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Hi Folks, I am dealing with data which have been presented as at each x_i, mean m_i of the y-values at x_i, sd s_i of the y-values at x_i number n_i of the y-values at x_i and I want to linearly regress y on x. There does not seem to be an option to 'lm' which can deal with such data directly, though the regression problem could be...

Greetings. For R gurus this may be a no brainer, but I could not find pointers to efficient computation of this beast in past help files. Background - I wish to implement a Cramer-von Mises type test statistic which involves double sums of max(X_i,Y_j) where X and Y are vectors of differing length. I am currently using ifelse pointwise in a vector, but have a nagging suspicion that there is a more efficient way to do this. Basically, I require three sums: sum1: \sum_i\sum_j max(X_i,X_j) sum2: \sum_i\sum_j max(Y_i,Y_j) sum3: \sum_i\sum_j ma...

Hi the list, According to what I know, the Canberra distance between X et Y is : sum[ (|x_i - y_i|) / (|x_i|+|y_i|) ] (with | | denoting the function 'absolute value') In the source code of the canberra distance in the file distance.c, we find : sum = fabs(x[i1] + x[i2]); diff = fabs(x[i1] - x[i2]); dev = diff/sum; which correspond to the formula : sum[ (|x_i - y_i...

Dear all, I try to conduct a SEM / two stage least squares regression with the following equations: First: X ~ IV1 + IV2 * Y Second: Y ~ a + b X therein, IV1 and IV2 are the two instruments I would like to use. the structure I would like to maintain as the model is derived from economic theory. My problem here is that I have trouble solving the equations to get the reduced form so I can run

...uld benefit from being written in C to avoid unnecessary memory allocations. I made this version using inline: require( inline ) sequence_c <- local( { fx <- cfunction( signature( x = "integer"), ' int n = length(x) ; int* px = INTEGER(x) ; int x_i, s = 0 ; /* error checking */ for( int i=0; i<n; i++){ x_i = px[i] ; /* this includes the check for NA */ if( x_i <= 0 ) error( "needs non negative integer" ) ; s += x_i ; } SEXP res = PROTECT(...

...91080581 -0.34774436 0.9552182 [9,] 0.19131383 0.14980569 -0.37458224 -0.09371273 -1.7667203 [10,] -0.85159276 -0.66679528 1.63019340 0.56920196 -2.4049600 And I define a boundary of X: The smallest "ball" that nests all the observations of X. I wish to check if a particular point x_i > x_i <- matrix(rnorm(5), 1, 5) > x_i [,1] [,2] [,3] [,4] [,5] [1,] -0.1525543 0.4606419 -0.1011011 -1.557225 -1.035694 is inside the boundary of X or not. I know it's easy to do it with 1-D or 2-D, but I don't knot how to manage it when the dimens...

...questions about clustering mixed numerical/categorical data. This time I am more into data mining. I am given a set of known statistical indexes {s_i}, i=1,2...N for a N countries. These indexes in general are a both numerical and categorical variables. For each country, I also have a property x_i whose value is known, but that I also would like to be able to predict correctly using a model. This is needed in order to assess the importance of the various indexes in determining {x_i}. There are two cases of interest (1) all the {x_i} are numerical variables, e.g. the average life expectanc...

Dear R users, I?m a graduate students and in my master thesis I must obtain the values of the parameters x_i which maximize this Multinomial log?likelihood function log(n!)-sum_{i=1]^4 log(n_i!)+sum_ {i=1}^4 n_i log(x_i) under the following constraints: a) sum_i x_i=1, x_i>=0, b) x_1<=x_2+x_3+x_4 c)x_2<=x_3+x_4 I have been using the ?ConstrOptim? R-function with the instructions I report be...

...omeone could help me anyway. Suppose, I have two random variables X and Y whose means are both known to be zero and I want to get an estimate of their covariance. I have n sample pairs (X1,Y1) (X2,Y2) . . . . . (Xn,Yn) , so that the covariance estimate is clearly 1/n *(sum from i = 1 to n of ( X_i*Y_i) ) But, suppose that it is know that the X_i are positively correlated with each other and that the Y_i are independent of each other. Then, does this change the formula for the covariance estimate at all ? Intuitively, I would think that, if the X_i's are positively correlated , then so...

Canberra distance is defined in function `dist' (standard library `mva') as sum(|x_i - y_i| / |x_i + y_i|) Obviously this is undefined for cases where both x_i and y_i are zeros. Since double zeros are common in many data sets, this is a nuisance. In our field (from which the distance is coming), it is customary to remove double zeros: contribution to distance is zero when bo...

Canberra distance is defined in function `dist' (standard library `mva') as sum(|x_i - y_i| / |x_i + y_i|) Obviously this is undefined for cases where both x_i and y_i are zeros. Since double zeros are common in many data sets, this is a nuisance. In our field (from which the distance is coming), it is customary to remove double zeros: contribution to distance is zero when bo...

