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Let's assume T=1 and let S be a geometric gaussian process with zero drift, i.e. $\ln(S_1/S_0)$ is normally distributed with mean $-1/2\times\mathrm{VEV}^2$ and volatility VEV. Then $$\ln(\mathrm{VaR}/S_0) = -1/2\mathrm{VEV}^2 - \mathrm{VEV} \times 1.96$$ with the VAR at $0.975$ quantile. This is a quadratic equation in VEV, with solutions \mathrm{VEV}... 6 Gordon's answer is spot on. Another way to see it though, would be using Bayes formula and a change of variable. \begin{align*} ES_X(p) &=E\left(X \mid X\le Q_X(1-p)\right)\\ &=\int_{-\infty}^{\infty} x\, \phi\left(x \mid x\le Q_X(1-p)\right) dx \\ &=\int_{-\infty}^{\infty} x\, \frac{\phi\left(x\le Q_X(1-p) \mid x \right)\phi(x)}{\int_{-\infty}^... 6 Simple example where sub-additivity fails Let there be four possible outcomes i=1,2,3,4 that occur with equal probability \frac{1}{4}. Payoffs for X, Y, and X + Y are given by: X = \begin{bmatrix}-1\\0\\1\\2 \end{bmatrix} \quad Y = \begin{bmatrix}0\\-1\\1\\2 \end{bmatrix} \quad X + Y = \begin{bmatrix}-1\\-1\\2\\4 \end{bmatrix}$$What's the ... 6 No reply has been given so I wanted to at least give a visualisation of the expectiles. Suppose the curvy dashed line in my picture represents a cumulative distribution function of some random variable X. Then blue part corresponds exactly to \mathbb{E}[(X-x)_+], while the orange surface corresponds to \mathbb{E}[(X-x)_-]. In the picture x=1. Now if ... 6 The 99.97% confidence is somtimes referred to as corresponding to the 1-year probability of default of 3 bps for AA-rated entities. (Here for example https://papers.ssrn.com/sol3/papers.cfm?abstract_id=963233 ) The normal approximation works better for general securities portfolios than for credit portfolios and might thus be seen as good enough from a ... 5 You don't really have a multivariate case: we can only define VaR (in its usual sense) for a one-dimensional output. Recall that$$ \operatorname{VaR}_\alpha(X) = \inf\{v:F_X(v)\geq \alpha\} and since in your case X = X_1+X_2 you just need to compute F_X in terms of X_1 and X_2. For the notation of partial derivatives, I denote the generic ... 5 If the loss distribution is normal with mean \mu and variance \sigma^2, then the Value-at-Risk and Expexted Shortfall (or CVaR) at level \alpha \in (0, 1) are \begin{align*} \mbox{VaR}_\alpha & = \mu + \sigma \Phi^{-1}(\alpha) , \\ \mbox{ES}_\alpha & = \mu + \sigma \frac{\phi\{\Phi^{-1}(\alpha)\}}{1 - \alpha} , \end{align*} where \phi ... 5 Yes, it is correct. Underestimation: you under-estimate the risk, so you have more VaR violations than what your model predicts. Ex: With 100 observations, and a 99% VaR, you expect 1 violation but you observe 5 violations. Overestimation: you over-estimate the risk, i.e the risk is less important that you expect. You observe less VaR violations that you ... 5 Note that Q_X is the pseudo-inverse of the distribution function F, and for any uniform random variable U over [0, 1], the random variable Q_X(U) has the same distribution as X. Moreover, since X is continuous, Q_X is strictly increasing. Proofs of these facts are purely mathematical, and can be discussed some other questions. Here, we ... 5 VaR is not sub-additive in general. Relying on Mark Joshi comment, there are particular cases where it can be. Such cases occur for portfolios containing elliptically distributed risk factors. Of course the normal distribution is among the elliptical distributions family. The latter can be helpful for analytical VaR modelling as an elliptical model is ... 5 Let the n-dimensional vector of returns \mathbf{r} have a multivariate t distribution with \nu degrees of freedom. The marginal distribution of any component r_i has a univariate t distribution also with \nu degrees of freedom. To see this, assuming mean returns have been subtracted, the multivariate t distribution decomposes as the distribution ... 4 As a short summary and adaption of the question: You better redefine \hat{r}_i= \frac{S_{i-1}}{S_1}-1 and \hat{S}_i = (1+\hat{r}_i)S_0. The above definition of \hat{S}_i yields a sample of potential values for S for the future day. This approach is usually applied in historical simulation. The aim here is to use information of the past about the ... 4 First, I am quite sure that this is a typo and it should be 0 < VaR_1 < VaR_0 $$then$$ -VaR_0 < -VaR_1 $$and the plot is correct. Second, the put strategy does not change only the expected profit but the whole distribution of the P&L. If you buy a put with strike K_1 = -VaR_1 then you get compensated for losses below K_1. But you ... 4 It depends on the method by which you calculate VaR. Some models (t-distributuion, normal) lead to a form of VaR such that it is just scaled volatility:$$ VaR = c \sigma $$with some proper c (e.g. q_{\alpha} in the case of normal, bit more complicated for the t-distribution). Then as \sigma scales with square-root-of-time so does VaR. If VaR is ... 4 The VaR of level \alpha a loss random variable (the bigger the worse) is the quantity q such that the loss is bigger with probability 1-\alpha. Thus we need a q such that$$ P[L>q] = 1-\alpha, $$where we can imagine \alpha=99\% and thus we need the starting point of the 1\% tail. Because we have a probability of a loss of size 0 of 75\%... 4 By definition, your loss cannot be positive, so you'd set the VaR to zero. But it really depends, on how you calculate your VaR. If you calculate your returns, sort them and look at the 5% quantile (which, as you say, may be positive), then you'd simply set your VaR to zero. But if you treat your returns as realizations of some (unknown) random variable, ... 4 Let u=t^{-1}_v(\alpha) and recall that g_v(u)=c_v(v+u^2)^{-\frac{v+1}2} for some constant c_v. By the formulas you provided,$$\begin{eqnarray*}\lim_{\alpha\to 1^-}\frac{\mathrm{ES}_\alpha(X)}{\mathrm{VaR}_\alpha(X)}&=&\lim_{\alpha\to 1^-} \frac{g_v(t^{-1}_v(\alpha))}{(1-\alpha)(v-1)\left(\frac{t^{-1}(\alpha)}{v+(t^{-1}(\alpha))^2}\right)}\\ =\...

