# Tag Info

10

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}^... 5 If z_\alpha is the so-called standard normal z-score of the significance level \alpha such that \frac 1 {\sqrt{2\pi}}\int_{-\infty}^{z_\alpha} e^{-\xi^2/2}d\xi=\alpha $$and we assume normality, (ignoring skewness and kurtosis,) then we can estimate the \alpha quantile of a distribution with cdf \Phi as$$\Phi^{-1}(\alpha)=\mu + \sigma z_\alpha....

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

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 ...

4

There is a formula for calculating ES from a normal distribution. There is also a formula for ES of arbitrary distributions using a Cornish-Fisher expansions (easy for univariate processes but frustrating for multivariate). However, the most common approach is a scenario representation of the distribution. This could include using the historical 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

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 ...

4

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 $\phi$ ...

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 $75\%... 4 By definition, your loss cannot be positive, so you'd set the VaR to zero. But it really depends, on how you calculate your VaR. If you calculate your returns, sort them and look at the 5% quantile (which, as you say, may be positive), then you'd simply set your VaR to zero. But if you treat your returns as realizations of some (unknown) random variable, ... 4 VaR is not sub-additive in general. Relying on Mark Joshi comment, there are particular cases where it can be. Such cases occur for portfolios containing elliptically distributed risk factors. Of course the normal distribution is among the elliptical distributions family. The latter can be helpful for analytical VaR modelling as an elliptical model is ... 4 Simple example where sub-additivity fails Let there be four possible outcomes$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 ... 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)}\\ =\... 4 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 ... 4 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 ... 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 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 ... 3 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 ... 3 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 ... 3 Skewness decays with time, but the rate of that skewness decay will vary based on the instruments and how they are traded, so a simple estimator such as the square root of time rule is not appropriate. I typically recommend that to scale VaR or ES it makes more sense to lower your confidence level (raise the alpha parameter) to one that makes sense for your ... 3 the risk neutral drift is needed for pricing of derivatives. For a 100\% equity portfolio you can take the real world drift - sometimes a good guess is a drift of zero. For fixed-income you could do the same and might need more sophistication for the variance term. If you have short-dated bonds then you will need a special model for the pull-to-par. For ... 3 VaR is not a good measure of risk taking, in my opinion. It suffers from inherent faulty assumptions (check out VaR Wiki to start) and it omits many other important aspects of risk measurement. When I evaluate an asset's risk and return I like to start looking at the following: Historical risk and returns of an asset. This leads to the Sharpe Ratio, ... 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 You got some things wrong: You don't have to devide sd by \sqrt{n}, the division is already part of the definition of sd. The t distribution has a parameter \nu, the degrees of freedom. The variance of a standard t distributed random variable T is$$ VAR(T) = \nu/(\nu-2).$$Thus you have to define$\sigma = sd * \sqrt{(\nu-2)/\nu}\$ and a random ...

3

1) You are computing the "actual" VaR, in the sense that you are not forecasting it to see if your VaR model is able to estimate it, but you are just computing the VaR that "has taken place". To obtain a volatility forecast (either in-sample or out-of-sample) you can use the "ugarchforecast" function. 2) I think you are estimating the VaR on the wrong side ...

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