I am reading this statistics book where they have mentioned that the attached top plot has no correlation between adjacent residuals. Whereas, the bottom most has correlation with p-0.9. Can anybody please provide some direction as to how to analyze this? Thank you very much for your time.
Correlated errors mean that the lag 1 correlation is p. That is, Cor(Yi, Yi-1) = p. This can be modelled using Yi = mu + p epsiloni-1 + epsiloni where epsiloni ~ N(0, 1) for all i. We can verify that the correlation between adjacent data points is p: Cov(Yi, Yi-1) = Cov(p epsiloni-1 + epsiloni, p epsiloni-2 + epsiloni-1) = Cov(p epsiloni-1, epsiloni-1) = p Var(epsiloni-1) = p. Code to demonstrate appears below:
set.seed(123)
epsilonX <- rnorm(100, 0, 1)
epsilonY <- rnorm(100, 0, 1)
epsilonZ <- rnorm(100, 0, 1)
X <- NULL
Y <- NULL
Z <- NULL
Y[1] <- epsilonY[1]
X[1] = epsilonX[1]
Z[1] = epsilonZ[1]
rhoX = 0
rhoY = 0.5
rhoZ = 0.9
for (i in 2:100) {
Y[i] <- rhoY * epsilonY[i-1] + epsilonY[i]
X[i] <- rhoX * epsilonX[i-1] + epsilonX[i]
Z[i] <- rhoZ * epsilonZ[i-1] + epsilonZ[i]
}
param = par(no.readonly = TRUE)
par(mfrow=c(3,1))
plot(X, type='o', xlab='', ylab='Residual', main=expression(rho*"=0.0"))
abline(0, 0, lty=2)
plot(Y, type='o', xlab='', ylab='Residual', main=expression(rho*"=0.5"))
abline(0, 0, lty=2)
plot(Z, type='o', xlab='', ylab='Residual', main=expression(rho*"=0.9"))
abline(0, 0, lty=2)
#par(param)
acf(X)
acf(Y)
acf(Z)
Note from the acf plots that the lag 1 correlation is insignificant for p = 0, higher for p = 0.5 data (~0.3), and still higher for p = 0.9 data (~0.5).
Related
I am having an issue with the ggplot code line where R doesn't like the "group = Year".
Here is what my data looks like:
> head(data.scores.pa)
NMDS1 NMDS2 NMDS3 Site Year Elevation Fire history
1 -0.737547 0.73473457 0.7575643 BF 2004 1710 Burnt
......
> head(spp.scrs2)
species MDS1 MDS2 pval
1 Acrothamnus.montanus 0.8383 -0.02382347 1e-04
........
> head(vec.sp.df.pa)
MDS1 MDS2 species pvals
Elevation 0.834847 0.747474 Elevation 0.005
Here is the code I am using:
>xy <- ggplot(data.scores.pa, aes(x = NMDS1, y = NMDS2, group = Year)) +
geom_point(size = 3, aes(shape = Fire history, colour = Year))+
stat_ellipse(mapping = NULL, data = NULL, geom = "path", position = "identity", type = "t", level = 0.95, segments = 51, na.rm = FALSE, show.legend = NA, inherit.aes = TRUE) +
geom_segment(data=vec.sp.df.pa, aes(x=0,xend=MDS1,y=0,yend=MDS2),
arrow = arrow(length = unit(0.5,"cm")),colour="grey")+
geom_text_repel(data=vec.sp.df.pa,aes(x=MDS1,y=MDS2,label=species),size=2)+
geom_segment(data=spp.scrs2,aes(x=0,xend=MDS1,y=0,yend=MDS2),
arrow = arrow(length = unit(0.5, "cm")),colour="black")+
geom_text_repel(data=spp.scrs2, aes(x=MDS1,y=MDS2,label=species),size=2)+
annotate("text", x = -1.6, y = 1, label = paste0("3D stress: ", format(ord.pa$stress, digits = 4)), hjust = 0) +
theme_cowplot() + scale_color_brewer(palette = "BrBG", direction = 1) +
theme(panel.border = element_rect(colour = "black"))+
ggtitle("All Sites - distance data using Bray-Curtis")+
labs(x = "NMDS1", y = "NMDS2")
> Error in FUN(X[[i]], ...) : object 'Year' not found
However, when I remove the geom_segment and geom_text_repel code lines it fixes the problem and I am able to plot the graph...
