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Conditional Value-at-Risk (CVaR) is one of the commonly used risk measures. The paper shows t hat the optimal estimator of CVaR is strong consistency if the first-order moment of the population exists. We subsequently carry out numerical simulations to test the conclusion. We use the results to make an empirical analysis of Shenzhen A shares.

Because of inherent market instability, the ability to assess risk is very important so as to hedge against catastrophic loss. Therefore, some scholars have put forward the theory of value at risk (VaR) and conditional value at risk (CVaR) theory as a form of risk mitigation.

To mitigate this risk, Markowitz [

The paper is organized as follows. In Section 2, we prove the strong consistency of CVaR optimal estimator under the finite first-order moment; we use R to simulate the correctness of the above conclusion in Section 3; in Section 4, we use the results to make an empirical analysis of Shenzhen A shares (A-shares trade on the two Chinese stock exchanges, the Shanghai Stock Exchange and the Shenzhen Stock Exchange. A-shares are shares of mainland China-based companies, and these shares were historically only available for purchase by mainland citizens because foreign investment was restricted).

Let Z be a random variable that represents a loss (or cost), F ( z ) = P ( Z ≤ z ) is the distribution function of Z. For α ∈ ( 0 , 1 ) , the α-quantile of Z is defined as

F − 1 ( α ) : = inf z ∈ R { z : F ( z ) ≥ α } .

In the context of monetary cost (loss), the α-quantile F − 1 ( α ) is a measure of risk commonly used in the finance industry, known as VaR and denoted VaR α ( Z ) = F − 1 ( α ) . At this point, if VaR stands for the maximum loss, it can be exceeded only in ( 1 − α ) 100 % of cases. And then CVaR, denoted by CVaR α ( Z ) , can be thought of as the conditional expectation of losses that exceed the VaR α ( Z ) level. In case Z has a continuous distribution, CVaR α ( Z ) is given by the expectation of the right ( 1 − α ) tail of the cost distribution, that is

CVaR α ( Z ) = E [ Z | Z ≥ VaR α ( Z ) ] .

Föllmer and Schied [

CVaR α ( Z ) = 1 1 − α ∫ α 1 VaR γ ( Z ) d γ ,

where is [ x ] + = max { 0 , x } , x ∈ R .

Let Z is a random variable, { Z n , n ≥ 1 } is a Simple random from Z. Based on the definition of CVaR optimization proposed by Plung [

θ ^ n = inf t ∈ R { t + 1 1 − α θ n ( t ) } ,

where is θ n ( t ) = n − 1 ∑ i = 1 n [ Z i − t ] + . In addition, we define that

θ * = CVaR α ( Z ) = inf t ∈ R { t + 1 1 − α E [ Z − t ] + } , θ ( t ) = E [ Z − t ] + .

It is obvious that θ n ( t ) and θ ( t ) are monotonically decreasing and Lipschitz continuous. If − ∞ < s < t < + ∞ , that is

0 ≤ θ n ( s ) − θ n ( t ) = 1 n ∑ i = 1 n ( [ Z i − s ] + − [ Z i − t ] + ) ≤ t − s ,

0 ≤ θ ( s ) − θ ( t ) = E ( [ Z − s ] + − [ Z − t ] + ) ≤ t − s .

Lemma 2.1. If E | Z | < + ∞ and − ∞ < a < b < + ∞ , then

lim n → + ∞ sup t ∈ [ a , b ] | θ n ( t ) − θ ( t ) | = 0 a .s .

Proof: For ∀ k ∈ R , the intervals [ a , b ] are divided into k parts equally, where is t j = j ( b − a ) / k + a , j = 0 , 1 , 2 , ⋯ , k . θ n ( t ) and θ ( t ) are monotonically decreasing and Lipschitz continuity, and t j − t j − 1 = ( b − a ) / k .

sup t ∈ [ a , b ] | θ n ( t ) − θ ( t ) | = max 1 ≤ j ≤ k sup t ∈ [ t j − 1 , t j ] | θ n ( t ) − θ ( t ) | = max 1 ≤ j ≤ k sup t ∈ [ t j − 1 , t j ] | θ n ( t ) − θ ( t ) − θ n ( t j ) + θ n ( t j ) − θ ( t j ) + θ ( t j ) | ≤ max 1 ≤ j ≤ k { sup t ∈ [ t j − 1 , t j ] | θ n ( t ) − θ n ( t j ) | + sup t ∈ [ t j − 1 , t j ] | θ ( t j ) − θ ( t ) | + | θ n ( t j ) − θ ( t j ) | } ≤ max 1 ≤ j ≤ k { ( θ n ( t j − 1 ) − θ n ( t j ) ) + ( θ ( t j − 1 ) − θ ( t j ) ) + | θ n ( t j ) − θ ( t j ) | } ≤ 2 ( b − a ) k + max 1 ≤ j ≤ k | θ n ( t j ) − θ ( t j ) |

Applying the strong law of large number [

lim n → + ∞ | θ n ( t j ) − θ ( t j ) | = 0 a .s .

for 1 ≤ j ≤ k . We can get

lim n → + ∞ max 1 ≤ j ≤ k | θ n ( t j ) − θ ( t j ) | = 0 a .s .

