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According to a few resources online the formula of iVaR is : VaR (after adding the new element) - VaR (before)

My question is how can this be correct given the lack of subadditivity of VaR? Meaning, we are not supposed to add VaR results of different portfolio but it's OK to subtract them?!?!

In general, what's the best way to calculate iVaR?

thanks

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IMHO I think this is just to get the feeling of having something additive.

This is just a telescopic serie.

Let $X_1,\dots,X_n$ be $n$ financial positions, then:

$$\text{Var}(\sum_{i=1}^nX_i)=\sum_{k=1}^n\underbrace{\text{Var}(\sum_{i=1}^kX_i)-\text{Var}(\sum_{i=1}^{k-1}X_i)}_{\stackrel{\rm def}{=}\text{iVar}(X_k)}$$

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  • $\begingroup$ Thanks for your reply - could you please elaborate on this? which method would you use to calculate the incremental VaR? $\endgroup$ – sen_saven Jul 31 '16 at 19:58
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I just thought I would add a formal example & discussion.

Firstly Incremental VaR is NOT Marginal or Conditional VaR.

The formula below illustrates an Additive Function has the below properties:

An additive function f(n) is said to be completely additive if f(ab) = f(a) + f(b)

Say $g(n)$ is a $VaR$ method. However, the $iVaR$ method is not $g(n)$ it is a new function $h(n)$.

Here: $$h(n) = g(\sum_{i=1}^{n} X_i) - g(\sum_{i=1}^{n-1} X_i)$$

where $X_{i}$ is a P&L vector of a trade/asset $i$

Firstly the key thing to note is incremental VaR or $iVaR$ is not actual $VaR$.

This statement can be confusing but the point is $iVaR$ is the difference between to already calculated $VaR$ values, thus the rules of additivity although relevant, are not needed to accept $iVaR$'s definition.

It would be like saying since $f(n) = n^2$ is not additive, therefore $g(x,y) = f(x) - f(y)$ is not a valid function.

This is why you can subtract them to get $iVaR$

Below is a simplified example using R

x=4
y=2
square <- function(n) { n^2 }

Clearly square is not an additive function.

However, say we want the impact on the square calculation of adding y to x my original variable.

We would need a new function to get this impact

diffSquare <- function(m,n) { square(m) - square(n) }

Now to get the impact of adding y to x on the square calculation we would run the below:

> diffSquare(y+x,x)
[1] 20

i.e. $6^2$ - $4^2$ $= 20$

This could be your "Incremental Square Value" $ISquare$

Obviously this is fairly redundant for such a simple mathematical expression.

Below is a VaR specific example using R

Here the we define a normally distributed set of returns for the portfolio, portfolio1, and the new trade we want to evaluate trade2.

portfolio1 <- rnorm(n=100, mean=3, sd=10)
trade2 <- rnorm(n=100, mean=0, sd=4)

Here we define a VaR function, where we can specify the alpha.

VaR <- function(timeSeries,alpha=0.05){
  sort(timeSeries)[round(length(timeSeries)*alpha)]
}

Now say to get the $iVaR$ of adding trade2 to our portfolio.

VaR(portfolio1+trade2) - VaR(portfolio1)

For me when I ran the code (will vary as rnorm generates random obs.)

> VaR(portfolio1+trade2) - VaR(portfolio1)
[1] -1.544173
> VaR(portfolio1+trade2)
[1] -13.80089
> VaR(portfolio1)
[1] -12.25672

The $iVaR$ will be -1.54, causing a more negative $VaR$ if I add the trade to the portfolio, thus increasing my portfolio's overall loss at a 95% confidence level.

Basically: $$iVaR = VaR_{portfolio + new trade} - VaR_{portfolio}$$

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