Questions on the Mathematics behind Expected Shortfall and Value at Risk
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I have a couple questions on the mathematics behind some general value at risk (VaR) and expected shortfall (ES) calculations.
$$
P(L > VaR(alpha))=alpha
$$
$$
ES(alpha)=E[L|L > VaR(alpha)]
$$
First, I need to find that, for distribution $R$ with pdf $f$ ,
$$
ES(alpha) = frac{-S_0}{alpha}int^{q_alpha}_{-infty}rf(r)dr
$$
I have done the following
$$
ES(alpha) = E[L|L>VaR(alpha)] = E[-S_0R|L>VaR(alpha)]
$$
$$
= int^{q_alpha}_{-infty}-S_0rfrac{P(R=r)}{P(L>VaR(alpha))}dr
= -S_0int^{q_alpha}_{-infty}rfrac{f(r)}{alpha}dr
= frac{-S_0}{alpha}int^{q_alpha}_{-infty}rf(r)dr
$$
I understand this for the most part, but I am somewhat confused as to how the $q_alpha$ comes into the fray?
Second, I need to find that, for distribution $Rsim N(mu, sigma^2)$,
$$
ES(alpha)=-S_0left(mu-frac{sigma}{alphasqrt{2pi}}e^{frac{-z^2_alpha}{2}}right)
$$
where $z_alpha$ is the $alpha$th quantile of $N(0,1)$. I assume that I follow from where the first part left off and use the normal distribution pdf in place of $f(r)$ such that
$$
ES(alpha)= frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr
$$
where $frac{(r-mu)^2}{sigma^2}=z^2$. However, how this would connect back to the given $ES(alpha)$ is very unclear to me.
integration probability-distributions normal-distribution
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I have a couple questions on the mathematics behind some general value at risk (VaR) and expected shortfall (ES) calculations.
$$
P(L > VaR(alpha))=alpha
$$
$$
ES(alpha)=E[L|L > VaR(alpha)]
$$
First, I need to find that, for distribution $R$ with pdf $f$ ,
$$
ES(alpha) = frac{-S_0}{alpha}int^{q_alpha}_{-infty}rf(r)dr
$$
I have done the following
$$
ES(alpha) = E[L|L>VaR(alpha)] = E[-S_0R|L>VaR(alpha)]
$$
$$
= int^{q_alpha}_{-infty}-S_0rfrac{P(R=r)}{P(L>VaR(alpha))}dr
= -S_0int^{q_alpha}_{-infty}rfrac{f(r)}{alpha}dr
= frac{-S_0}{alpha}int^{q_alpha}_{-infty}rf(r)dr
$$
I understand this for the most part, but I am somewhat confused as to how the $q_alpha$ comes into the fray?
Second, I need to find that, for distribution $Rsim N(mu, sigma^2)$,
$$
ES(alpha)=-S_0left(mu-frac{sigma}{alphasqrt{2pi}}e^{frac{-z^2_alpha}{2}}right)
$$
where $z_alpha$ is the $alpha$th quantile of $N(0,1)$. I assume that I follow from where the first part left off and use the normal distribution pdf in place of $f(r)$ such that
$$
ES(alpha)= frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr
$$
where $frac{(r-mu)^2}{sigma^2}=z^2$. However, how this would connect back to the given $ES(alpha)$ is very unclear to me.
integration probability-distributions normal-distribution
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a couple questions on the mathematics behind some general value at risk (VaR) and expected shortfall (ES) calculations.
$$
P(L > VaR(alpha))=alpha
$$
$$
ES(alpha)=E[L|L > VaR(alpha)]
$$
First, I need to find that, for distribution $R$ with pdf $f$ ,
$$
ES(alpha) = frac{-S_0}{alpha}int^{q_alpha}_{-infty}rf(r)dr
$$
I have done the following
$$
ES(alpha) = E[L|L>VaR(alpha)] = E[-S_0R|L>VaR(alpha)]
$$
$$
= int^{q_alpha}_{-infty}-S_0rfrac{P(R=r)}{P(L>VaR(alpha))}dr
= -S_0int^{q_alpha}_{-infty}rfrac{f(r)}{alpha}dr
= frac{-S_0}{alpha}int^{q_alpha}_{-infty}rf(r)dr
$$
I understand this for the most part, but I am somewhat confused as to how the $q_alpha$ comes into the fray?
