# Tag Info

## Hot answers tagged estimation

15

Let me add two points to Quantoisseur's answer. Standard Errors The difference between estimating variances and means is that the standard error of the variance estimator depends on the size of the sample (number of observations), whereas the standard error of the mean depends on the length (or duration) of the sample, see here. So, if you use finer and ...

9

The use of kernels to estimate volatility using intraday data is "nothing more" than combining: intraday volatility estimation kernel smoothing Thus you have to take care about the "usual pits" of these two approaches. Intraday volatility estimation. I hope you know the "signature plot" effect. Of course if you use the proper estimation method, it should ...

7

Very interesting question. I am not an expert on the subject, however, I was able to find a collection of papers on the subject that should get you started. Here is a good and very informative paper that walks you through several tick by tick volatility estimators that seek to reduce the volatility imposed by market micro-structure: Efficient estimation of ...

7

I personally use the simple Garch(1,1) for volatility filtering in the risk management area. In fact in most cases I don't even estimate the parameters, I stick 0.94 for mean reversion, 0.04 for the squared error and I get the constant by matching the series variance. My experience is that there is no point pretending to finetune parameters when vol is ...

6

There is no standard approach to this problem to the best of my knowledge. Different approaches exist and each has its own pros and cons as usual. To mention a few: Information-based methods: these aim at "risk-neutralising" the distribution observed under the physical measure relying on some information criterion e.g. minimising the KL divergence, or ...

6

To answer, the assertion that volatility is easier to predict than expected return requires clarification. The phrase "easier to predict" is particularly ambiguous. To me this means that the estimation of volatility from a sample of returns is more robust than the estimation of expected return in the context of relative sampling error. Suppose over a time ...

6

The answer is not statistical. In almost every other area of statistics, estimating the mean is easier (i.e. it can be estimated with higher precision) and estimating higher moments like variance (and thus volatility), skewness, kurtosis, etc. is harder -- sometimes much harder. The key points that make financial statistics (or financial econometrics, if you ...