Is anyone able to provide some insight into this issue?
Thank you!
I tried to obtain MLEs of the Vasicek function using the following function.
I am running into into the following error constantly and I have no way to solve it. Please help me. Thanks!
Error in if (!all(lower[isfixed] <= fixed[isfixed] & fixed[isfixed] <= :
missing value where TRUE/FALSE needed
Here is the background:
Likelihood function
likehood.Vasicek<-function (theta, kappa, sigma, rt){
n <- NROW(rt)
y <- rt[2:n,] # Take rates other than r0
dt <- 1/12 # Simulated data is monthly
mu <- rt[1:(n-1),]* exp(-kappa*dt) + theta* (1- exp(-kappa*dt)) #Take prior rates for mu calculation
sd <- sqrt((sigma^2)*(1-exp(-2*kappa*dt))/(2*kappa))
pdf_yt <- dnorm(y, mu, sd, log = FALSE)
- sum(log(pdf_yt))
}
Simulating scenarios
IRModeling.Vasicek = function(r0, theta, kappa, sigma, T, N){
M <- T*12 # monthly time step
t <- 1/12 # time interval is monthly
rt = matrix(0, M+1, N) # N sets of scenarios with M months of time steps
rt[1,] <- r0 # set the initial value for each of the N scenarios
for (i in 1:N){
for (j in 1:M){
rt[j+1,i] = rt[j,i] + kappa*(theta - rt[j,i])*t + sigma*rnorm(1,mean=0,sd=1)*sqrt(t)
}
}
rt # Return the values
}
MLE
r0 = 0.03
theta = 0.03
kappa = 0.3
sigma = 0.03
T = 5 # years
N = 500
rt = IRModeling.Vasicek (r0, theta, kappa, sigma, T, N)
theta.est <- 0.04
kappa.est <- 0.5
sigma.est <- 0.02
parameters.est <- c(theta.est, kappa.est, sigma.est)
library(stats4)
bound.lower <- parameters.est*0.1
bound.upper <- parameters.est*2
est.mle<-mle(likelihood.Vasicek, start= list(theta = theta.est, kappa = kappa.est, sigma = sigma.est),
method="L-BFGS-B", lower=bound.lower, upper= bound.upper, fixed = list(rt = rt))
summary(est.mle)
Error
Error in if (!all(lower[isfixed] <= fixed[isfixed] & fixed[isfixed] <= :
missing value where TRUE/FALSE needed
I'm trying to implement RGB to HSV conversion from opencv in pure numpy using formula from here:
def rgb2hsv_opencv(img_rgb):
img_hsv = cv2.cvtColor(img_rgb, cv2.COLOR_RGB2HSV)
return img_hsv
def rgb2hsv_np(img_rgb):
assert img_rgb.dtype == np.float32
height, width, c = img_rgb.shape
r, g, b = img_rgb[:,:,0], img_rgb[:,:,1], img_rgb[:,:,2]
t = np.min(img_rgb, axis=-1)
v = np.max(img_rgb, axis=-1)
s = (v - t) / (v + 1e-6)
s[v==0] = 0
# v==r
hr = 60 * (g - b) / (v - t + 1e-6)
# v==g
hg = 120 + 60 * (b - r) / (v - t + 1e-6)
# v==b
hb = 240 + 60 * (r - g) / (v - t + 1e-6)
h = np.zeros((height, width), np.float32)
h = h.flatten()
hr = hr.flatten()
hg = hg.flatten()
hb = hb.flatten()
h[(v==r).flatten()] = hr[(v==r).flatten()]
h[(v==g).flatten()] = hg[(v==g).flatten()]
h[(v==b).flatten()] = hb[(v==b).flatten()]
h[h<0] += 360
h = h.reshape((height, width))
img_hsv = np.stack([h, s, v], axis=-1)
return img_hsv
img_bgr = cv2.imread('00000.png')
img_rgb = cv2.cvtColor(img_bgr, cv2.COLOR_BGR2RGB)
img_rgb = img_rgb / 255.0
img_rgb = img_rgb.astype(np.float32)
img_hsv1 = rgb2hsv_np(img_rgb)
img_hsv2 = rgb2hsv_opencv(img_rgb)
print('max diff:', np.max(np.fabs(img_hsv1 - img_hsv2)))
print('min diff:', np.min(np.fabs(img_hsv1 - img_hsv2)))
print('mean diff:', np.mean(np.fabs(img_hsv1 - img_hsv2)))
But I get big diff:
max diff: 240.0
min diff: 0.0
mean diff: 0.18085355
Do I missing something?
