[R-SIG-Finance] A question on portfolio value calculation
david.ardia at unifr.ch
Thu Jan 6 13:50:46 CET 2011
Have a look at this note:
This may help
On 01/06/2011 01:28 PM, Guy Green wrote:
> In any realistic portfolio you will have some starting equity, and you would
> also have the costs/proceeds of your long & short positions. The portfolio
> value would then be:
> Starting equity + $(m1-m2-m3) -cost(pos1) +proceeds(pos2) +proceeds(pos2)
> or to put it another way:
> Starting equity +/- unrealised gains/losses on your positions.
> Your position sizes will be linked to your starting equity, both in the
> sense that your starting equity is a real-world constraint on the sizes of
> the positions that your broker will allow you to enter into, and also in the
> more theoretical (but still real-world) sense that the relative sizes of
> your starting equity and your positions contribute to the likelihood of your
> strategy exhausting all your equity at some point in the future, even if it
> is a winning one over the long term.
> Megh Dal wrote:
>> Hi all, can somebody suggest me on what is the correct way to calculate
>> value of a portfolio (i.e. mark-to-market value) with having both long and
>> short position? For example, suppose I have 3 positions in my portfolio
>> pos1, po2, and pos3 and type of transaction is long, short, short
>> Say, m2m value of those 3 positions are m1, m2 and m3 in money term. Then
>> should m2m value of this portfolio be $(m1-m2-m3)?
>> If this is correct I feel there are some practical problem with this
>> approach. Let say I calculated the volatility of this portfolio assuming
>> some normal distribution of return, let say it is $X. Then if I want to
>> answer, what is the volatility for per unit value of my entire portfolio
>> answer would be : $X/$(m1-m2-m3). However if it happenes that $(m1-m2-m3)
>> 0 then above calculation becomes undefined.
>> This approach also may be problametic if I have all short, in this case
>> SD for my portfolio becomes obviously negative.
>> Or should I go with $(abs(m1)+abs(m2)+abs(m3)) to avoid above scenario?
>> Any explanation would be highly appreciated.
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