Second, I need to find that, for distribution $Rsim N(mu, sigma^2)$,
$$
ES(alpha)=-S_0left(mu-frac{sigma}{alphasqrt{2pi}}e^{frac{-z^2_alpha}{2}}right)
$$
where $z_alpha$ is the $alpha$th quantile of $N(0,1)$. I assume that I follow from where the first part left off and use the normal distribution pdf in place of $f(r)$ such that
$$
ES(alpha)= frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr
$$
where $frac{(r-mu)^2}{sigma^2}=z^2$. However, how this would connect back to the given $ES(alpha)$ is very unclear to me.
integration probability-distributions normal-distribution
I have a couple questions on the mathematics behind some general value at risk (VaR) and expected shortfall (ES) calculations.
$$
P(L > VaR(alpha))=alpha
$$
$$
ES(alpha)=E[L|L > VaR(alpha)]
$$
First, I need to find that, for distribution $R$ with pdf $f$ ,
$$
ES(alpha) = frac{-S_0}{alpha}int^{q_alpha}_{-infty}rf(r)dr
$$
I have done the following
$$
ES(alpha) = E[L|L>VaR(alpha)] = E[-S_0R|L>VaR(alpha)]
$$
$$
= int^{q_alpha}_{-infty}-S_0rfrac{P(R=r)}{P(L>VaR(alpha))}dr
= -S_0int^{q_alpha}_{-infty}rfrac{f(r)}{alpha}dr
= frac{-S_0}{alpha}int^{q_alpha}_{-infty}rf(r)dr
$$
I understand this for the most part, but I am somewhat confused as to how the $q_alpha$ comes into the fray?
Second, I need to find that, for distribution $Rsim N(mu, sigma^2)$,
$$
ES(alpha)=-S_0left(mu-frac{sigma}{alphasqrt{2pi}}e^{frac{-z^2_alpha}{2}}right)
$$
where $z_alpha$ is the $alpha$th quantile of $N(0,1)$. I assume that I follow from where the first part left off and use the normal distribution pdf in place of $f(r)$ such that
$$
ES(alpha)= frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr
$$
where $frac{(r-mu)^2}{sigma^2}=z^2$. However, how this would connect back to the given $ES(alpha)$ is very unclear to me.
integration probability-distributions normal-distribution
integration probability-distributions normal-distribution
asked Nov 16 at 1:03
strwars
346
346
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For the first part, you are conditioning on on $L>VaR(alpha),$ which means $R<q_alpha.$ The density of $R,$ conditional on $R<q_alpha$ is zero for $R>q_alpha,$ so when you integrate over it to get the conditional expectation, the integral cuts off at $q_alpha.$ You are correct that the conditional density is given by $frac{f(r)}{alpha}$ for $r<q_alpha,$ and your line of reasoning to derive the formula looks mostly right, except that $P(R=r)$ is not the same thing as the density $f(r)$ (in fact $P(R=r)=0$ for any $r$ assuming the distribution is continuous).
For the second part, you are right that you just need to plug the normal distribution. As for where to go from there, you have to compute the integral. To do this, substitute $u=(r-mu)/sigma$ and get $$ frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr = frac{-S_0}{alpha}int^{z_alpha}_{-infty}(mu+sigma u)frac{1}{sqrt{2pi}}e^{frac{-u^2}{2}}du\=-frac{S_0mu}{alpha}int_{-infty}^{z_alpha}frac{1}{sqrt{2pi}}e^{-u^2/2}du -frac{S_0sigma}{alphasqrt{2pi}}int_{-infty}^{z_alpha} ue^{-u^2/2}du$$ where we used the fact that $z_alpha=(q_alpha-mu)/sigma.$
The first integral is $alpha$ by definition and the second can be done with a substitution.
So would you say it is more accurate to use $= frac{int^{infty}_{-infty}-S_0rf(r)dr}{P(L>VaR(alpha))}$ than the $P(R=r)$ above?
