“Liquidity Alpha”
Nov 27th, 2007 | Filed under: CAIA Alternative Viewpoints Columns, CAPM / Alpha Theory, Guest PostsALTERNATIVE VIEWPOINTS …powered by CAIA
As part of our on-going series of features written by holders of the Chartered Alternative Investment Analyst (CAIA) designation, we are pleased to bring you this piece by Ranjan Bhaduri, Ph.D., CFA, CAIA. Dr. Bhaduri is Vice President in the Graystone Research group at Morgan Stanley. Prior to this, he was with a multi-billion dollar capital management firm where he was involved in all aspects of its fund of hedge funds and structured finance business. He has also held advisory roles at the East-West Center, a leading think tank on the Asia-Pacific region and has taught finance and mathematics at several universities. He is the author of several articles on advanced risk management techniques and hedge fund issues and is a member of the American Mathematical Society, the Mathematical Association of America and the Global Association of Risk Professionals. Dr. Bhaduri also serves on the Advisory Council of the World Trade University.
Dr. Bhaduri has just returned from a speaking tour that took him from Chicago to London and Beijing where he addressed audiences on the role of liquidity in hedge fund returns.
“Liquidity Alpha”
By: Ranjan Bhaduri, special to AllAboutAlpha.com
The word liquidity gets bandied about quite a lot, but it is surprising how many portfolio managers take a naïve approach to liquidity. It is well known that one should be compensated for investing in less liquid instruments (liquidity premium), but how much? What is the value of liquidity?
It is dangerous in merely trust one’s intuition on the value of liquidity. Consider the following one-person game:
The “Balls in the Hat Game”
The game consists of a hat that contains 6 black balls and 4 white balls. The player picks balls from the hat and gains $1 for each white ball, and loses $1 for each black ball. The selection is done without replacement. At the end of each pick, the player may choose to stop or continue. The player has the right to refuse to play (i.e. not pick any balls at all). Given these rules, and a hat containing 6 black balls and 4 white balls, would you play? (Why?)
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I’m missing something on the hat/ball game…how do you get a positive expected value when there are more black balls (losses) than white balls (gains)? Is there a Level III hedge hidden somewhere or is my stat training off?
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The game consists of a hat that contains 6 black balls and 4 white balls. The player picks balls from the hat and gains $1 for each white ball, and loses $1 for each black ball. The selection is done without replacement. At the end of each pick, the player may choose to stop or continue. The player has the right to refuse to play (i.e. not pick any balls at all). Given these rules, and a hat containing 6 black balls and 4 white balls, would you play? (Why?)
Mathematically one can prove that there is a POSITIVE expected value (of 1/15) in playing this game, so one SHOULD
Suppose you start with one black and one white ball. If you draw the black ball, you then keep playing and draw the white ball, and so end up even–the fact that the odds for the next draw move in your favor after a bad outcome provides you with your hedge. On the other hand, if you draw the white ball, you are up 1 and you stop.
Thus the expected value of the game (1,1) is 1/2 even though you might think initially that the game is fair.
This generalizes, and this is what gets you the positive value for the game (4,6)
Hi Cedric,
Thanks for your comment. It’s easy to make an error in the stats here (probability calculations can often be delicate). I did not give the calculation in the piece that I did, so your question is a good one. Below I give some more details:
The best way to calculate the expected winnings from a given hat is to realize that it can be done so in a recursive nature. Consider, a hat with m black balls and n white balls. Suppose you pick a white ball, then you are left with m black balls and (n-1) white balls, and the decision of whether or not to play is determined by the expected winnings from this “new, smaller” hat. Indeed,
E (winnings from m black balls, n white balls) =
max { m/(m+n) [-1 + E(n,m-1)] + n/(m+n) [1 + E(n-1,m)] , 0 }
The above formula shows that we should calculate the expected winnings from the “small” hats and build our way up. This recursion lends itself well to a computer language, and can be done on excel in minutes. I will email you the excel sheet separately later.
The full details of the solution of this game, including exploring other strategies, will appear in the January 2008 issue of Wilmott Magazine.
