Simulated expected return for a hedge fund
forgot to copy the list
On Mon, Dec 26, 2011 at 10:19 AM, G See <gsee000 at gmail.com> wrote:
David, The problem is that it takes longer to win than to lose. Imagine a game where there is a 50% chance you will lose $100 on day 1, and a 50% chance you'll make $100 on day 100. Ignoring the time value of money (that a dollar today is worth more than a dollar in the future), The expected value of the game is 0 (=0.5*-100 + 0.5*100). However, if you lose, you will lose $100 per day on average, but if you win, you will only make $1 per day on average. ?You may think that if you lose, you lose $100 on day one and 0 for each of the next 99 days. ?But, that isn't the case, because if you lose (and have more money), you could play again the second day. Our Hedge Fund problem is similar because it takes less time to lose 50k than it takes to make 100k. That said, I found copy of the book, and you are correct that the author is asking for the expected profit from the whole game. Garrett On Sun, Dec 25, 2011 at 6:31 PM, Afshartous, David <d.afshartous at vanderbilt.edu> wrote:
Hi Garrett, Thank you very much for your reply. ? ?Fixing the typo with the negative sign, my updated expected profit calculation yields: ## expected profit
mean(above)*mean(profit.above) + mean(below)*mean(profit.below) + mean(between)*mean(profit.between)
[1] 9433.936 This value seems to make sense given that the probability of crossing the upper boundary is .39 = mean (above), and the probability of the crossing the lower boundary is .58 = mean(below), and these events yield gains/losses of approximately 100k and -50k. ? ? I'm not sure I see the need to account for how many days it took to earn/lose the money and calculate the expected profit per day, since the above calculation determines the expected amount of money that this strategy will yield? ? And isn't this what is ultimately of interest? ? At any rate, assuming this is correct, it's somewhat troubling that this number is positive while the expected profit per day is negative. Cheers, David
________________________________________
From: G See [gsee000 at gmail.com]
Sent: Sunday, December 25, 2011 2:25 PM
To: Afshartous, David
Cc: r-sig-finance at r-project.org
Subject: Re: [R-SIG-Finance] Simulated expected return for a hedge fund
Hi David,
2 things. ?First, the second term of your expected profit expression
should have a positive sign. ?i.e.
mean(above)*mean(profit.above) + mean(below)*mean(profit.below) +
mean(between)*mean(profit.between)
Second, your expected profit calculation does not take into account
how many days it took you to earn/lose that money. ?If you do that,
our answers agree
prof.above <- profit.above/above.ending.day
prof.below <- profit.below/below.ending.day
prof.between <- profit.between/100
mean(above)*mean(prof.above) + mean(below)*mean(prof.below) + mean(between)*mean(prof.between)
[1] -1098.853
However, just because our answers agree doesn't mean they are right.
:-) ?I would be interested to learn what answer the author came up
with.
Cheers,
Garrett
On Sun, Dec 25, 2011 at 1:45 PM, Afshartous, David
<d.afshartous at vanderbilt.edu> wrote:
Hi Garrett,
Thanks for looking into the problem. ? You are correct, one has to be careful to note that portfolio value for early sells will not be exactly equal to the boundary values that trigger the early sells (and also that -100% loss conversion to daily return is an issue).