Muy buenas, Tengo la siguiente duda/problema, He optimizado con éxito un problema de este tipo: \sum f(x_i) donde f es una curva exponencial (función no lineal) sujeto a: a_i < x_i < b_i y \sum f(x_i) < Presupuesto Vamos, es repartir un presupuesto forzando a que inviertas como poco a_i y como mucho b_i para cada i Esto lo hecho correctamente usando el paquete: http://cran.r-project.org/w...

I am trying to test out several mgcv::gam models in a scalar-on-function regression analysis. The following is the 'hierarchy' of models I would like to test: (1) Y_i = a + integral[ X_i(t)*Beta(t) dt ] (2) Y_i = a + integral[ F{X_i(t)}*Beta(t) dt ] (3) Y_i = a + integral[ F{X_i(t),t} dt ] equivalents for discrete data might be: 1) Y_i = a + sum_t[ L_t * X_it * Beta_t ] (2) Y_i = a + sum_t[ L_t * F{X_it} * Beta_t ] (3) Y_i = a + sum_t[ L_t * F{X_it,t} ] where Y_i are sc...

...iones de una distribucion binomial bivariada en la que hay _cierto_ grado de correlacion (denotemoslo rho). Podria por favor alguien indicarme como hacerlo en R? Este es el contexto. Supongamos que se tienen dos experimentos en los que la variable respuesta _sigue_ una distribucion binomial, i.e., X_i ~Binomial(n_i, p_i), i=1,2 y que, por ahora, n_i y p_i son conocidos. El interes principal es calcular T = P(X_1=1, X_2=1). Si las X_i''s fueran independientes (caso 1), seria suficiente generar binomiales con los parametros correspondientes a cada tipo de experimento (via rbinom) y calcul...

Dear All, Here is what I am trying to achieve: I would like to plot some data in 3D. Usually, one has a matrix of the kind y_1(x_1) , y_1(x_2).....y_1(x_i) y_2(x_1) , y_2(x_2).....y_2(x_i) ........................................... y_n(x_1) , y_n(x_2)......y_n(x_i) where e.g. y_2(x_1) is the value of y at time 2 at point x_1 (see that the grid in x is the same for the y values at all times). Instead, in my case, the quantity y is observed at each...

Suppose one wanted to consider random variables X_1,...X_n and from each subtract off the piece which is correlated with the previous variables in the list. i.e. make new variables Z_i so that Z_1=X_1 and Z_i=X_i-cov(X_i,Z_1)Z_1/var(Z_1)-...- cov(X_i,Z__{i-1})Z__{i-1}/var(Z_{i-1}) I have code to do this but I keep getting a "non-conformable array" error in the line with the covariance. Does anyone have any suggestions? Here is my code: gov=read.table(file.choose(), sep="\t",header=T)...

hello. I have a question on the interpretation of a logistic model. is it helpful to exponentiate the coefficients (estimates)? I think I once read something about that, but I cannot remember where. if so, how would be the interpretation of the exp(estimate) ? would there be a change of the interpretation of the ANOVA table (or is the ANOVA table not really helpful at all?). thanks for your

Hi, I'm trying to get maximum likelihood estimates of \alpha, \beta_0 and \beta_1, this can be achieved by solving the following three equations: n / \alpha + \sum\limits_{i=1}^{n} ln(\psihat(i)) - \sum\limits_{i=1}^{n} ( ln(x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} 1/(psihat(i)) - (\alpha+1) \sum\limits_{i=1}^{n} ( 1 / (x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} ( i / \psihat(i) ) - (\alpha + 1) \sum\limits_{i=1}^{n} ( i / (x_i + \psihat(i)) ) = 0 where \psihat=\beta_0 + \beta_1 * i. Now i want t...

I tried lrm in library(Design) but there is always some error message. Is this function really doing the weighted logistic regression as maximizing the following likelihood: \sum w_i*(y_i*\beta*x_i-log(1+exp(\beta*x_i))) Does anybody know a better way to fit this kind of model in R? FYI: one example of getting error message is like: > x=runif(10,0,3) > y=c(rep(0,5),rep(1,5)) > w=rep(1/10,10) > fit=lrm(y~x,weights=w) Warning message: currently weights are ignored in model valida...

...e Hosmer and Lemeshow text on Applied Survival Analysis to that of the help that comes with the survival package. I am trying to back out the values for the baseline hazard, h_o(t_i), for each event time or observation time. Now survfit(fit)$surv gives me the value of the survival function, S(t_i|X_i,B), using mean values of the covariates and the coxph() object provides me with the estimate of the linear predictors, exp(X'B). If S(t_i|X_i,B)=S_o(t_i)^exp(X_iB) is the expression for the survival function And -ln(S_o(t_i) ) is the expression for the cumulative baseline hazard function, H_o(...