– strwars
Nov 16 at 14:55
@strwars Sure, just replace $P(R=r)$ with $f(r).$
– spaceisdarkgreen
Nov 16 at 15:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For the first part, you are conditioning on on $L>VaR(alpha),$ which means $R<q_alpha.$ The density of $R,$ conditional on $R<q_alpha$ is zero for $R>q_alpha,$ so when you integrate over it to get the conditional expectation, the integral cuts off at $q_alpha.$ You are correct that the conditional density is given by $frac{f(r)}{alpha}$ for $r<q_alpha,$ and your line of reasoning to derive the formula looks mostly right, except that $P(R=r)$ is not the same thing as the density $f(r)$ (in fact $P(R=r)=0$ for any $r$ assuming the distribution is continuous).
For the second part, you are right that you just need to plug the normal distribution. As for where to go from there, you have to compute the integral. To do this, substitute $u=(r-mu)/sigma$ and get $$ frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr = frac{-S_0}{alpha}int^{z_alpha}_{-infty}(mu+sigma u)frac{1}{sqrt{2pi}}e^{frac{-u^2}{2}}du\=-frac{S_0mu}{alpha}int_{-infty}^{z_alpha}frac{1}{sqrt{2pi}}e^{-u^2/2}du -frac{S_0sigma}{alphasqrt{2pi}}int_{-infty}^{z_alpha} ue^{-u^2/2}du$$ where we used the fact that $z_alpha=(q_alpha-mu)/sigma.$
The first integral is $alpha$ by definition and the second can be done with a substitution.
So would you say it is more accurate to use $= frac{int^{infty}_{-infty}-S_0rf(r)dr}{P(L>VaR(alpha))}$ than the $P(R=r)$ above?
– strwars
Nov 16 at 14:55
@strwars Sure, just replace $P(R=r)$ with $f(r).$
– spaceisdarkgreen
Nov 16 at 15:09
add a comment |
up vote
1
down vote
accepted
For the first part, you are conditioning on on $L>VaR(alpha),$ which means $R<q_alpha.$ The density of $R,$ conditional on $R<q_alpha$ is zero for $R>q_alpha,$ so when you integrate over it to get the conditional expectation, the integral cuts off at $q_alpha.$ You are correct that the conditional density is given by $frac{f(r)}{alpha}$ for $r<q_alpha,$ and your line of reasoning to derive the formula looks mostly right, except that $P(R=r)$ is not the same thing as the density $f(r)$ (in fact $P(R=r)=0$ for any $r$ assuming the distribution is continuous).
For the second part, you are right that you just need to plug the normal distribution. As for where to go from there, you have to compute the integral. To do this, substitute $u=(r-mu)/sigma$ and get $$ frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr = frac{-S_0}{alpha}int^{z_alpha}_{-infty}(mu+sigma u)frac{1}{sqrt{2pi}}e^{frac{-u^2}{2}}du\=-frac{S_0mu}{alpha}int_{-infty}^{z_alpha}frac{1}{sqrt{2pi}}e^{-u^2/2}du -frac{S_0sigma}{alphasqrt{2pi}}int_{-infty}^{z_alpha} ue^{-u^2/2}du$$ where we used the fact that $z_alpha=(q_alpha-mu)/sigma.$
The first integral is $alpha$ by definition and the second can be done with a substitution.
So would you say it is more accurate to use $= frac{int^{infty}_{-infty}-S_0rf(r)dr}{P(L>VaR(alpha))}$ than the $P(R=r)$ above?
– strwars
Nov 16 at 14:55
@strwars Sure, just replace $P(R=r)$ with $f(r).$
– spaceisdarkgreen
Nov 16 at 15:09
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For the first part, you are conditioning on on $L>VaR(alpha),$ which means $R<q_alpha.$ The density of $R,$ conditional on $R<q_alpha$ is zero for $R>q_alpha,$ so when you integrate over it to get the conditional expectation, the integral cuts off at $q_alpha.$ You are correct that the conditional density is given by $frac{f(r)}{alpha}$ for $r<q_alpha,$ and your line of reasoning to derive the formula looks mostly right, except that $P(R=r)$ is not the same thing as the density $f(r)$ (in fact $P(R=r)=0$ for any $r$ assuming the distribution is continuous).