I corrected my code and now obtain the same expected return as your code to within 5 decimal places. ? However, I still obtain a positive value for the expected profit. ?My calculation is a conditional probability calculation whereas your calculation is direct. ? I doubt that such big difference in expected profit could be due to rounding error of such a process but at the same time we obtain the same expected return. ? ?Code below:
niter = 10000 ? ? ## number of iterations
below = rep(0, niter) ? ## set up storage
above = rep(0, niter)
between = rep(0, niter)
between.ending.value = NULL
below.ending.day = NULL
above.ending.day = NULL
profit.above = NULL
profit.below = NULL
profit.between = NULL
set.seed(2009)
for (i in 1:niter) {
?r = rnorm(100, mean = .05/253, sd = .23/sqrt(253))
?logPrice = log(1e6) + cumsum(r) ?## 1e6 = initial price
?minLogP = min(logPrice) ?## min price over next 45 days
?maxLogP = max(logPrice) ?## min price over next 45 days
?crossBelowLogP.day = ifelse( minLogP < log(950000), ?min(which(logPrice < log(950000))), 101) ## 101 means doesn't cross
?crossAboveLogP.day = ifelse( maxLogP > log(1100000), ?min(which(logPrice > log(1100000))), 101)
?if ( (minLogP >= log(950000)) & (maxLogP <= log(1100000)) ) { ? ## Neither cross boundary
? ? between[i] = 1
? ? between.ending.value = append(between.ending.value, exp(logPrice[100]))
? ? profit.between = ? append(profit.between, between.ending.value - 1000000)
? ? }
?if ?( (minLogP < log(950000)) & (maxLogP > log(1100000))) { ## Both cross
? ? ? ? ? ? ifelse (crossBelowLogP.day < crossAboveLogP.day, below[i] <- 1, above[i] <-1)
? ? ? ? ? ? ifelse (crossBelowLogP.day < crossAboveLogP.day, below.ending.day <- append(below.ending.day, crossBelowLogP.day),
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?above.ending.day <- append(above.ending.day, crossAboveLogP.day))
? ? ? ? ? ? ifelse (crossBelowLogP.day < crossAboveLogP.day, profit.below <- ? append(profit.below, exp(logPrice[crossBelowLogP.day]) - 1000000),
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?profit.above ?<- ? ? ?append(profit.above, exp(logPrice[crossAboveLogP.day]) - 1000000))
? ? ? ? ? }
?if ( (minLogP < log(950000)) ?& (maxLogP <= log(1100000)) ?) { ? ?## Only cross below
? ? ? ? ? ? below[i] = 1
? ? ? ? ? ? below.ending.day = append(below.ending.day, crossBelowLogP.day)
? ? ? ? ? ? profit.below = ? append(profit.below, exp(logPrice[crossBelowLogP.day]) - 1000000)
? ? ? ? ? }
?if ( (minLogP >= log(950000)) ?& (maxLogP > log(1100000)) ?) { ? ?## Only cross above
? ? ? ? ? ? above[i] = 1
? ? ? ? ? ? above.ending.day = append(above.ending.day, crossAboveLogP.day)
? ? ? ? ? ? profit.above = ?append(profit.above, exp(logPrice[crossAboveLogP.day]) - 1000000)
? ? ? ? ? }
}
## expected profit
mean(above)*mean(profit.above) - mean(below)*mean(profit.below) + mean(between)*mean(profit.between)
[1] 77227.87
ret.above = (profit.above/50000)/above.ending.day ?## daily return, return wont' be exactly 1 since P_t won't be exactly equal to 100,000
ret.below = (profit.below/50000)/below.ending.day ?## daily return
ret.between = (profit.between/50000)/100 ?## daily return
## exp.return:
mean(above)*mean(ret.above) ? + mean(below)*mean(ret.below) ? + mean(between) * mean(ret.between)
[1] -0.02197705
________________________________________
From: G See [gsee000 at gmail.com]
Sent: Sunday, December 25, 2011 12:44 PM
To: Afshartous, David
Subject: Re: [R-SIG-Finance] Simulated expected return for a hedge fund
BTW,
I get a negative value for expected Profit also
niter = 10000 ? ? ## number of iterations
set.seed(2009)
PnLs <- rep(0, niter)
for (i in 1:niter) {
+ ? ? r = rnorm(100, mean = .05/253, sd = .23/sqrt(253))
+ ? ? SV <- 1e6 * exp(cumsum(r)) #StockValue
+ ? ? daily.PnL <- if (min(SV) > 950000 && max(SV) < 1100000) {
+ ? ? ? ? #between
