the linked article that might state what you did in the first post a little better: "Talent in baseball is not normally distributed. It is a pyramid. For every player who is 10 percent above the average player, there are probably twenty players who are 10 percent below average.")

[rluzinski]

According to Tom Tango (tangotiger.net), teams pay linearly for wins above replacement. A guy who's 4 wins over a AAAA player is twice as valuable as a 2 win player. While the 4 win player is much more rare, it doesn't appear to translate into exponentially more value. I think there's just enough slots to fill that a 0 - 2 WAR upgrade is always available. If a team had only 3 slots to fill, I think the 4 win player would be worth non-linearly more.

[sbrylski]

I think there's just enough slots to fill that a 0 - 2 WAR upgrade is always available. If a team had only 3 slots to fill, I think the 4 win player would be worth non-linearly more.

 

Can you clarify or reword this?

 

I have read that value is paid linearly. I don't know the why, because intuitively I'm with TLB. You appear to be stating the why here, but I don't follow. (Maybe because its 3 AM.)

[rluzinski]

A 4 WAR player is more rare but you can just use two 2 WAR players to get the same value. It's rare to find a team that already has all 2 WAR players. If a starting lineup was just 2 batters and 1 pitcher, I'm speculating that a 4 WAR player would be exponentially more valuable than a 2 WAR player.

 

Now, it still gives you a little more flexability to have one 4 WAR player over two, 2 WAR players, so it's reasonable to conclude the 4 WAR player is worth more than twice a 2 WAR. Tango has concluded that teams don't pay more than twice. Perhaps injury related risks of having so much value tied to one player compensates for the small flexibility advantage?

 

Not sure if I explained that any better the second time. http://forum.brewerfan.net/images/smilies/smile.gif

[sbrylski]

No, that makes more sense. Thanks.

[TooLiveBrew]

Thanks baldkin, sbrylski, & rluz. I knew this was the right place to ask this question. I guess what I really was trying to ask was answered by baldkin (the bell curve), but he also raised an interesting angle that I hadn't really considered (player compensation). A short thread, but a great one. Thanks again.

[rluzinski]

This might be obvious but I'll point it out, regardless. The MLB distribution of something like OBP, might look roughly like a bell curve but it's more complicated than that. In terms of skill, the pool of players MLB draws from is from one extreme tail of the bell curve. Even bad major leaguers are 3+ standard deviations away from the average skill level of the population of adults playing organized baseball. So it's really more like a pyramid. It's just that the AAAA type major leaguers don't get as many PA's, which somehwat cuts down the fat end of the pyramid.

 

I'm not sure what the resulting distribution looks like exactly. Can we use BR.com to estimate the # of PA in 2009 that was between a certain OBP?

[rluzinski]

This might be obvious but I'll point it out, regardless. The MLB distribution of something like OBP, might look roughly like a bell curve but it's more complicated than that. In terms of skill, the pool of players MLB draws from is from one extreme tail of the bell curve. Even bad major leaguers are 3+ standard deviations away from the average skill level of the population of adults playing organized baseball. So it's really more like a pyramid. It's just that the AAAA type major leaguers don't get as many PA's, which somehwat cuts down the fat end of the pyramid.

 

I'm not sure what the resulting distribution looks like exactly. Can we use BR.com to estimate the # of PA in 2009 that was between a certain OBP?

[rluzinski]

Just did the 2009 AL, non-pitchers. Used 1.8OBP+SLG. Looks like a bell curve but it's not! http://forum.brewerfan.net/images/smilies/smile.gif I'd like to see what the distribution looks like for actual skill (not just sample stats).

[endaround]

Why wouldn't it be? Just because you draw from a given population doesn't mean the conditional distribution wouldn't be approximately normal. And that's what you are looking at.

[Baldkin]

We're drawing from the set of MLB players for the data represented....not the set of every person on the planet.

 

Those two data sets should look different.

[rluzinski]

We start with a tail of a normal distribution (the most skilled players in the world). The players that are better get more PA, but injuries are proportional to playing time. The end result is something that approximates a normal distribution. Just thought that was odd/interesting.

[endaround]

But you have enough observations that you're in CLT territory.

[rluzinski]

Please elaborate.

[endaround]

Central limit theorem. Since you're basically dealing with a type of binomial distribution (or a series or bernoulli experiments) it will end up asymptotically normal.

[rluzinski]

I guess that I will need a more in-depth explanation to wrap my brain around this one. I don't understand how this ends up as a kind of bernouli experiement. I see it as more of an illustration of selective sampling. That said, all I know about probability theory is what I learned from basic stat courses in college and Google. An endaround 1 sentance explanation isn't going to be enough, sadly http://forum.brewerfan.net/images/smilies/smile.gif