Cheers,
Ranjan
to Ranjan Bhaduri:
I wonder if you know of the work of Lawrence Shepp (Rutgers University) on this problem. See page 999 of:
Explicit Solutions to Some Problems of Optimal Stopping
The Annals of Mathematical Statistics, Vol. 40, No. 3. (Jun., 1969), pp. 993-1010
Roger Purves
Ok, for your calculations on the expected value of the game for the 10 balls, how do you get a positive expected value of 1/15. I saw your reply mentioning the case where you have m+n balls, but i just want to see how you get the positive expected value for just 10 balls, 4 white and 6 black ones. I have the impression that the expected value should be -2 and for risk averse person should never play this game since it has a negative payoff. Even a risk neutral individual would not play this game as it appears to be pure gambling than actually “investing”. And ok, maybe if you play it once you might be lucky, but if you play 10 or 100 times this game you are guaranted to lose on average -2 $ per game. I also don’t get why there are no negative values in the matrix that you present. In each step you have 60% chance of losing a $, but it seems that you don’t include that possibility in your figures. I had a brief look in your paper and it seems that it makes sence when you mention that someone should play the game when there is a positive payoff or if there are more white balls than black, both of which make sence, but on the other hand neither of these conditions makes sence when you have 4 white balls and 6 black. I might be wrong as my background is in economics and not in probability theory.
I just realised that you mentioned that there is no substitution, so actually the probability of getting a white or black ball are not independent, so the probability of losing 60% applies only to the first step and not in each step, but even with that, in the second step you have 4 white and 5 black, so the odds are again against you. In any case, would you play this game if you lose 10000$ if you draw a black ball and win 10000$ if you get a white one and play this game 10 times?
Hi Roger,
Thanks for your note. I’m not familiar with Shepp’s work, but I’ll be sure to check out the reference that you gave. Just from the title it looks interesting. Thanks!
Best Regards,
Ranjan
I applaud the analysis, and wish that more hedge fund investors correctly valued the liquidity option afforded by funds which invest in more liquid instruments. In practice however all too many hedge fund investors prefer funds with high liquidity but where the underlying instruments are illiquid (the worst of all worlds in aggregate, although attractive for the marginal investor who can be an “early redeemer” and exploit the fact that the positions in the portfolio are marked to market at a price that is not necessarily indicative of the full size of the position). The best solution for a portfolio of illiquid but leveragable securities may be imposing a redemption penalty for redemption on short notice that is then paid to the remaining investors in the funds.
One minor quibble: In the essay you write “Hedge funds that have the opposite type of liquidity mismatch (trade liquid instruments but have onerous redemption terms) often claim that they are trying to avoid “hot money†of fund of funds. However hedge fund managers always retain the right to refuse a potential investor, so that does not really wash.”
In practice, there is significant information ineffeciency, and while many investors often profess to not be “hot money” the manager never really knows for sure until the investor redeems. A better solution to this problem (rather than lockup) is some sort of redemption fee for redemptions on shorter notice.
There are also some special cases worth considering such as newly launched funds, where lockup is required to secure the stability of the business (although often in these cases the early investors will appropriately trade longer lockup for lower fees).
Good going, Ranjan. Love the dialogue your article has triggered. A newbie.
Ranjan,
Thanks for sharing this with us.
It seems as though the balls in the hat example could be related to contrarian trading too.
Whenever you remove a ball, you can think of this as a time step so you get a stochastic process with some mean reversion built in. E.g., when you take out a black ball at t, the probability of a white ball goes up at t+1, etc.
This is similar to the situation where there is a fixed no. of potential buyers and sellers in a security, of fixed size. Whenever someone buys, they drop out so that the no. of sellers is proportionately larger. In theory this may lead to a mean reverting process.
Thanks again,
Hari
I am lost here. I mentioned some time ago that your calculations are wrong. You maximize the expected payoff, instead of using the expected payoff. That way you find the maximum payoff possible, but not the average payoff. You use the maximum payoff that you could ever get, but usually you will not get a maximum payoff, no matter if you want to stop or not. I don’t know if my thinking is wrong, but lets take a typical book in probabilities like Grimmett and Stirzaker, or David Williams’s book. Can you show me where they maximize the expected payoff???