+ ? ? ? ? SellPrice <- tail(SV, 1) # Sell on last day (100 returns in
the future)
+ ? ? ? ? PnL <- (SellPrice - 1000000)
+ ? ? ? ? PnL / 100 ?# average daily PnL
+ ? ? } else {
+ ? ? ? ? # Sell at first value that is less than 950000 or greater than 1100000
+ ? ? ? ? SellPrice <- head(SV[SV < 950000 | SV > 1100000], 1)
+ ? ? ? ? days.held <- head(which(SV == SellPrice), 1) # first place
that StockValue equals SellPrice
+ ? ? ? ? #if (SellPrice < 950000) SellPrice <- 950000
+ ? ? ? ? PnL <- (SellPrice - 1000000)
+ ? ? ? ? PnL / days.held
+ ? ? }
+ ? ? PnLs[i] <- daily.PnL
+ }
mean(PnLs)
[1] -1098.661
mean(PnLs) / 50000
[1] -0.02197322
On Sat, Dec 24, 2011 at 10:31 PM, G See <gsee000 at gmail.com> wrote:
David,
I attempted to work the problem myself, and have concluded that it is
a poor question. :-P
The question says that you will sell for a profit of _at least_
$100,000, if the value of the stock
rises to at least $1,100,000 or sell for a loss if the value falls
below $950000.
I interpret that to mean that on the first day that the stock value is
at or above 1.1 million, I sell
at the prevailing price (which may, and probably will, be higher than
1.1 million). ?Likewise, I think
it says that I will sell on the first day that the value of the stock
is below $950,000. ?But, if the
stock gets to $950,000, I'm out of capital. ?It is unlikely that the
value of the stock will be
exactly $950,000, which means that I would be losing more than 100
percent of my capita. ?I
don't know how to convert an N-day return that is less than negative
100 percent to either a daily
or annual return. ?So, I just divided it by the number of days. ?i.e.
if you lose 110% over 10 days
then you're losing 11% per day on average. (But, I'm aware that that
doesn't make sense)
My attempt is below; I came up with an "answer" of -0.02197322
niter = 10000 ? ? ## number of iterations
set.seed(2009)
returns <- rep(0, niter)
for (i in 1:niter) {
? ? r = rnorm(100, mean = .05/253, sd = .23/sqrt(253))
? ? SV <- 1e6 * exp(cumsum(r)) #StockValue
? ? daily.ret <- if (min(SV) > 950000 && max(SV) < 1100000) {
? ? ? ? #between
? ? ? ? SellPrice <- tail(SV, 1) # Sell on last day (100 returns in the future)
? ? ? ? ret <- (SellPrice - 1000000) / 50000
? ? ? ? ret / 100 ?# average daily return
? ? } else {
? ? ? ? # Sell at first value that is less than 950000 or greater than 1100000
? ? ? ? SellPrice <- head(SV[SV < 950000 | SV > 1100000], 1)
? ? ? ? days.held <- head(which(SV == SellPrice), 1) # first place
that StockValue equals SellPrice
? ? ? ? #if (SellPrice < 950000) SellPrice <- 950000
? ? ? ? ret <- (SellPrice - 1000000) / 50000
? ? ? ? ret / days.held
? ? }
? ? returns[i] <- daily.ret
}
mean(returns)
# -0.02197322
HTH,
Garrett
On Sat, Dec 24, 2011 at 5:51 PM, Afshartous, David
<d.afshartous at vanderbilt.edu> wrote:
Thanks Garrett. ?I fixed that typo and also put the conditional returns in terms of daily returns and the answer now seems reasonable. ? The changes to the code are:
ret.above = 1/above.ending.day ?## daily return
ret.below = -1/below.ending.day
ret.between = net/100
which now yields:
exp.return = mean(above)*mean(ret.above) ? + mean(below)*mean(ret.below) ? + mean(between) * mean(ret.between)
exp.return
[1] -0.03455923
Interesting, since the expected profit is positive:
exp.profit = mean(above)*100000 - mean(below)*50000 + mean(between)*(mean(between.ending.value) - 1000000)
exp.profit
[1] 7530.114
________________________________________
From: G See [gsee000 at gmail.com]
Sent: Saturday, December 24, 2011 1:27 PM
To: Afshartous, David
Cc: r-sig-finance at r-project.org
Subject: Re: [R-SIG-Finance] Simulated expected return for a hedge fund
I'm not at a computer, but it looks like if the account value falls from 50,000 to zero, you are counting that as a 0% return instead of -100%?
