In the 2 and 3 vector case it is possible do define a fairly simple sampling
space where this is possible.
Consider the unit square where the sample space is the area where x+y <1.
It generalizes to 3 dimensions with no difficulty.
x= (0:100)/100
y= (0:100)/100
z=outer(x,y, function(x,y) 1-
# fixed formula part:
f <- dat ~ a0 * exp(-S*(x - 250)) + K
# convert to character
f <- as.character(f)
# component:
C <- "(p0*exp(-0.5*((x-p1)/p2)^2))"
# number of components (defined randomly):
n <- sample(1:3, 1)
C <- rep(C, n)
# collapse:
C <- paste(C, collapse = "+")
# combine
f <- paste(f[2],
> xx <- as.factor(c("AL", "AK", "CA", "FL"))
> xx
[1] AL AK CA FL
Levels: AK AL CA FL
> as.character(xx)
[1] "AL" "AK" "CA" "FL"
I hope this helps.
Chel Hee Lee
On 14-11-22 01:12 AM, Aditya Singh wrote:
Dear Boris and R-Experts,
I have a variable my_state which is a 2-letter character string
Hi Boris/ David
Many thanks for the kind assistance. I will try following your codes.
Time has always been a slippery subject to me!
Cheers
On Sat, Nov 22, 2014 at 12:59 AM, David Winsemius
wrote:
>
> On Nov 21, 2014, at 3:19 PM, David Winsemius wrote:
>
>>
>> On Nov 21, 2014, at 2:55 PM, Ragh
Ok, thanks for the suggestions. I will look into that. And you are absolutely
right that I should have been more clear about what type of weighting I want.
So to clarify: I run time series regressions of returns of company i on two
different sets of explanatory variables. Then I extract the resp
On 21 Nov 2014, at 19:18, David Winsemius wrote:
>
> On Nov 21, 2014, at 6:52 AM, ivan wrote:
>
>> I am aware of the fact that bootstrapping produces different CIs with every
>> run. I still believe that there is a difference between both types of
>> procedures. My understanding is that sett
Of course they are random. But they can't all be randomly picked from [0,1).
By scaling them, one is effectively scaling the interval from which they are
picked.
B.
Nb: the scaling procedure will work for any probability distribution.
On Nov 22, 2014, at 10:54 AM, Ranjan Maitra
wrote:
> I d
I don't understand this discussion at all.
n random numbers constrained to have sum <=1 are still random. They are not all
independent.
That said, the original poster's question is ill=formed since there can be
multiple distributions these random numbers come from.
best wishes,
Ranjan
On Sa
These are contradictory requirements: either you have n random numbers from the
interval [0,1), then you can't guarantee anything about their sum except that
it will be in [0,n). Or you constrain the sum, then your random numbers cannot
be random in [0,1). You could possibly scale the random num
(Hit send key by accident before I was finished ...)
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Sat, Nov 22, 2014 at 7:14 AM, Bert Gunter wrote:
> Well, if th
Well, if their sum must be < 1 they ain't random...
But anyway... given n
randnums <- function(n)
{
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Sat, Nov 22,
Dear all,
I use R 3.1.1 for Windows.
kindly how can I generate n number of random numbers with probability from [0,1]
and their sum must not be more than one
thanks in advance
Ragia
[[alternative HTML version deleted]]
> On 22 Nov 2014, at 08:12 , Aditya Singh wrote:
>
> Dear Boris and R-Experts,
>
> I have a variable my_state which is a 2-letter character string telling which
> American state the user inputs. This I am do a if(identical(database
> entry,my_state)) to check for occurrences in the database.
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