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J.L. Kelly
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? |
#2
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Re: J.L. Kelly
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. |
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Re: J.L. Kelly
[ 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... |
#4
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Re: J.L. Kelly
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. |
#5
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Re: J.L. Kelly
[ 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. |
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