Therefore let n → + ∞ and then let k → + ∞ , the conclusion can be get.

Lemma 2.2. If E | Z | < + ∞ and b > 0 , then

lim ¯ n → + ∞ sup t > b | θ n ( t ) − θ ( t ) | ≤ 2 E | Z | I { Z > b } a .s .

Proof: θ n ( t ) and θ ( t ) is monotonically decreasing, we can get

sup t > b | θ n ( t ) − θ ( t ) | ≤ sup t > b θ n ( t ) + sup t > b θ ( t ) = sup t > b 1 n ∑ i = 1 n [ Z i − t ] + + sup t > b E [ Z − t ] + = 1 n sup t > b ∑ i = 1 n [ Z i − t ] + + sup t > b E [ Z − t ] + = 1 n ∑ i = 1 n ( Z i − b ) I { Z i > b } + E ( Z − b ) I { Z > b } ≤ 1 n ∑ i = 1 n | Z i | I { Z i > b } + E | Z | I { Z > b }

Applying the strong law of large number, it is easy to show that

lim n → ∞ 1 n ∑ i = 1 n | Z i | I { Z i > b } = E | Z | I { Z > b } a .s .

So we can get the conclusion.

Lemma 2.3. If E | Z | < + ∞ and a < 0 , then

lim ¯ n → + ∞ sup t < a | θ n ( t ) − θ ( t ) | ≤ 2 E | Z | I { Z < a } a .s .

Proof:

sup t < a | θ n ( t ) − θ ( t ) | = sup t < a | 1 n ∑ i = 1 n [ Z i − t ] + − E [ Z − t ] + | = sup t < a | 1 n ∑ i = 1 n ( Z i − t ) I { Z i ≥ t } − E ( Z − t ) I { Z ≥ t } | = sup t < a | 1 n ∑ i = 1 n ( Z i − t ) ( 1 − I { Z i < t } ) − E ( Z − t ) ( 1 − I { Z < t } ) |

= sup t < a | 1 n ∑ i = 1 n Z i − E Z − 1 n ∑ i = 1 n ( Z i − t ) I { Z i < t } + E ( Z − t ) I { Z < t } | ≤ | 1 n ∑ i = 1 n Z i − E Z | + sup t < a 1 n ∑ i = 1 n | Z i − t | I { Z i < t } + sup t < a E | Z − t | I { Z < t } ≤ | 1 n ∑ i = 1 n Z i − E Z | + sup t < a 1 n ∑ i = 1 n | Z i | I { Z i < t } + sup t < a E | Z | I { Z < t } = | 1 n ∑ i = 1 n Z i − E Z | + 1 n ∑ i = 1 n | Z i | I { Z i < a } + E | Z | I { Z < t }

Applying the strong law of large number, it is easy to show that

n − 1 ∑ i = 1 n Z i → E Z a .s . ,

n − 1 ∑ i = 1 n | Z i | I { Z i < a } → E | Z | I { Z < a } a .s .

So we can get the conclusion.

Proposition 2.1 If E | Z | < + ∞ , then

lim n → + ∞ sup t ∈ R | θ n ( t ) − θ ( t ) | = 0 a .s .

Proof: Given a and b satisfied − ∞ < a < 0 < b < + ∞ , we can get

sup t ∈ R | θ n ( t ) − θ ( t ) | = max { sup t < a | θ n ( t ) − θ ( t ) | , sup t ∈ [ a , b ] | θ n ( t ) − θ ( t ) | , sup t > b | θ n ( t ) − θ ( t ) | } .

By applying Lemma 1, Lemma 2 and Lemma 3, we can get

lim ¯ n → ∞ sup t ∈ R | θ n ( t ) − θ ( t ) | ≤ max { 2 E | Z | I { Z < a } , 0 , 2 E | Z | I { Z > b } } .

And let a → − ∞ and b → + ∞ , we can get the conclusion.

