Question on logarithmic scale

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Let me start by saying my understanding of logarithms is relatively limited, with that out of the way on to my question.


The question I have is in a situation where you have 2 items that are inversely related and both have a starting point of 100%. How would you calculate the point where one begins to increase in value faster then the other is able to decrease ?
 

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  • #2
haruspex
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... 2 items that are inversely related and both have a starting point of 100%. How would you calculate the point where one begins to increase in value faster then the other is able to decrease ?
Do you mean literally that y = 1/x, or y = x-n, or is the relationship only approximate?
What exactly do you mean by rates of increase and decrease here? E.g. are they both functions of some parameter, y = y(t), x = x(t), and you want the point where the derivatives are different? Or do you mean rates of increase in more of a geometric sense, [itex]\dot{y}/y[/itex] versus [itex]\dot{x}/x[/itex]?
 
  • #3
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This is one of those is one of those areas of understanding where I am almost out of my depth so I will try to better explain what I am after ...but still keep it simple.

Assume we are working with 2x - $1 bills. One is increasing in value inversly to the other side that is decreasing in value.

What I am trying to understand is at which point the $1 that could theoretically increase to infinity would begin to increase in value faster relative to the other $1 that could only devalue to zero would be able to decrease.


My interest is in relative terms not absolute hence the mention of logarithmic scale.

Hopefully you can see where I am going with this.
 
  • #4
haruspex
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Ok, so you mean strictly y = 1/x.
In relative terms, they always change at the same rate. Every time x goes up 1%, y goes down 1%. This is entirely consistent with the fact that one may tend to infinity as the other tends to zero.
 
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At some point though wouldn't the losing side be losing 1¢ or less for every 1% change while the other side was gaining 2¢ or more?

I guess where I am going with this I understand that a 50% loss requires a 100% gain to recover the loss. I just wasn't sure if that was the point of no return where they crossed .
 
  • #6
haruspex
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Make up your mind - do you care about relative change or absolute change?
In terms of absolute change, the higher value will change by more each time than the lower one will.
But I have misled you in one regard. My comment about 1% changes wasn't quite true. I should have said infinitesimal changes, i.e. for smooth continuous change. If you make a step change then the increasing value undergoes the greater change. The bigger the step change in percentage the bigger the difference. As you note, if one value goes down 50% the other goes up 100%.
 
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Make up your mind - do you care about relative change or absolute change?
..Ah the perils of my ignorance. :wink:

I do appreciate you making the effort to help me sort this out.



I figured at some point from an efficiency perspective the amount lost reached a point where the percent gained required recover became unrealistic. That is what I was trying to find an answer for. For most people it may be a subjective decision, but I was just hoping there was a more definitive way to determine it mathematically.

Percent decrease / Percent gain required to get you back to even
5% - 5%
10% - 11%
20% - 25%
25% - 33%
30% - 43%
35% - 54%
40% - 67%
45% - 82%
50% - 100%
75% - 300%
90% - 900%
 
  • #8
haruspex
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I should have asked for practical situation that gave rise to the question. This is gambling /investment, yes?
Consideration of what has been lost is generally not useful. It's known as the fallacy of sunk cost. What matters is whether and how much you expect the stock to go up or down from where it is now.
In some other contexts, looking at percentage changes is not the most appropriate. E.g. in changing gears, the ratios give a clearer picture. 1:3 is inverse of 3:1. There are usually several options for how to turn real world processes into numbers, and some choices are more helpful than others.
 
  • #9
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Your right this is for an investment and is related to a delta neutral hedged position that basically is put into place to mitigate small variations in value change with the assumption that a large movement is coming, and you want to be in position to capitalize on it ...you just don't know which direction its going before it happens.

So in this situation I know ahead of time that one side of the position is going to lose money as the other increases in value. If structured properly the winning side will increase in value greater than the expense of the losing side. I was attempting to minimize my loses on the losing side by finding a logical point to close the losing position... a variation of cut your losses and let your winners run. I just couldn't quite ascertain mathematically where that crossover point was.

Until a decision-maker irreversibly commits resources, the prospective cost is an avoidable future cost and is properly included in any decision-making processes. For example, if one is considering preordering movie tickets, but has not actually purchased them yet, the cost remains avoidable. If the price of the tickets rises to an amount that requires him to pay more than the value he places on them, he should figure the change in prospective cost into the decision-making and re-evaluate his decision.

Your right it is similar to a sunk cost but just slightly different because I am trying to avoid them beforehand ...

Sunk costs are retrospective (past) costs that have already been incurred and cannot be recovered. Sunk costs are sometimes contrasted with prospective costs, which are future costs that may be incurred or changed if an action is taken.

Surprisingly this is kind of thing doesn't seem to be as widely discussed as I would have thought even in a lot of the financial forums. I thought if I asked in a math forum I might be more likely to get a cut and dry response.
 
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  • #10
HallsofIvy
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With y= 1/x, dy/dt= -1/x^2 (dx/dt). y will be increasing faster than x is decreasing as soon as 1/x^2 is greater than one. That is, when x is less than one.
 

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