Also maybe it's possible to write numpy code more efficient, for example without flatten?
Also I have hard time finding original C++ code for cvtColor function, as I understand it should be actually function cvCvtColor from C code, but I can't find actual source code with formula.
From the fact that the max difference is exactly 240, I'm pretty sure that what's happening is in the case when both or either of v==r, v==g are simultaneously true alongside v==b, which gets executed last.
If you change the order from:
h[(v==r).flatten()] = hr[(v==r).flatten()]
h[(v==g).flatten()] = hg[(v==g).flatten()]
h[(v==b).flatten()] = hb[(v==b).flatten()]
To:
h[(v==r).flatten()] = hr[(v==r).flatten()]
h[(v==b).flatten()] = hb[(v==b).flatten()]
h[(v==g).flatten()] = hg[(v==g).flatten()]
The max difference may start showing up as 120, because of that added 120 in that equation. So ideally, you would want to execute these three lines in the order b->g->r. The difference should be negligible then (still noticing a max difference of 0.01~, chalking it up to some round off somewhere).
h[(v==b).flatten()] = hb[(v==b).flatten()]
h[(v==g).flatten()] = hg[(v==g).flatten()]
h[(v==r).flatten()] = hr[(v==r).flatten()]
I'd like to compare a betareg regression vs. the same regression using rjags
library(betareg)
d = data.frame(p= sample(c(.1,.2,.3,.4),100, replace= TRUE),
id = seq(1,100,1))
# I am looking to reproduce this regression with jags
b=betareg(p ~ id, data= d,
link = c("logit"), link.phi = NULL, type = c("ML"))
summary(b)
Below I am trying to do the same regression with rjags
#install.packages("rjags")
library(rjags)
jags_str = "
model {
#model
y ~ dbeta(alpha, beta)
alpha <- mu * phi
beta <- (1-mu) * phi
logit(mu) <- a + b*id
#priors
a ~ dnorm(0, .5)
b ~ dnorm(0, .5)
t0 ~ dnorm(0, .5)
phi <- exp(t0)
}"
id = d$id
y = d$p
model <- jags.model(textConnection(jags_str),
data = list(y=y,id=id)
)
update(model, 10000, progress.bar="none"); # Burnin for 10000 samples
samp <- coda.samples(model,
variable.names=c("mu"),
n.iter=20000, progress.bar="none")
summary(samp)
plot(samp)
I get an error on this line
model <- jags.model(textConnection(jags_str),
data = list(y=y,id=id)
)
Error in jags.model(textConnection(jags_str), data = list(y = y, id = id)) :
RUNTIME ERROR:
Invalid vector argument to ilogit
Can you advise
(1) how to fix the error
(2) how to set priors for the beta regression
Thank you.