5

You can't have a precise argument without a precise definition. In general, the appropriate notion of integral here is the Lebesgue-Stieltjes integral. In a fairly general setup, let $F: \mathbb R \to \mathbb R$ be a right-continuous function that is of locally bounded variation, that is $$V_F([a,b]) := \sup\lbrace \sum_{i=1}^n \vert F(x_{i+1}) - F(x_i ) \... 4 Thanks @Aksakal for suggesting Kalman Filter. Here I provide more details. We will view it as a state-space model:$$ \begin{split} z_t &= A_t z_{t-1} + B_t u_t + \epsilon_t, \\ y_t &= C_t z_t + D_t u_t + \delta_t, \\ \epsilon_t &\sim \mathcal{N}(0, Q_t),\ \delta_t \sim \mathcal{N}(0, R_t), \end{split} $$where z_t is the latent variable, y_t... 4 Did you try solving for w_k?$$\bar{r}_t = \sum_{k=0}^p w_k r_{t-k}\bar R = W R$$Since you probably have t>>k, you can solve for W using OLS$$\bar R = W R +\varepsilon$$-- UPDATE You can try applying Kalman filter. Here, your state evolution is$$r_t=\mu+\varepsilon_t$$. You introduce new vector x_t=(r_t, r_{t-1}, \dots, r_{t-p+1})... 4 If \log{(|R_t|)} is your first term, I'm not sure why this is a matrix. Modulus (determinant herein) applied to a matrix R_t gives a scalar. If your implementation in python produces a matrix, that's likely because modulus is treated as an element-wise abs() function for each element of a matrix. It may be easier and faster to use rugarch (univariate ... 4 The \lambda value used in the original paper is arbitrary, but you can estimate that by assuming (in the simplest case) 2 assets and running the following model: \sigma^2_{12,t+1} = \lambda$$*$$\sigma^2_{12,t-1}$$+$$(1-\lambda)$$r_{1,t}$$*$$r_{2,t}$; given$r_{1,t}$and$r_{2,t}$respectively as the returns for the asset 1 and 2 and$\sigma^2_{12,t}$... 4 The answer to the original question is simple: the Chopra-Ziemba paper is highly flawed and unreliable. Note that the framework is in-sample and based on a utility function. It has nothing to do with out-of-sample behavior of the mean vs. the covariance in an optimization. Estimation error grows linearly in the mean but quadratically in the covariance. At ... 4 Since you're asking on a quant finance forum, the mathematical approach would be Decide on a model that the stock price follows, and Compute the expected value of the price, conditional on the most recent price. A famous model, made ubiquitous by Black, Scholes and Merton, is a geometric Brownian motion. Under this model, the stock price$S_T$at time$T$... 4 Not really. For infinite maturity bonds we have$Price = coupon/yield$so your approximation is actually correct. However for short dated bonds it is not a good approximation. For example , a 1 year annual pay bond gives$Price=(1+coupon)/(1+yield)$which is very poorly approximated by$coupon/yield$. 4 Based on an my updated understanding of your problem you have a portfolio consisting of$N$illiquid assets. Valuations are not real time and usually lagged, by say, upto 3 months (or slightly longer), but at least valuations correspond to a consistent timestamp (or otherwise you interpolate a consistent timestamp). You want to construct a predictive model ... 4 Expected returns are very difficult to estimate reliably without incurring estimation error as found out by Merton (1980) "On estimating the expected return on the market". This is why estimating volatility/the covariance matrix has become the default approach in the mean-variance model because volatility is easier to predict than returns. Even the global ... 4 The time step typically depends on the context. Due to the self-similarity of Brownian motion the mathematics should work similarly on any time scale, although the resultant estimates might vary greatly (as you mention). Since the cited article assumes a "high-frequency market maker," the implied time step seems to be the shortest time step available or ... 3 Unfortunately, financial markets are not like physical measures, where you know the "true" value of a physical variable but you just access to it thanks to noised sensors. We do not know the "true" volatility, just because there is not such one value... In statistics you have two kinds of modelling procedures: the ones dedicated to estimate the unknown ... 3 Using a realized kernel for calculating volatility will give you results in the same resolution as the data you feed them. So if you feed them minute-by-minute data, then the volatility will be calculated minute-by-minute. What that really means is that only once per minute will you have a good estimate of the volatility of whatever asset you're looking at. ... 3 There are many techniques, but I would begin with Stambaugh Analyzing Investments Whose Histories Differ in Lengths. The full information maximum likelihood approach he describes basically involves regressing the short history series against the long history series to obtain the covariance with the longer history securities and adding back the covariance of ... 3 Interesting question, as All the answers (including mine) could not be generalized unfortunately. As far as I am concerned, I use a univariate EGARCH for risk modelling purposes (Filtered Historical Simulation (FHS), etc.). 1 - EGARCH, merely because GARCH models do not take into account so-called leverage effects, which is crucial to me for skewed and ... 3$\alpha=0$does not imply constant volatility. Consider just a simple Garch(1,1): $$\sigma^2_t = \omega + \alpha \eta_t^2 + \beta \sigma^2_{t-1}$$ Note that: $$\sigma^2_t = \omega + (\alpha + \beta) \eta_t^2 - \beta (\eta_t^2- \sigma^2_{t-1})$$ Now add$\eta_{t+1}^2\$ to both sides: $$\eta_{t+1}^2 = \omega + (\alpha + \beta) \eta_t^2 - \beta (\eta_t^2-... 3 If you estimate your model via Maximum Likelihood method, you are forced to re-estimate the full model. This is due to the fact that estimates are values which maximize the full likelihood, the latter being based on a recursive algorithm which use all observations (including the new one) and implies that a new observation may also impact likelihood values of ... 3 First, you should use an exponential moving average, since the amount of state you need to keep is much smaller than for a simple moving average. Second the well known estimator of volatility,$$ \hat{\sigma} = \sqrt{\frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2} $$is not very robust, since the squaring amplifies the contribution of outliers (which is why ... 3 The sample variance and standard deviation (volatility) formulas are: If your question is why is volatility easier to predict than returns, the intuitive answer is because the numerator is squared and thus has only positive values. This simplifies the problem as now I don't have to worry about predicting the sign of the return, only the size. 3 A simpler answer is thus. There are known historical values for the past year for the mean. It's simply the end of year value divided by the beginning value. However, we can't improve the estimate of the mean by looking at, say, the daily returns and aggregating them up to 250 days of trading to make a better estimate of the mean (return): it will simply ... 3 EWMA (and other sort of moving averages) introduces positive autocorrelation into otherwise uncorrelated returns. The fitted values of EWMA are linear combinations of past returns, and the constituent elements of these combinations overlap. Therefore, positive autocorrelation arises. If you have autocorrelated returns to begin with, they would in all ... 3 It makes no sense to write C^h \to e^{\delta C} as T \to \infty when C = I_K +\Lambda/T since e^{\delta C} on the right-hand side depends on T. What can be confirmed is (C^h)_{k,k} \to e^{\delta \lambda_k} as T \to \infty with \delta fixed. Note that$$\log (C^h)_{k,k} = \lfloor \delta T\rfloor \log\left(1 + \frac{\lambda_k}{T}\right) = \...

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