Estimating VaR with Normally Distributed Arithmetic Returns
- Thread starter Stuti
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Hi,
We suppose the returns(rt) are normally distributed with mean μr and standard deviation σr,now the Var is the worst possible loss at a given confidence level(CL=95%) we shall have a set of returns in this CL the worst possible return shall lie at the border that is at when cumulative probability=CL%,start with the highest return and continue till the worst return is reached and 95% of the returns are covered,we shall get the worst return possible out of these 95% returns observations,beyond this worst return are α=1-CL=1-0.95=5% of returns(cumulative probability,).This worst return is nothing but r* only=μr-σr*z this worst return lies on the left of the mean μr at z standard deviations away so that to the left of r* there is α=5% cumulative probability(5% of all returns observations).Thus we get the worst return r*=μr-σr*z possible out of a set of 95% observations or to say at 95% CL .This is the worst return that the portfolio can attain at 95% CL.Therefore the Var which is nothing but worst possible %loss at a given confidence level(CL=95%) is r*=μr-σr*z.In dollar terms this equals starting portfolio value(Pt-1)*r* what the worst $ loss that the portfolio suffered =Pt-1*r*=Pt-1*(μr-σr*z).
Hope it helps,
thanks
We suppose the returns(rt) are normally distributed with mean μr and standard deviation σr,now the Var is the worst possible loss at a given confidence level(CL=95%) we shall have a set of returns in this CL the worst possible return shall lie at the border that is at when cumulative probability=CL%,start with the highest return and continue till the worst return is reached and 95% of the returns are covered,we shall get the worst return possible out of these 95% returns observations,beyond this worst return are α=1-CL=1-0.95=5% of returns(cumulative probability,).This worst return is nothing but r* only=μr-σr*z this worst return lies on the left of the mean μr at z standard deviations away so that to the left of r* there is α=5% cumulative probability(5% of all returns observations).Thus we get the worst return r*=μr-σr*z possible out of a set of 95% observations or to say at 95% CL .This is the worst return that the portfolio can attain at 95% CL.Therefore the Var which is nothing but worst possible %loss at a given confidence level(CL=95%) is r*=μr-σr*z.In dollar terms this equals starting portfolio value(Pt-1)*r* what the worst $ loss that the portfolio suffered =Pt-1*r*=Pt-1*(μr-σr*z).
Hope it helps,
thanks
ok I am not sure if I understand fully but I will try to elaborate my understanding. Please correct me where I am wrong. The returns are arithmetic returns and follow a normal distribution with mean μr and standard deviation σr. Now in order to calculate the critical value which is r*, and this return basically means that for our portfolio we want returns greater than this particular value r*. Now this value r* can be calculated at any confidence value and since volatility affects the returns at a particular confidence level a loss will be calculated to the left of the mean which is basically ur-σr*z. Right?
Everything on the right of this return on the normal curve is what we want when we say r>r*. Since we want to calculate the dollar value of this return we take the value of the asset. My question here is why should be the value of asset a day before? and not P0?
Everything on the right of this return on the normal curve is what we want when we say r>r*. Since we want to calculate the dollar value of this return we take the value of the asset. My question here is why should be the value of asset a day before? and not P0?
I have the following question as well:
1. Weighted Historical Simulation - A stock market crash might have no effect on VaRs except at a vey high confidence level, so we could have a situation where everyone might agree that risk had suddenly increased and yet that increase in risk would be missed by most HS VaR estimates. This increase in risk would only show up later in VaR estimates if the stock market continued to fall in subsequent days. That said, the increase in risk would show up in ES estimates just after the first shock occurred - which is incidentally a good example of how ES can be a more informative risk measure than VaR. Please explain this as I am unable to get this scenario
1. Weighted Historical Simulation - A stock market crash might have no effect on VaRs except at a vey high confidence level, so we could have a situation where everyone might agree that risk had suddenly increased and yet that increase in risk would be missed by most HS VaR estimates. This increase in risk would only show up later in VaR estimates if the stock market continued to fall in subsequent days. That said, the increase in risk would show up in ES estimates just after the first shock occurred - which is incidentally a good example of how ES can be a more informative risk measure than VaR. Please explain this as I am unable to get this scenario
Hi,
As per my understanding,under HS the Var reports the 5th worst loss(95%CL) or the (1-CL)th loss for (99% CL) as there is market crash the large negative return would now become the worst loss ever so that the previous 4th worst loss would become the 5th worst loss(Var) but the impact of the market crash is not captured by the 4th worst loss which becomes Var after the crash. Therefore Var shall miss the worst loss until there are several worst losses for subsequent days such that the worst loss due to the crash would become the 5th worst loss and thus the Var captures the risk due to the crash.
