Exponential Growth !! and Logarithmic Growth ?? (long)


Posted by Richard Reid on July 01, 1997 at 21:33:57:

Greetings:

I think that a clear explanation about how some of the basic growth formulas are arrived at would be very helpful, not only to clear things for myself, but perhaps for others as well.

What I mean by detailed explanation and underlying rationale is as follows:

Let's take the basic formula for "exponential growth", Bf = ((1+f)^W)*((1-f)^L) * Bi ,
Where Bf is the final bankroll,
Bi is the initial bankroll,
f is the fraction of the bankroll that is bet,
W is the number of wins, and
L is the number of losses

How was this "exponential growth" formula arrived at?

The (1+f) part is what happens after a win. In other words, after one win the bankroll would be increased by the fraction (f) that one bets. So, with one win, Bf would equal the original bankroll plus the fraction of the original bankroll that has been bet. Mathematically this can be written as Bi + Bi*f. Factoring Bi out of the equation gives us, Bi * (1+f). Let's call this amount B1. With a second win, the new bankroll (B2) would be again increased by the fraction (f) times B1. So we would have won an amount equal to the B1 * (1+f) which in turn equals Bi * (1+f)(1+f) = Bi*(1+f)^2. For every win, we multiply the current bankroll by (1+f). Therefore, it is quite reasonable for the formula Bf = ((1+f)^W)*((1-f)^L) * Bi to have the first term (1+f)^W as part of the equation.

How about the losses?

After one loss the initial bankroll (Bi) would be decreased by the fraction (f) that a person bets. So, with one loss, the resulting bankroll (Bf) would equal the original bankroll (Bi) minus the fraction (f) of the original bankroll that has been bet. Mathematically this can be written as Bi - Bi*f. Factoring Bi out of the equation gives us, Bi * (1-f). Let's call this amount Ba. With a second loss, the new bankroll (Ba) would be again decreased by the fraction (f) times B1. So we would have lost an amount equal to Ba - Ba*f = Ba * (1-f) which in turn equals Bi * (1-f)(1-f) = Bi*(1-f)^2. For every win, we multiply the current bankroll by (1-f). Therefore, it is quite reasonable for the formula Bf = ((1+f)^W)*((1-f)^L) * Bi to have the (1-f)^L as part of the equation.

Why do we multiply the two terms ((1+f)^W)*((1-f)^L) instead of taking their difference ((1+f)^W) - ((1-f)^L)?

Let's see why. After one win, we know that we end up with Bf = Bi * (1+f). Let's call this Bx. Now, if we have a loss instead of a win on our next play, we have to subtract the fraction (f) that we bet of the current bankroll (Bx). In other words, we get Bf = Bx - Bx * f . Factoring out Bx gives us Bf = Bx (1-f). But Bx = Bi * (1+f). So we can say that Bf = Bi * (1+f) * (1-f). This is why the terms are multiplied. Now, as the wins and losses accumulate, the formula becomes Bf = ((1+f)^W)*((1-f)^L) * Bi .

I hope this explanation is clear for the basic formula for "exponential growth." The point of all my rambling is that somewhere I have missed a similar logic to the formula for "logarithmic growth."

So I ask, "What is the general formula for logarithmic growth?" and can someone please give a similarly detailed explanation of its rationale?

Sincerely,
Richard Reid


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