Yves

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bio website location Chambery, France age 51 member for 1 year, 8 months seen 11 hours ago profile views 296

Statistical consultant

 Dec3 comment How to extract covariance (error) matrix of independent parameters x,y from covariance matrix of dependent parameters x,y,z? Sorry, it is much simpler than I suggested. Your matrix is simply the $2 \times 2$ submatrix of the whole $3 \times 3$. Hopefully the submatrix is non-degenerated while the large one has rank $\leq 2$. Dec3 comment How to extract covariance (error) matrix of independent parameters x,y from covariance matrix of dependent parameters x,y,z? You could begin with a Gaussian random vector and linear $f$, i.e. $f= ax + b y+ cz$, which will help you to clarify the assumptions: you need $c \neq 0$. In this case, block matrix computations will give you the answer. You may then carefully extend your result for an $f$ is close enough to a linear function by relying on a linear approximation. Nov20 comment How do I check if my data fits an exponential distribution? Tests for the exponential distribution have been discussed in question 32061. The 'best' test strongly depends on the alternative hypothesis: gamma, Weibull, Pareto, ... Log-normal and Pareto seem to be good choices for salaries. Sep6 comment What are the standard statistical tests to see if data follows exponential or normal distributions? I came across some more recent and extensive ressources on testing for exponentiality. 1) An article: A Henze, N. and Meintanis, S.G. (2005): 'Recent and classical tests for exponentiality: a partial review with comparisons'. Metrika, vol. 61, pp. 29–45. 2) A CRAN R package named 'exptest' implementing the tests of the mentioned article. Aug13 comment Comparing multiple incidence rates In this LR test, I think that the number of degrees of freedom to use is 3 = 4 - 1, since the resticted 'null' model has 1 df while the unrestricted has 4 df, one by region. Aug13 comment Comparing multiple incidence rates This is a Likelihood-Ratio test. A LR test works also for the Poisson model where the expectation of $S$ is assumed to be $\lambda \times \text{Pop}$. The null hypothesis is then $\lambda_{\text{A}} = \lambda_{\text{B}} = \lambda_{\text{C}} = \lambda_{\text{D}}$. LR could also work for overdispersed the Negative Binomial case, but more data would be required to estimate the two parameters (using a non-canonical glm link). Jul3 comment Bayesian estimation of tridiagonal covariance Sorry @najmeh, the notations of my previous comment are very confusing. If the distribution is $p$-dimensional, each $\mathbf{X}_i$ in a sample of size $N$ is a timeseries $X_{i,t}$ with length $p$. So we can have $X_{i,t} = \mu_t + \varepsilon_{i,t} - \theta \varepsilon_{i,t-1}$. Anyway, the idea was to deal with $\boldsymbol{\Sigma}$ through its Cholesky LDL' decomposition and a simple parametrisation of it. This does not imply a timeseries interpretation. Jul3 comment Bayesian estimation of tridiagonal covariance With $X_t = \mu + \varepsilon_t - \theta \varepsilon_{t-1}$ you will have a mean $\mu$ and thus $3$ parameters. The covariance matrix has a simple expression, and embeds $2$ free parameters only, instead of $2N-1$ for a general tridiagonal covariance. Bayesian inference for this model is possible through MCMC. However to get a more flexible covariance description, you can let $\theta$ vary smoothly with $t$. Jul3 comment Bayesian estimation of tridiagonal covariance A moving average MA(1) model for a timeseries $X_t$ leads to a specific tridiagonal covariance matrix. This reads as $X_t = \varepsilon_t - \theta \varepsilon_{t-1}$ where $\varepsilon_t$ is Gaussian White Noise and $\theta$ the MA coefficient. The two parameters $\theta$ and the variance $\sigma^2_\varepsilon$ can have independent priors, e.g. uniform and gamma. By specifying a slowly varying $\theta$ and/or variance, you will get more general tridiagonal matrices. This may be a hint for a hierarchical formulation of your model if the timeseries formulation makes sense in your context. May30 comment Question about squares of the coefficients of variation Sorry. In my last comment, the 2nd question should be : is it true that the CV of $Z^a$ is always an decreasing function of $a$? May30 comment Question about squares of the coefficients of variation I would rephrase the question as: given a r.v. $Z \geqslant 1$, is it possible to have $\textrm{CV}(Z^b) > \textrm{CV}(Z^a)$ with $b >a$? Or: is it true that the CV of $Z^a$ is always an increasing function of $a$? May9 comment 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? May9 comment 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. May9 comment 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? May2 comment Fitting a Poisson distribution from missing observations So why not simply omit the missing observations in the estimation e.g., using Maximum Likelihood? May1 comment 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. Apr19 comment Visualizing high dimensional data A very good tool indeed, which also works with R. Mar29 comment 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, ...). Mar26 comment 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. Feb24 comment 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.