J.L. Kelly

[five]

is smarter than me. I was reading his 1956 paper, and I cannot figure out some of his log derivations.

For example,
"Let us maximize G with respect to l. The maximum value with respect to the Yi of a quantity of the form Z = (Sigma)Xi log Yi, subject to the constraint (Sigma)Yi = Y , is obtained by putting
Yi = (Y/X)*Xi
where X = (Sigma) Xi."

How? What? Who? Huh?

[bigpooch]

Kelly may have been smart, but surely Euler, Gauss,
Leibniz and Riemann were geniuses! (And Ramanujan was
extremely talented as Hardy noted.)

Note that the (Sigma) in your post is just the summation
from j=1 to j=n for some fixed positive integer n. Also
note that the statement is trivial for n=1 for then there is
just one term. (I've changed indices from i to j because if
I bracket an "i", this is an editor command to put the text
in italics!)


Okay, first of all, the statement to maximize Z is true for
positive X[j]. ( I'll use this for X "subscript" j ) and
positive Y[j] and that is almost certainly what J.L. Kelly
meant. It's easier to look at n=2
(when the summation denoted by sigma is for j=1 to n=2):

[ Just so I won't have to type out so many brackets, let
X1=X[1], X2=X[2], ... ]

The maximum with respect to the Y[j] of

Z = X1 log Y1 + X2 log Y2

subject to the constraint Y1+Y2=Y

is obtained by letting

Y1 = (Y/X)*X1, Y2 =(Y/X)*X2

where X=X1+X2.


For the maximum of Z with respect to Y1, by calculus, the
derivative of Z with respect to Y1 is zero. (Well, there's
a bit more to that, such as Z is almost everywhere
differentiable, ...)

Now, Z = X1 log Y1 + X2 log (Y-Y1) since Y1+Y2=Y.

Thus, the derivative

dZ/dY1 = X1/Y1 - X2/(Y-Y1)

since d(ln(a+bx))/dx = b/(a+bx).

For any extrema (maxima or minima), dZ/dY1 = 0 implies

X1(Y-Y1) = X2*Y1 or X1*Y = (X1+X2)*Y1 = X*Y1
or Y1 = (Y/X)*X1 which is the desired result.

Admittedly, it is MUCH MORE DIDACTIC to say in the
statement

Y[j] = (X[j]/X) * Y
where X = sigma X[j] = X[1] + X[2] +...+ X[n]

To check that this is indeed a maxima, it is enough to note
that the second derivative of Z with respect to Y1 is

d(dZ/dY1)/dY1 = -X1/(Y1^2) -X2/(Y-Y1)^2 which is clearly
negative since the X[j] are positive and the denominators
are positive. Hence, there is a maximum at Y1 = (Y/X)*X1.

For n>2, the argument is similar and you can use
mathematical induction.

The intuitive aspect of this statement can be grasped when
you are choosing how to maximize e^Z which is just

(Y1^X1) * (Y2^X2) * ... * (Y[n]^X[n]) when the X[j]'s are
given and the sum of the Y[j]'s is Y.

For each Y[j], you pick a "proportion" of Y that is
proportional to how big the exponent is relative to the sum
of all of the exponents.


EXAMPLES
--------

1) How do you maximize the n-dimensional volume of a box in
n-dimensions if the constraint is that the sum of the
length, width, height, ... is fixed?

Here, e^Z = Y1*Y2*...*Y[n], so the "weights" should be the
same for each of the Y[j], so the answer is a square, a
cube or that n-dimensional box with edges of equal length.

2) You've got some metal to build a cylindrical can. How
do you maximize the volume? Let r be the radius and h be the
height.

Well, V = (Pi)(r^2)(h) and A = 2(pi)(r^2) + 2(pi)rh =
(2*pi)(r^2 + rh). The "trick" is to use the constraint as

r^2 + rh = A/(2*pi)

since the RHS is a constant and let y1=r^2 and y2=rh. Also,
the volume is just pi(r^2)h = pi*(y1^(1/2))*y2, so if we
are maximizing volume, we are just maximizing y1^(1/2)*y2
which will be the case when y2 has "twice the weight" as y1
or rh = 2(r^2) or h = 2r.