Lemma 2.4. If E [ Z ] + < + ∞ , then

| θ ^ n − θ * | ≤ 2 1 − α sup t ∈ R | θ n ( t ) − θ ( t ) | .

Proof:

| θ ^ n − θ * | = ( θ ^ n − θ * ) I { θ ^ n > θ * } + ( θ * − θ ^ n ) I { θ ^ n < θ * } = ( inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + } − inf t ∈ R { t + 1 1 − α E [ Z − t ] + } ) I { θ ^ n > θ * } + ( inf t ∈ R { t + 1 1 − α E [ Z − t ] + } − inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + } ) I { θ ^ n < θ * } = inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + − inf t ∈ R { t + 1 1 − α E [ Z − t ] + } } I { θ ^ n > θ * } + inf t ∈ R { t + 1 1 − α E [ Z − t ] + − inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + } } I { θ ^ n < θ * } : = I n 1 + I n 2

First, we can get

I n 1 ≤ inf t ∈ R | t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + − inf t ∈ R { t + 1 1 − α E [ Z − t ] + } | ≤ inf t ∈ R | t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + − ( t + 1 1 − α E [ Z − t ] + ) + ( t + 1 1 − α E [ Z − t ] + ) − inf t ∈ R { t + 1 1 − α E [ Z − t ] + } | = inf t ∈ R | 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + − 1 1 − α E [ Z − t ] + + ( t + 1 1 − α E [ Z − t ] + ) − inf t ∈ R { t + 1 1 − α E [ Z − t ] + } |

≤ sup t ∈ R | 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + − 1 1 − α E [ Z − t ] + | + inf t ∈ R | t + 1 1 − α E [ Z − t ] + − inf t ∈ R { t + 1 1 − α E [ Z − t ] + } | = sup t ∈ R | 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + − 1 1 − α E [ Z − t ] + | = 1 1 − α sup t ∈ R | θ n ( t ) − θ ( t ) | .

And then, we can get

I n 2 ≤ inf t ∈ R | t + 1 1 − α E [ Z − t ] + − inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + } | ≤ inf t ∈ R | t + 1 1 − α E [ Z − t ] + − ( t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + ) + ( t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + ) − inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + } | = inf t ∈ R | 1 1 − α E [ Z − t ] + − 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + + ( t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + ) − inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + } |

≤ sup t ∈ R | 1 1 − α E [ Z − t ] + − 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + | + inf t ∈ R | t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + − inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + } | = sup t ∈ R | 1 n ( 1 − α ) ∑ i = 1 n [ Z i − t ] + − 1 1 − α E [ Z − t ] + | = 1 1 − α sup t ∈ R | θ n ( t ) − θ ( t ) | .

Hence, it is easy to show I n 1 + I n 2 ≤ 2 sup t ∈ R | θ n ( t ) − θ ( t ) | / ( 1 − α ) . And we can obtain this conclusion.

Theorem 2.1. If E | Z | < + ∞ , then θ ^ n → θ * a.s.

Proof: By applying Lemma 2.4 and Propostion 2.1, it is obvious that theorem is right.

We carry out numerical simulations by using the standard t-distribution with 2 degrees of freedom (denoted t ( 2 ) ), because the first-order moment of the distribution exists, and the second-order moment does not exist. Let Z ∼ t ( 2 ) and { z n , n ≥ 1 } be observations of Z. Andreev et al. [

CVaR α = 2 f ( q ) F ( − q ) ( 1 + q 2 2 ) = 1 2 ( 1 − α ) ( 1 + q 2 2 ) − 1 / 2

where is that q is the a-quantile of F ( x ) . In addition, Min et al. [

θ ^ n = inf t ∈ R { t + 1 n ( 1 − α ) ∑ i = 1 n [ z i − t ] + } = min 1 ≤ j ≤ n { z j + 1 n ( 1 − α ) ∑ i = 1 n [ z i − z j ] + }

Next, we use R and the above two equations to calculate the true value of CVaR to compare it with the value of the optimized estimator.

size is N = 20 000 , α is 0.001, 0.005, 0.01, 0.025, 0.05 respectively and the simulated step length is 100. In

There are real valuations and estimations (obtained by numerical simulation) of CVaR under different α and n. From this table, we can see that difference between CVaR real valuations and estimations is 0.0312, 0.025, 0.0254, 0.0255, 0.0249 respectively when n = 10000 and α =0.001, 0.005, 0.01, 0.025, 0.05; but when n = 200000 , that difference is −0.0024, −0.0034, −0.0038, −0.0045, −0.0047 respectively.

