Value at risk oct 2010 q4(ii)

[John H]

In this question it states that log returns are distributed ~N(0,0.15)

So to calculate VaR I would calculate the rerun at the 0.05 point in the distribution and multiply by the fund value.

I.e P(R<x) =0.05

(Log(R) -0) / 0. 15*sqrt (1/250) =-1.6449
Log(R) = -0.0156
R = exp(-0.0156) = 0.98

So VaR = 20000*0.98 = 19690

The solution seem to ignore the log returns and just multiply the -0.0156 by the fund value to get the loss.

Am I making a mistake here and over complicating the answer?

 

[John Potter]

Hi John H,

You're not over-complicating it, indeed the official solution is NOT taking into account the "log" in the question. Well done for spotting this! However, I think your calculations have gone slightly astray:

ln [S(t) / 20,000] = ln S(t) - ln 20,000 ~ N[0, 0.15^2 t]

We need x in

P[S(1/250) < x] = 0.05

P[{S(1/250) - ln 20,000}/0.15sqrt(1/250) < {ln x - ln 20,000}/0.15sqrt(1/250)] = 0.05

Phi[{ln x - ln 20,000}/0.15sqrt(1/250)] = 0.05

{ln x - ln 20,000}/0.15sqrt(1/250) = -1.6449

x = 19,960.32

So, VaR = £309.68, not that different from the official solution.

Good luck!
John

 

[John H]

Thanks John.

This means that log (1+return) is normally distributed and not log(returns) as stated in the question.

Can I assume for future VaR questions any reference to return is actually (1+return)?

Thanks for correcting my calls below!!

 

[Colin McKee]

log returns

Hi,
I would say that, in most ST5 situations, returns can be used directly and you do not need to convert to log returns unless asked. The examiners would probably give marks for both so long as you explained what it was you were doing and why you were converting it.