[five]

[ QUOTE ]
J.L. Kelly and bigpooch are smarter than me.

[/ QUOTE ]

FMP. But seriously, thanks for all the help Pooch. I constructed a basic gambling model to help me try to understand this all more, but ended up getting over my head.

Suppose we are betting on horses in a race. There are n horses competing. There are two probabilities associated with the race. P is the probability each horse will win, so P1 is the chance horse 1 wins, P2 for horse 2, ... Pn for horse n. Q is the probability associated with the payouts, that is, if you wager T dollars on horse 1, you win T/Q1 dollars, and zero otherwise. Also assume you can choose not to wager some of your money, and there is a small reward for that. Like your wife will give you a small amount for each dollar (call this r) that you don't wager on a horse, and call this amount T0.

Thus your strategy consists of (T0,T1,T2,..Tn), and I would like to find the optimal strategy, and was wondering if an optimal strategy could be created without T0 (or if this is a necessity).

So it appears the value of our strategy is:

T0(r) + (Sigma j=1 to n) Pj * log(Tj/Qj)

However, the sum of (Tj/Qj) no longer equals T (actually T/Q but Q = 1).

I tried the method used in your post to determine a formula for this, but I got that

Z = T0(r) + (Sigma j=1 to n) Pj * log(Tj/Qj) where n=2 can be written as:

Z = T0(r) + P1 * log(T1/Q1) + P2 * log(T2/Q2).

Z = T0(r) + P1 * log(T1/Q1) + P2 * log(T2/(Q-Q1)).

dz/dQ1 = (P1/(T1/Q1))*T1 + (P2/(T2/(Q-Q1))*(-T2) = 0

which reduces to Q1 = (Q/P)*P2.

I think I should be taking the derivative with respect to T, or somehow get that out of this derivative...

[bigpooch]

It seems intuitive that you find those j for which P[j]>Q[j]
and pick the j with the biggest value of P[j]/Q[j] since
you on average net P[j]/Q[j]-1 dollars for every dollar you
bet on the jth horse.

Obviously, if for all j, P[j]<=Q[j], you sit on the
sidelines and wait for the next race.

If there are two (or more) indices for which the ratio of
the P[j]/Q[j] is identical, you can pick weights in
accordance to how much risk you want to take: do you go for
a home run or take a likely win? If you want to minimize
the chance of ruin, you obviously take the largest P[j].

In any case, to optimize, you use your full bankroll to bet
if there is at least one horse for which P[j]>Q[j]. This
gives you the maximum +EV, but also you will be "ruined"
with probability 1-P[j], so maybe the "value" you give in
your post is the risk-adjusted value (I merely skimmed your
post!).

Practically, if the edge is much too small in accordance to
how risk averse you are, you may decide to sit on the
sidelines and wait for another opportunity.

I am not sure how the "value" is derived and why it was
necessary to have T0.

[five]

[ QUOTE ]

(I merely skimmed your post!).

[/ QUOTE ]

I don't blame you [img]/images/graemlins/smile.gif[/img]

[ QUOTE ]
I am not sure how the "value" is derived and why it was
necessary to have T0.

[/ QUOTE ]

The post you made makes sense if you know all the p's and q's. I guess I threw you off by not telling you all the information. Basically, I want to get into investing/finanace in some sort of capacity, and I wanted to ground myself in theory first. The real basic model I want to make is very similar to the horse racing model above, but may make more sense now.

You have a certain starting amount of money, Wo. The model will be a single period model. There are m different states the world can end up in. You can purchase m different assets, each with initial price 1.

The final value of the asset will be
Vi(Sj) = {Qi^-1 if i=j
0 otherwise }
where Vi(Sj) refers to the value of the asset in state j.

Here, all we know about p and q is that they are probability measure on the states of the world. (Does it makes sense to say that?)

Suppose you are an investor with a logarithmic utility function, and the portfolio you choose is H =(H0, H1, H2.. HM), where H0 is the amount you put into a risk free asset (at a base rate of interest) and H1, H2 etc are the amounts of the risky assets you purchase (risky because, you don't what state you end up in, and thus you don't know the terminal value).

Is it possible to construct an optimal portfolio here? If so, is it possible to construct it with H0 =0?

Thanks for the help bigpooch.