Therefore we can get the conclusion that the larger the value is, the closer the estimation is to the real value. So we test the conclusion that the CVaR estimator has strong consistency.

In this Section, we use the results to make an empirical analysis of Shenzhen A shares, which included SWYA and Vanke A (two typical shares). We select a data range from November 27, 2006 to November 27, 2016 by R. Let the daily opening price of a single stock be P O , and the daily closing one be P C . We calculate the daily logarithmic yield of the stock V, i.e., V = ln P C − ln P O . In addition, let L be the daily logarithmic loss rate of the stock. CVaR α ( V ) is CVaR of the stock’s daily logarithmic yield rate at α level ; correspondingly, CVaR 1 − α ( L ) is the CVaR of the stock’s logarithmic loss rate at 1 − α level.

n | α | ||||
---|---|---|---|---|---|

0.001 | 0.005 | 0.01 | 0.025 | 0.05 | |

10,000 | 0.0759 | 0.1258 | 0.1675 | 0.2520 | 0.3493 |

20,000 | 0.0466 | 0.0993 | 0.1416 | 0.2274 | 0.3272 |

30,000 | 0.0584 | 0.1088 | 0.1501 | 0.2355 | 0.3339 |

40,000 | 0.0399 | 0.0962 | 0.1384 | 0.2226 | 0.3202 |

50,000 | 0.0234 | 0.0805 | 0.1230 | 0.2078 | 0.3051 |

60,000 | 0.0721 | 0.1265 | 0.1676 | 0.2503 | 0.3475 |

70,000 | 0.0273 | 0.0830 | 0.1257 | 0.2110 | 0.3099 |

80,000 | 0.0291 | 0.0878 | 0.1303 | 0.2141 | 0.3123 |

90,000 | 0.0525 | 0.1084 | 0.1503 | 0.2341 | 0.3321 |

100,000 | 0.0555 | 0.1106 | 0.1524 | 0.2375 | 0.3358 |

120,000 | 0.0489 | 0.1026 | 0.1445 | 0.2290 | 0.3265 |

140000 | 0.0439 | 0.0995 | 0.1413 | 0.2257 | 0.3241 |

160000 | 0.0332 | 0.0911 | 0.1339 | 0.2181 | 0.3163 |

180000 | 0.0504 | 0.1048 | 0.1463 | 0.2312 | 0.3293 |

200000 | 0.0423 | 0.0969 | 0.1383 | 0.2220 | 0.3197 |

Real valuation | 0.0447 | 0.1003 | 0.1421 | 0.2265 | 0.3244 |

The CVaR estimations of daily yield are calculated by

CVaR α ( V ) = min 1 ≤ j ≤ n { V j + ∑ i = 1 n [ V i − V j ] + n ( 1 − α ) } .

Plung [

E V = ( 1 − α ) CVaR α ( V ) − α CVaR ( 1 − α ) ( L ) .

In this Section, we analyze the data of Vanke A and SWYA deeply.

In this figure, the circle represents the real logarithmic yield of Vanke A, and the blue horizontal line represents the average logarithmic yield of the corresponding day. We can get the conclusion that real logarithmic yield of Vanke A always fluctuate around the average ones from this figure.

In addition, we can apply this property to invest. For example, we take Vanke A and SWYA two stocks as an example to analyze. We get CVaR 1 − α ( L ) estimations of SWYA and Vanke A in

time/year | α | ||||
---|---|---|---|---|---|

0.001 | 0.005 | 0.01 | 0.025 | 0.05 | |

10 | 0.1224 | 0.0986 | 0.0894 | 0.0762 | 0.0654 |

5 | 0.0924 | 0.0795 | 0.0699 | 0.0562 | 0.0487 |

time/year | α | ||||
---|---|---|---|---|---|

0.001 | 0.005 | 0.01 | 0.025 | 0.05 | |

10 | 0.1601 | 0.1152 | 0.1018 | 0.0860 | 0.0769 |

5 | 0.1891 | 0.1306 | 0.1150 | 0.0951 | 0.0823 |

Time/year | SWYA | Vanke A |
---|---|---|

10 | 0.0027 | 0.00209 |

5 | 0.0030 | 0.00207 |

In this article, we have proved that the optimal estimator of CVaR is strong consistency if the first-order moment of the population exists, subsequently carrying out numerical simulations to test the conclusion. Further, we will study the complete convergence and convergence rate of optimal estimator of CVaR.

Li, X.L. (2018) Strong Consistency of CVaR Optimal Estimator. Open Journal of Statistics, 8, 416-426. https://doi.org/10.4236/ojs.2018.83027