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A linear relationship between expected returns and covariance with a risk factor is a necessary consequence of a linear asset pricing function In theory, a CAPM relationship can be derived when a pricing kernel $S$ is affine in the return of the market portfolio. Different sets of assumptions lead to this affine relationship. Be aware that the CAPM is an ...

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What are the underlying assumptions for doing this Assumption: Historical returns are lognormally distributed with no autocorrelation. can those assumptions be tested statistically Testing: $\sqrt{xy} = \sqrt{x} \sqrt{y}$ Substitute time $t$ and variance $\sigma^2$ for $x$ and $y$ respectively $\sqrt{t\sigma^2} = \sqrt{t} \sqrt{\sigma^2} = \sigma\... 4 I would suggest to start with Euler capital allocation as a first step to dive into the subject, here is an example of introductory paper (Capital Allocation to Business Units and Sub-Portfolios: the Euler Principle, Dirk Tasche). In general, I don't think that you will find any satisfactory theory on the subject for practical uses beyond the case of an ... 4 This problem can be addressed efficiently by linear programming. An (in my opinion) even better reference than the original paper by Uryasev, Rockafeller provided by noob2 is "PORTFOLIO OPTIMIZATION WITH CONDITIONAL VALUE-AT-RISK OBJECTIVE AND CONSTRAINTS" by Pavlo Krokhmal, Jonas Palmquist, and Stanislav Uryasev in The Journal of Risk, V. 4, # 2, ... 4 When returns follow an elliptical distribution (e.g. the Gaussian distribution), then minimising VaR and ES is equivalent to minimising variance. See https://people.math.ethz.ch/~embrecht/ftp/pitfalls.pdf. Then, the frontiers will be the same. 3 These are identical definitions of ES. It's just a matter of expressing losses as negatives or positives. First definition Notice the integral bounds are$a$and$1$: losses are positive; this is so-called Loss(+)/Profit(-). Here alpha might be 95%, as in 95% confidence VaR or ES. Second definition Losses are negative, and the corresponding quantile is ... 3 The standard approach is to multiply by the square root of the number of trading days in a year. If you assume there are 250 trading days in the year, you multiply by$\sqrt{250}$. Investopedia is one source explaining this approach. 3 One of the easiest ways is described in Duffie, Pan (1997) "Bootstrapped Simulation from Historical Data" p.55.$R$is the set of all risk factors (a time series)$C_{norm}$is the Covariance Matrix during normal times.$C_{stressed}$is the Covariance Matrix from a period of stress. You can update$R$in the following way.$R_{i,stressed}=C_{stressed}^{1/...

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Value at risk is quoted by absolute value. This is the amount of money you can lose, so everyone knows the sign by default. For the second question, the last line explains it. Probability of at least one of the assets losing money is ~9.6%. Probability of both losing money is pretty small and is ignored. So, since 9.6% > 5%, it means that you lose on one of ...

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It is correct! You can also see it this way: $$\text{CVaR}_\alpha(X)=\mathbb{E}(X|X\leq \text{VaR}_\alpha(X)) = \frac{\int_{\mathbb{R}} x\cdot 1_{X\leq \text{VaR}_\alpha(X)}dF(x)}{\int_\mathbb{R}1_{X\leq \text{VaR}_\alpha(X)}dF(x)} = \frac{1}{\alpha} \int_{-\infty}^{\text{VaR}_\alpha(X)}xdF(x)$$ The sign problem still remains (in both versions). If you ...

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I think Cholesky on correlation matrix is better because it makes code apply more generally in case we don't have full rank. For example, suppose we want to simulate three correlated normals with covariance matrix [[a^2,0,0], [0,b^2,0], [0,0,c^2]] i.e. variables are uncorrelated and have vols a, b, and c. Because this is positive definite, we can do ...

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You can use the either, as both necessarily are symmetric positive definite; covariance is a personal preference. It's really just a matter of scaling, as $\mathcal{N}(0,\Sigma)$ is distributionally $\sqrt{\Sigma} \mathcal{N}(0,1)$. Correlation would require additional scaling (i.e. multiplication of every $\mathcal{N}(0,\rho)$ element by its respective ...

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