On Dec 24, 2011 12:58 PM, "G See" <gsee000 at gmail.com<mailto:gsee000 at gmail.com>> wrote:
Sorry, please disregard. I didn't look at your code closely.
On Dec 24, 2011 10:55 AM, "G See" <gsee000 at gmail.com<mailto:gsee000 at gmail.com>> wrote:
Hi David,
Here is my hint:
1> set.seed(2009)
1> ?r = rnorm(100, mean = .05/253, sd = .23/sqrt(253))
1> ?logPrice = log(1e6) + cumsum(r) ?## 1e6 = initial price
1> logPrice
?[1] 13.80340 13.80755 13.80516 13.80164 13.82105 13.79453 13.79750 13.80408 13.78316 13.77246 13.77529
?[12] 13.79100 13.78908 13.79427 13.79268 13.76960 13.78917 13.79897 13.80548 13.80673 13.83131 13.82956
?[23] 13.82829 13.83970 13.84860 13.88746 13.90191 13.89439 13.88666 13.89621 13.92104 13.91894 13.92787
?[34] 13.91217 13.90804 13.91452 13.92865 13.95498 13.98419 13.98820 13.99140 14.00165 13.97519 13.96530
?[45] 13.95029 13.95800 13.94906 13.94807 13.91299 13.90617 13.90202 13.90779 13.89517 13.88682 13.89177
?[56] 13.88537 13.89815 13.87475 13.86916 13.88294 13.87048 13.86705 13.87571 13.86564 13.85744 13.87504
?[67] 13.88990 13.88075 13.87261 13.86023 13.85191 13.85510 13.84572 13.84649 13.85447 13.83746 13.82143
?[78] 13.80111 13.78505 13.79430 13.81151 13.81456 13.80200 13.79527 13.80288 13.76598 13.79024 13.81018
?[89] 13.80240 13.77962 13.78402 13.78493 13.80514 13.79839 13.77328 13.78443 13.78491 13.77488 13.75898
[100] 13.74589
That's probably not what you intended ;-)
Garrett
On Fri, Dec 23, 2011 at 2:43 PM, Afshartous, David <d.afshartous at vanderbilt.edu<mailto:d.afshartous at vanderbilt.edu>> wrote:
All,
Below is an R-lab problem from Ch2 of Statistics and Data Analysis for Financial Engineering (Ruppert, 2010). ? The R code runs okay but the answer seems strange. ? Any hints much appreciated.
Cheers,
David
David Afshartous, Ph.D.
Research Associate Professor
PSTAT?: ASA Accredited Professional Statistician
Department of Biostatistics
Vanderbilt University Medical Center
## suppose Hedge fund owns $1,000,000 of stock and used $50,000 capital and
## $950,000 borrowed money for the purchase.
## Daily log returns on the stock have a mean of 0.05/year and SD of 0.23/year,
## which can be converted to per trading day by dividing by 253 and sqrt(253)
## the hedge fund will sell the stock for a profit of at least $100,000 if the value
## of the stock rises to at least $1,100,000 at the end of one of the first 100 trading days,
## sell it for a loss if the value falls below $950,000 at the end of one of the first
## 100 trading days, or sell after 100 trading days if the closing price has stayed between
## $950,000 and $1,1000,000.
## PROBLEM 7: What is the expected return? When answering this question, remember
## that only $50,000 was invested. ?Also, the units of return are time, e.g.,
## one can express a return as a daily return or a weekly return. ?Therefore,
## one must keep track of how long the hedge fund holds its position before selling
The code below simulates whether value falls below 950,000, goes above 1,1000,000, or stays between. The amount lost/gained in the 1st/2nd scenario is a constant, whereas the amount lost/gained in the third scenario is random.
