[R-SIG-Finance] Simulated expected return for a hedge fund

[Thomas Schwiertz]

Hi David,

I am not quite sure, but I think the solution can be calculated without any Computer simulation by using the "brownian reflection principle". 
Nevertheless one assumes the stock price is a Brownian motion, and it is well known historical stock returns have "fat tails". In my diploma thesis I used "truncated Lėvy-flights" to cover this issue. I examined various questions for random walks: what's the maximum/minimum, what's the probability of a recurrence (the stock price after 100 days is at its starting Price again)... Unfourtunately the thesis is in German, but let me know if it might help... Anyway my hint: brownian reflection principle.

Merry Christmas,
Thomas


Am 23.12.2011 um 21:44 schrieb "Afshartous, David" <d.afshartous at vanderbilt.edu>:

> 
> 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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