ES would immediately reflect on the crash,as ES is the average of the worst losses above the Var therefore after the crash the worst % loss due to crash would become the large negative return as the 1st or the 2nd worst loss so ES=average of 1st+2nd+3rd+4th+5th worst losses would reflect the worst % loss due to crash which can be 1st worst return and would affect the ES, therefore ES reflecting upon the crash.The effect of 1st worst return(the loss due to crash ) is being shown up in the ES estimate.
thanks
As per my understanding,under HS the Var reports the 5th worst loss(95%CL) or the (1-CL)th loss for (99% CL) as there is market crash the large negative return would now become the worst loss ever so that the previous 4th worst loss would become the 5th worst loss(Var) but the impact of the market crash is not captured by the 4th worst loss which becomes Var after the crash. Therefore Var shall miss the worst loss until there are several worst losses for subsequent days such that the worst loss due to the crash would become the 5th worst loss and thus the Var captures the risk due to the crash.
ES would immediately reflect on the crash,as ES is the average of the worst losses above the Var therefore after the crash the worst % loss due to crash would become the large negative return as the 1st or the 2nd worst loss so ES=average of 1st+2nd+3rd+4th+5th worst losses would reflect the worst % loss due to crash which can be 1st worst return and would affect the ES, therefore ES reflecting upon the crash.The effect of 1st worst return(the loss due to crash ) is being shown up in the ES estimate.
thanks
Hi @Stuti The derivation above (which is Dowd 3.11 to 3.14) is very interesting; for me, it helps greatly to illustrate with an example in order to understand his distinction between normal VaR and lognormal VaR.
For example, assume drift µ = 1.0%, σ = 2.0%, VaR confidence is 99.0% (so normal deviate is 2.326), and yesterday's P(t-1) = $200.00
The critical ("cutoff") r_* = 0.01 - 0.02*2.325 = -3.65270%
As arithmetic returns imply r_* = [P*(t) - P(t-1)]/P(t-1), P*(t) = P(t-1) + P(t-1)*r(*) = 200.00 + 200.00*(-3.65270%) = $192.69, and the VaR($) = 200.00 - 192.69 = $7.31;
put another way, per the final formula, r_*= VaR/P(t) = 7.31/200 = 3.6520%.
It's circular, what's the point? Just to show the relationship between the critical r_* and arithmetic VaR: given these assumptions, critical r_* is 3.65270% such that arithmetic VaR is very easy, we simply multiply by the price to get VaR($) = P(t-1)*r_* = $200*3.6520% = $7.31.
It's useful to contrast with the lognormal VaR, where the critical r_* is exactly the same, r_* = 0.01 - 0.02*2.325 = -3.65270%, but instead of assuming arithmetic returns, we assume lognormal returns. In the case of lognormal returns, the implied "cutoff" P(t) is slightly higher (the lognormal is non-negative), P(t) = 200*exp(-3.65270%) = 192.8264, such that the lognormal VaR is slightly less at 200*[1-exp(-3.65270%)] = $7.1736. I hope that's helpful!
For example, assume drift µ = 1.0%, σ = 2.0%, VaR confidence is 99.0% (so normal deviate is 2.326), and yesterday's P(t-1) = $200.00
The critical ("cutoff") r_* = 0.01 - 0.02*2.325 = -3.65270%
As arithmetic returns imply r_* = [P*(t) - P(t-1)]/P(t-1), P*(t) = P(t-1) + P(t-1)*r(*) = 200.00 + 200.00*(-3.65270%) = $192.69, and the VaR($) = 200.00 - 192.69 = $7.31;
put another way, per the final formula, r_*= VaR/P(t) = 7.31/200 = 3.6520%.
It's circular, what's the point? Just to show the relationship between the critical r_* and arithmetic VaR: given these assumptions, critical r_* is 3.65270% such that arithmetic VaR is very easy, we simply multiply by the price to get VaR($) = P(t-1)*r_* = $200*3.6520% = $7.31.
It's useful to contrast with the lognormal VaR, where the critical r_* is exactly the same, r_* = 0.01 - 0.02*2.325 = -3.65270%, but instead of assuming arithmetic returns, we assume lognormal returns. In the case of lognormal returns, the implied "cutoff" P(t) is slightly higher (the lognormal is non-negative), P(t) = 200*exp(-3.65270%) = 192.8264, such that the lognormal VaR is slightly less at 200*[1-exp(-3.65270%)] = $7.1736. I hope that's helpful!
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