This error occurs because you have supplied the id vector to the scalar function logit. In Jags inverse link functions cannot be vectorized. To address this, you need to use a for loop to go through each element of id. To do this I would probably add an additional element to your data list that denotes how long id is.
d = data.frame(p= sample(c(.1,.2,.3,.4),100, replace= TRUE),
id = seq(1,100,1), len_id = length(seq(1,100,1)))
From there you just need to make a small edit to your jags code.
for(i in 1:(len_id)){
y[i] ~ dbeta(alpha[i], beta[i])
alpha[i] <- mu[i] * phi
beta[i] <- (1-mu[i]) * phi
logit(mu[i]) <- a + b*id[i]
}
However, if you track mu it is going to be a matrix that is 20000 (# of iterations) by 100 (length of id). You are likely more interested in the actual parameters (a, b, and phi).
I need to write a program to generate and display a piecewise quadratic Bezier curve that interpolates each set of data points (I have a txt file contains data points). The curve should have continuous tangent directions, the tangent direction at each data point being a convex combination of the two adjacent chord directions.
0.1 0,
0 0,
0 5,
0.25 5,
0.25 0,
5 0,
5 5,
10 5,
10 0,
9.5 0
The above are the data points I have, does anyone know what formula I can use to calculate control points?
You will need to go with a cubic Bezier to nicely handle multiple slope changes such as occurs in your data set. With quadratic Beziers there is only one control point between data points and so each curve segment much be all on one side of the connecting line segment.
Hard to explain, so here's a quick sketch of your data (black points) and quadratic control points (red) and the curve (blue). (Pretend the curve is smooth!)
Look into Cubic Hermite curves for a general solution.
From here: http://blog.mackerron.com/2011/01/01/javascript-cubic-splines/
To produce interpolated curves like these:
You can use this coffee-script class (which compiles to javascript)
class MonotonicCubicSpline
# by George MacKerron, mackerron.com
# adapted from:
# http://sourceforge.net/mailarchive/forum.php?thread_name=
# EC90C5C6-C982-4F49-8D46-A64F270C5247%40gmail.com&forum_name=matplotlib-users
# (easier to read at http://old.nabble.com/%22Piecewise-Cubic-Hermite-Interpolating-
# Polynomial%22-in-python-td25204843.html)
# with help from:
# F N Fritsch & R E Carlson (1980) 'Monotone Piecewise Cubic Interpolation',
# SIAM Journal of Numerical Analysis 17(2), 238 - 246.
# http://en.wikipedia.org/wiki/Monotone_cubic_interpolation
# http://en.wikipedia.org/wiki/Cubic_Hermite_spline
constructor: (x, y) ->
n = x.length
delta = []; m = []; alpha = []; beta = []; dist = []; tau = []
for i in [0...(n - 1)]
delta[i] = (y[i + 1] - y[i]) / (x[i + 1] - x[i])
m[i] = (delta[i - 1] + delta[i]) / 2 if i > 0
m[0] = delta[0]
m[n - 1] = delta[n - 2]
to_fix = []
for i in [0...(n - 1)]
to_fix.push(i) if delta[i] == 0
for i in to_fix
m[i] = m[i + 1] = 0
for i in [0...(n - 1)]
alpha[i] = m[i] / delta[i]
beta[i] = m[i + 1] / delta[i]
dist[i] = Math.pow(alpha[i], 2) + Math.pow(beta[i], 2)
tau[i] = 3 / Math.sqrt(dist[i])
to_fix = []
for i in [0...(n - 1)]
to_fix.push(i) if dist[i] > 9
for i in to_fix
m[i] = tau[i] * alpha[i] * delta[i]
m[i + 1] = tau[i] * beta[i] * delta[i]
#x = x[0...n] # copy
#y = y[0...n] # copy
#m = m
interpolate: (x) ->
for i in [(#x.length - 2)..0]
break if #x[i] <= x
h = #x[i + 1] - #x[i]
t = (x - #x[i]) / h
t2 = Math.pow(t, 2)
t3 = Math.pow(t, 3)
h00 = 2 * t3 - 3 * t2 + 1
h10 = t3 - 2 * t2 + t
h01 = -2 * t3 + 3 * t2
h11 = t3 - t2
y = h00 * #y[i] +
h10 * h * #m[i] +
h01 * #y[i + 1] +
h11 * h * #m[i + 1]
y