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The total equity is then equal to $1,300 + $2,700 = $4,000 The short sell action 2 is split into ( 1, 1) where the first 1 offsets the long position The offsetting action causes a charge of the $50 transaction fee and, at the same time, credits the trade profit of $2,700 to the current cash balance giving $1,300 $50 + $2,700 = $3,950 This is the balance before entering the next trade, a short sale of an additional one contract, corresponding to taking the short position 1 at the point 3 Of course, both contracts are sold at the same price 455 However, with initial margin of $1,350 and the cash balance of $3,950, we have enough funds to sell two (the total action of 3 liquidates one and enters two new trades) but not one (total action 2) contract at point 3.

This is equivalent to saying that there is no directional preference in the uctuating eld This restriction leads to isotropic turbulence, which by its de nition is always homogeneous To illustrate the difference between the two types of turbulence, consider the RMS velocity uctuations In homogeneous turbulence, the three components of the RMS velocity can all be different, but each value must be constant over the entire turbulent eld In isotropic turbulence, spherical symmetry requires that the uctuations be independent of the direction of reference, or that all the RMS values be equal A bowl of peanuts or pretzels is isotropic in a two dimensional sense: It is the same no matter how you look at it or where you place the reference axis The same is true (in three dimensions) for isotropic turbulence.

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This is the case of the first reserve where the excess P&L at the point of new transaction is used for adding a contract The r- or l-algorithm points are used, but we sell short more contracts because the total equity has grown and the unused funds exceed the additional margin requirement The strategy that recognizes the first P&L reserve case is (0, 1, 0, 3, 2) Adding money management, the ability to change the number of contracts, to the potential profit strategy (0, 1, 0, 2, 1) creates the new strategy (0, 1, 0, 3, 2) which outperforms the parent.

The branches and leaves on a tree, on the other hand, have a speci c arrangement, so moving the axis changes the image The tree is highly anisotropic True isotropic homogeneous ow requires that there be no directional preference in the three dimensional ow There can be no mean velocity gradients, thus no shearing stresses All three normal stresses must be equal, and all nonnormal stresses (uv, vw) must be equal to zero If the ow has no directional preference and no coherent organized structures, there can be no correlation between components of the uctuating velocity The normal components (uu, vv, ww), on the other hand, will always be positive because they are squared terms Experimentally, such a ow can be obtained approximately in the turbulence developed behind a properly designed grid.

Indeed, P&L = $50 [cost of initiating long position 1] + $100 (455 428) [profit offsetting long position 1] $50 [cost of offsetting long position 1] 2 $50 [cost of initiating short position 2] 2 $100 (449 455) [profit offsetting short position 2] 2 $50 [cost of offsetting short position 2] = $50 + $2,700 $50 $100 + $1,200 $100 = $3,600 This is $3,600 $3,100 = $500 greater The extra gain comes from one more additional contract sold short at the point 3 as 1 $100 (449 455) = $600 The last amount is reduced by the initiating cost of $50 and the offsetting cost of $50 for this extra position In order to see where the second reserve case may be applied, we check the total equity at point 2 The current open position equity is equal to $100 (443 428) = $1,500.

This restriction excludes consideration of the trailing vortices in mixing vessels, which have a clearly de ned orientation; it also excludes ow anywhere in the tank where velocity gradients exist There is no possibility of seeing truly isotropic turbulence in a stirred tank While the fully isotropic assumption is not a good match to physical reality, the implications of isotropy are profound for turbulence modeling and measurements Isotropy allows the entire turbulent spectrum to be de ned from one component of uctuating velocity, because the ow is perfectly without directional preference It allows simpli cation of the equations to include only the normal stresses It also allows one to make spectral arguments to simplify the measurement of the dissipation.

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