### simulation code; written for readability as opposed to speed
niter = 10000 ? ? ## number of iterations
below = rep(0, niter) ? ## set up storage
above = rep(0, niter)
between = rep(0, niter)
between.ending.value = NULL
below.ending.day = NULL
above.ending.day = NULL
set.seed(2009)
for (i in 1:niter) {
?r = rnorm(100, mean = .05/253, sd = .23/sqrt(253))
?logPrice = log(1e6) + cumsum(r) ?## 1e6 = initial price
?minLogP = min(logPrice) ?## min price over next 45 days
?maxLogP = max(logPrice) ?## min price over next 45 days
?crossBelowLogP.day = ifelse( minLogP < log(950000), ?min(which(logPrice < log(950000))), 101) ## 101 means doesn't cross
?crossAboveLogP.day = ifelse( maxLogP > log(1100000), ?min(which(logPrice > log(1100000))), 101)
?if ( (minLogP >= log(950000)) & (maxLogP <= log(1100000)) ) { ? ## Neither cross boundary
? ?between[i] = 1
? ?between.ending.value = append(between.ending.value, exp(logPrice[100]))
? ?}
?if ?( (minLogP < log(950000)) & (maxLogP > log(1100000))) { ## Both cross
? ? ? ? ? ?ifelse (crossBelowLogP.day < crossAboveLogP.day, below[i] <- 1, above[i] <-1)
? ? ? ? ? ?ifelse (crossBelowLogP.day < crossAboveLogP.day, below.ending.day <- append(below.ending.day, crossBelowLogP.day),
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? above.ending.day <- append(above.ending.day, crossAboveLogP.day))
? ? ? ? ?}
?if ( (minLogP < log(950000)) ?& (maxLogP <= log(1100000)) ?) { ? ?## Only cross below
? ? ? ? ? ?below[i] = 1
? ? ? ? ? ?below.ending.day = append(below.ending.day, crossBelowLogP.day)
? ? ? ? ?}
?if ( (minLogP >= log(950000)) ?& (maxLogP > log(1100000)) ?) { ? ?## Only cross above
? ? ? ? ? ?above[i] = 1
? ? ? ? ? ?above.ending.day = append(above.ending.day, crossAboveLogP.day)
? ? ? ? ?}
}
mean(above) ?## this will give the probability of the value crossing the above threshold
When annualizing returns, the day in which a boundary needs to be accounted for. ? ?The following code calculates the overall expected return by conditioning on each scenario:
# cross above: always make constant 100,000 based on initial investment of 50,000
# P_t/P_t-1 = 100,000/50,000 = 2 = gross return --> net return = 2 - 1 = 1
ret.above = (1 + 1)^(365/(above.ending.day)) ?- 1 ? ?### annualized
# cross below: always lose constant 50,000 based on initial investment of 50,000
# P_t/P_t-1 = 50,000/50,000 = 1 = gross return --> net return = 1 - 1 = 0
ret.below = 0
## betwen: final value depends on realized value on 100th day
# P_t/P_t-1 = between.ending.value/50,000 = --> net return = gross return ?- 1
net = ifelse(between.ending.value < 1000000,
? ? ? ? ? ? ? ? ? (1000000 - between.ending.value)/50000 - 1, ?## net return, gain
? ? ? ? ? ? ? ? ? (between.ending.value ?- 1000000)/50000 - 1) ? ## net return, loss
ret.between = (1 + net)^(365/100) - 1
exp.return = mean(above)*mean(ret.above) ? + mean(below)*mean(ret.below) ? + mean(between) * mean(ret.between)
8.669104e+50???
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