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You have the correct approach. (1) The simulation generates sampled portfolio values, $P_1,P_2, \dots, P_n$ at time $t=T$. VaR is specified as a left-tail percentile. Order the sample as $$P_{(1)} \leq P_{(2)} \leq \dots \leq P_{(n)}.$$ If you are considering $VaR_\alpha$ at the $100(1-\alpha) \%$ confidence level , then choose the smallest integer $k$ ...

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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 $$\mathrm{VEV}... 9 Values of VaR are just the inverses of the cumulative distributions. CVaR is not a very commonly used term, its more frequently used synonym is Expected Shortfall. See http://www.maths.manchester.ac.uk/~saralees/chap17.pdf for the list of Expected Shortfall values for more than 20 distributions. 7 Regression analysis, as a minimization of the sum of squared errors, does not require normality of the error term. The requirements are that errors are homoscedastic and uncorrelated. And these are the fundamental assumptions (together with exogeneity). Then estimators are unbiased, optimal (exhibit the minimum variance within the class of unbiased ... 7 The answer to your question is no. Value at Risk is not additive in the sense that \text{VaR}(X+Y) \neq \text{VaR}(X) + \text{VaR}(Y). But I guess your question is more to aimed at finding a formula for your investments than to look at the property itself. I think the only way to get a nice formula for this is to assume that both assets are: Normally ... 7 Simple example where sub-additivity fails Let there be four possible outcomes i=1,2,3,4 that occur with equal probability \frac{1}{4}. Payoffs for X, Y, and X + Y are given by:$$ 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}$$What's the ... 7 This issue is incredibly important and I agree there is little practical information about it. To me, the key idea is to find the right matrix completion algorithm that best suits your needs. I work mostly with equity time series and there are substantial missing values issues due to, e.g., as you cite, IPOs with limited history. Recently I have had good ... 7 Suppose X\sim N(\mu_X,\sigma_X^2) and Y\sim N(\mu_Y,\sigma_Y^2) are correlated jointly normal random variables. Then,$$X+Y\sim N(\mu_X+\mu_Y,\sigma^2_X+\sigma_Y^2+2\rho\sigma_X\sigma_Y).Suppose X and Y denote the profit of your portfolio returns (so negative values for X,Y mean losses). Then, the 5% value at risk is the 0.05 quantile of the ... 6 As discussed, banks do use VaR for risk management. They will have something modified for the specific use (i.e. probably not your VaR from a fitted normal distribution), it's likely more sophisticated but the underlying idea is the same. VaR is used for reporting/ceremonial business decisions as much as (or perhaps even more than) it is for trading ... 6 I know this was asked almost two years ago, but I thought I'd answer the question. It appears that the H that you want to estimate is identical to the values you received from the Johansen test, with the exception of rows 1:4 and columns 2:4. You only need to set those values to zeroes and ones, which is fairly easy considering that the diagonal is (very ... 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 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 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 First, I am quite sure that this is a typo and it should be 0 < VaR_1 < VaR_0 $$then$$ -VaR_0 < -VaR_1 $$and the plot is correct. Second, the put strategy does not change only the expected profit but the whole distribution of the P&L. If you buy a put with strike K_1 = -VaR_1 then you get compensated for losses below K_1. But you ... 5 You don't really have a multivariate case: we can only define VaR (in its usual sense) for a one-dimensional output. Recall that$$ \operatorname{VaR}_\alpha(X) = \inf\{v:F_X(v)\geq \alpha\} $$and since in your case X = X_1+X_2 you just need to compute F_X in terms of X_1 and X_2. For the notation of partial derivatives, I denote the generic ... 5 The most important difference is that the calculations are based on a "stressed" historical period in the markets as opposed to the most recent X number of years. 5 The VaR of level \alpha a loss random variable (the bigger the worse) is the quantity q such that the loss is bigger with probability 1-\alpha. Thus we need a q such that$$ 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\%... 5 Depending of \lambda, pasts observations will be weighted differently, if you compute the volatility at time t , the t-1 observation will be weighted by (1-\lambda)*\lambda^{0}, the t-2 observation by (1-\lambda)*\lambda^{1} and so on so forth. For \lambda= 0.94  : The first observation is weighted by = (1-0.94) * 0.94^0 =0.06% The second ... 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 There are a few different ways to calculate VaR. Historical Method For this method, you calculate the return of your portfolio each day, and get a list of daily returns over your calibration period. Once you have this, then you find the 5th percentile to give the 95% VaR. The advantage of this method is that it is the most straightforward to compute, and ... 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 Do N MC simulations of M samples, calculating your estimate of VaR for each one \{\widehat{VaR}_i\}_{i=1}^N and you now have an IID sample! Take the sample (or unbiased) standard deviation for your estimate of VaR (this is probably what you mean by error) SD(\widehat{VaR})=\sqrt{\frac{1}{N-1} \sum_{i=1}^N (\widehat{VaR}_i - \overline{VaR})^2} and of ... 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 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 ...

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EVT has pluses and minuses, but (under certain conditions) provides the best estimate of extreme quantile returns in a portfolio given the data available. Probably the simplest and easiest way to do this is to use the peak over threshold method and fit the Generalized Pareto Distribution (GPD). The GPD is very convenient for calculating VaR and ES. A good ...

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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, ...

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