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| bio | website | |
|---|---|---|
| location | Chambery, France | |
| age | 51 | |
| visits | member for | 1 year, 1 month |
| seen | 8 hours ago | |
| stats | profile views | 236 |
Statistical consultant
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May 9 |
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Fitting GEV to non-stationary time series of extremes (general stationarity question?) @rbatt. Yes, the Likelihood maximisation may fail, even for $\mu$ constant. A good precaution is to scale the data. You can easily produce non-convergence with extreme values fits by making the data larger, say multiplying them by 1000. The shape parameter $\xi$ has considerable impact and has no gaussian equivalent. Many estimators of $\xi$ exist for the stationary case, and maybe these can be used as initial values, first ignoring covariates? |
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May 9 |
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Fitting GEV to non-stationary time series of extremes (general stationarity question?) I changed the notations in the answer to be closer to classical ones. |
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May 9 |
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Fitting GEV to non-stationary time series of extremes (general stationarity question?) The variance of the gaussian regression error $e \sim N(0,\,1)$ is necessarily unity? |
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May 2 |
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Fitting a Poisson distribution from missing observations So why not simply omit the missing observations in the estimation e.g., using Maximum Likelihood? |
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May 1 |
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Fitting a Poisson distribution from missing observations If there is an underlying Poisson process or a Negative Binomial Lévy process, the distribution of the number of events within a period can be related to the effective duration which may be known, possibly with error. For instance, if you know that within a month the process was only observed during 20 days, this can be used in a GLM with appropriate link. |
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Apr 19 |
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Visualizing high dimensional data A very good tool indeed, which also works with R. |
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Mar 29 |
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the combination of two independent continuous time Markov chains Since the two MC $x_t$ and $y_t$ are time homogeneous so is the MC $s_t$. The time spent by $s_t$ in the state $11$, which is the time spent by $z_t$ in state $1$, can be computed using continuous MC theory. This imply: first writing the $4 \times 4$ generator matrix for $s_t$ using the two rates $\lambda$ and $\mu$ for transitions $0 \rightarrow 1$ and $1 \rightarrow 0$, then find the stationnary distribution. From a more more statistical point a view, if only $z_t$ is observed, inference on $x_t$ and $y_t$ can be drawn using HMM techniques (Baum-Welch, Viterbi, ...). |
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Mar 26 |
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the combination of two independent continuous time Markov chains Happy if this helps... Yes, at $t_0$ the time spent in state $z=0$ contains information about how quickly we can leave it. One can draw a graph with the four states of $s_t$ as nodes and transitions as arcs. Then $z_t=0$ means that $s_t$ is in the group of states $\{00,\,01, 10\}$. Within this group, $s_t$ can be one step away from the exit $z =0 \rightarrow z=1$, or two steps away. |
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Feb 24 |
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How to generate a non-integer amount of consecutive Bernoulli successes? Thank you @cardinal. I agree with all your comments except perhaps (3). I actually made an error since $c$ is $-1/\Gamma(-a)$ (edited), but the exponent of $n$ seems the right one. I used the representation of $\Gamma(z)$ as found e.g. on Wikipedia page on infinite product and took $z:=-a$ which gives an equivalent for the product $\prod_{k=1}^{n-1}$. I would be more confident if you could check this. |
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Feb 24 |
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How to generate a non-integer amount of consecutive Bernoulli successes? A related idea would be to make the rvs $X_k$ dependent with extremal index $\theta$ $(0 < \theta < 1)$, meaning that $M_n$ has PZ $q^{n\theta}$ rather than $q^n$. Taking $n\theta = a$ would do the job for any $a>0$. Given a sequence of iid rv.s $X_n$ following a standard Frechet, there are known methods to generate a dependent sequence $X_n^\star$ with standard Frechet margin and the prescribed extremal index $\theta$. However, what happens if we replace standard Frechet'' by Bernoulli''? |
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Oct 11 |
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Relations between distributions @Miroslav Sabo. Sorry, I missed your comment while searching for the title. Yet a nice graph! |
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Sep 13 |
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Computing the mode of data sampled from a continuous distribution As a possible property of a distribution, log-concavity is stronger than unimodality (proved to be strong unimodality by Ibragimov) and more appealing in many situations. Fitting a log-concave density can be an option, e.g. using the 'logconcdens' R package. |
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Sep 12 |
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Combining a linear Kalman Filter with additional linear constraints? Thank you, Bogdanovist. I soon edit the text and turn it into a answer. |
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Sep 9 |
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Combining a linear Kalman Filter with additional linear constraints? Among other spatio-temporal techniques, the Kriged Kalman filter can be a solution. The spatial smoothing captures the spatial low frequencies thanks to the Principal Kriging Functions and seems comparable to that of thin plate splines. RBF are used if the covariance kernel is square exponential. southampton.ac.uk/~sks/research/papers/sahumardia05.pdf |
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Aug 27 |
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Appropriate clustering techniques for temporal data? Searches with "time series segmentation" or "regime switching models" keywords may help you. |
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Aug 23 |
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Determine range that contains given ratio of data points Yes, at least if the distribution seems unimodal. You can also use a nonparametric estimation such as given by the 'density' R function in the stats package. |
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Aug 19 |
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L1 regression estimates median whereas L2 regression estimates mean? A quick and dirty version of the algebra given by muratoa exists for the L1 case. Observe that except when $\beta = y_i$, the derivative of $| y_i -\beta |$ w.r.t $\beta$ is $-\mathrm{sgn}(y_i-\beta)$, that is $-1$ if $\beta < y_i$ and $+1$ if $\beta > y_i$. So $\frac{\mathrm{d}}{\mathrm{d}\beta} \,\frac{1}{n}\sum_i | y_i -\beta | = -\frac{1}{n}\,\sum_i \mathrm{sgn}(y_i-\beta)$, except when $\beta$ is an $y_i$. The derivative vanishes when there is the same number of positive and negative terms among the $y_i-\beta$, which roughly speaking arises when $\beta$ is the median of the $y_i$. |
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Jul 1 |
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Parameter estimation for the sum of two Independent (not necessarily i.d.) Gamma RVs As Procrastinator said, you will face identifiability problems and need inequality constraints on the parameters. In the frequentist context, you can use the Expectation-Maximisation (EM) algorithm. This needs here to compute numerical densities and conditional expectations by using the Discrete Fourier Transform (DFT). It will require writing a program (R, octave or Matlab), and also testing it carefully. You can also use similar ideas in a bayesian framework. |
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Jun 28 |
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What is the name of this continuous univariate distribution? @Xi'an Of course you are right: what is in a name? Yet although the exponential is only a special case of gamma, it is useful to have it in R under this name. |
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Jun 27 |
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What is the name of this continuous univariate distribution? @Macro: my motivation is to find a nice name for a R implementation of this specific 2-parameters distribution. It is only a very special case within the two families and it would be misleading to call it, say, rescBeta. |