For the second part, you are right that you just need to plug the normal distribution. As for where to go from there, you have to compute the integral. To do this, substitute $u=(r-mu)/sigma$ and get $$ frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr = frac{-S_0}{alpha}int^{z_alpha}_{-infty}(mu+sigma u)frac{1}{sqrt{2pi}}e^{frac{-u^2}{2}}du\=-frac{S_0mu}{alpha}int_{-infty}^{z_alpha}frac{1}{sqrt{2pi}}e^{-u^2/2}du -frac{S_0sigma}{alphasqrt{2pi}}int_{-infty}^{z_alpha} ue^{-u^2/2}du$$ where we used the fact that $z_alpha=(q_alpha-mu)/sigma.$
The first integral is $alpha$ by definition and the second can be done with a substitution.
For the first part, you are conditioning on on $L>VaR(alpha),$ which means $R<q_alpha.$ The density of $R,$ conditional on $R<q_alpha$ is zero for $R>q_alpha,$ so when you integrate over it to get the conditional expectation, the integral cuts off at $q_alpha.$ You are correct that the conditional density is given by $frac{f(r)}{alpha}$ for $r<q_alpha,$ and your line of reasoning to derive the formula looks mostly right, except that $P(R=r)$ is not the same thing as the density $f(r)$ (in fact $P(R=r)=0$ for any $r$ assuming the distribution is continuous).
For the second part, you are right that you just need to plug the normal distribution. As for where to go from there, you have to compute the integral. To do this, substitute $u=(r-mu)/sigma$ and get $$ frac{-S_0}{alpha}int^{q_alpha}_{-infty}rfrac{1}{sqrt{2pisigma^2}}e^{frac{-(r-mu)^2}{2sigma^2}}dr = frac{-S_0}{alpha}int^{z_alpha}_{-infty}(mu+sigma u)frac{1}{sqrt{2pi}}e^{frac{-u^2}{2}}du\=-frac{S_0mu}{alpha}int_{-infty}^{z_alpha}frac{1}{sqrt{2pi}}e^{-u^2/2}du -frac{S_0sigma}{alphasqrt{2pi}}int_{-infty}^{z_alpha} ue^{-u^2/2}du$$ where we used the fact that $z_alpha=(q_alpha-mu)/sigma.$
The first integral is $alpha$ by definition and the second can be done with a substitution.
edited Nov 16 at 4:54
answered Nov 16 at 2:01
spaceisdarkgreen
31.6k21552
31.6k21552
So would you say it is more accurate to use $= frac{int^{infty}_{-infty}-S_0rf(r)dr}{P(L>VaR(alpha))}$ than the $P(R=r)$ above?
– strwars
Nov 16 at 14:55
@strwars Sure, just replace $P(R=r)$ with $f(r).$
– spaceisdarkgreen
Nov 16 at 15:09
add a comment |
So would you say it is more accurate to use $= frac{int^{infty}_{-infty}-S_0rf(r)dr}{P(L>VaR(alpha))}$ than the $P(R=r)$ above?
– strwars
Nov 16 at 14:55
@strwars Sure, just replace $P(R=r)$ with $f(r).$
– spaceisdarkgreen
Nov 16 at 15:09
So would you say it is more accurate to use $= frac{int^{infty}_{-infty}-S_0rf(r)dr}{P(L>VaR(alpha))}$ than the $P(R=r)$ above?
– strwars
Nov 16 at 14:55
So would you say it is more accurate to use $= frac{int^{infty}_{-infty}-S_0rf(r)dr}{P(L>VaR(alpha))}$ than the $P(R=r)$ above?
– strwars
Nov 16 at 14:55
@strwars Sure, just replace $P(R=r)$ with $f(r).$
– spaceisdarkgreen
Nov 16 at 15:09
@strwars Sure, just replace $P(R=r)$ with $f(r).$
– spaceisdarkgreen
Nov 16 at 15:09
add a comment |
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