Hi,
I am trying to use dunif and runif
however, I have two problems:
if I do
dunif(1:10, min=1, max=10)
I get 10 values, which summed give me 1.1111
I understand that the probability is computed as f(x) = 1 / (max-min)
but in this case it looks wrong: I have 10 values, each one
equiprobable, and the probability for each one should be 0.1 and not
0.11111 (which is, consistently with the definition, 1/9)
It looks like one of the extremes is not considered in the computation
of the probability, but then it's assigned a probability anyway.
Similar problem with punif.
if I do
punif(1, min=1, max=10)
I get 0 as result, as if the lower extreme is not considered, which is
not consistent with the description where min <= x <= max
If the lower extreme is not considered because cdf(x) = p(X<x) {and
not p(X<=x)} the problem stands in p(X<11) which should be the sum of
everything. ( P(1) + P(2) + ... + P(10) )
What is happening here?
Dunif and Punif
4 messages · michele donato, R. Michael Weylandt, Peter Langfelder
In short, the unif() distribution corresponds to the continuous
uniform distribution, not the discrete.
Longer: dDIST() doesn't give a pmf so summing it isn't what you are
looking for: it gives a pdf. For punif() consider P\{X <= 1\} when X
is distributed on [1, 10]. Clearly this has probability zero because
it can only occur for one out of uncountably many values -- though
this notion should be made more precise using a little bit of measure
theory.
Michael
On Mon, Nov 7, 2011 at 11:49 AM, michele donato
<michele.donato at wayne.edu> wrote:
Hi,
I am trying to use dunif and runif
however, I have two problems:
if I do
dunif(1:10, min=1, max=10)
I get 10 values, which summed give me 1.1111
I understand that the probability is computed as f(x) = 1 / (max-min)
but in this case it looks wrong: I have 10 values, each one
equiprobable, and the probability for each one should be 0.1 and not
0.11111 (which is, consistently with the definition, 1/9)
It looks like one of the extremes is not considered in the computation
of the probability, but then it's assigned a probability anyway.
Similar problem with punif.
if I do
punif(1, min=1, max=10)
I get 0 as result, as if the lower extreme is not considered, which is
not consistent with the description where min <= x <= max
If the lower extreme is not considered because cdf(x) = p(X<x) ? {and
not p(X<=x)} the problem stands in p(X<11) which should be the sum of
everything. ( P(1) + P(2) + ... + P(10) )
What is happening here?
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
A point of clarification: dDIST() sometimes gives a pmf, e.g., dpois(). But dunif() is a pdf. Sorry for the typo. Michael On Mon, Nov 7, 2011 at 12:08 PM, R. Michael Weylandt
<michael.weylandt at gmail.com> wrote:
In short, the unif() distribution corresponds to the continuous
uniform distribution, not the discrete.
Longer: dDIST() doesn't give a pmf so summing it isn't what you are
looking for: it gives a pdf. For punif() consider P\{X <= 1\} when X
is distributed on [1, 10]. Clearly this has probability zero because
it can only occur for one out of uncountably many values -- though
this notion should be made more precise using a little bit of measure
theory.
Michael
On Mon, Nov 7, 2011 at 11:49 AM, michele donato
<michele.donato at wayne.edu> wrote:
Hi,
I am trying to use dunif and runif
however, I have two problems:
if I do
dunif(1:10, min=1, max=10)
I get 10 values, which summed give me 1.1111
I understand that the probability is computed as f(x) = 1 / (max-min)
but in this case it looks wrong: I have 10 values, each one
equiprobable, and the probability for each one should be 0.1 and not
0.11111 (which is, consistently with the definition, 1/9)
It looks like one of the extremes is not considered in the computation
of the probability, but then it's assigned a probability anyway.
Similar problem with punif.
if I do
punif(1, min=1, max=10)
I get 0 as result, as if the lower extreme is not considered, which is
not consistent with the description where min <= x <= max
If the lower extreme is not considered because cdf(x) = p(X<x) ? {and
not p(X<=x)} the problem stands in p(X<11) which should be the sum of
everything. ( P(1) + P(2) + ... + P(10) )
What is happening here?
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
On Mon, Nov 7, 2011 at 8:49 AM, michele donato <michele.donato at wayne.edu> wrote:
Hi,
I am trying to use dunif and runif
however, I have two problems:
if I do
dunif(1:10, min=1, max=10)
I get 10 values, which summed give me 1.1111
I understand that the probability is computed as f(x) = 1 / (max-min)
but in this case it looks wrong: I have 10 values, each one
equiprobable, and the probability for each one should be 0.1 and not
0.11111 (which is, consistently with the definition, 1/9)
It looks like one of the extremes is not considered in the computation
of the probability, but then it's assigned a probability anyway.
Similar problem with punif.
if I do
punif(1, min=1, max=10)
I get 0 as result, as if the lower extreme is not considered, which is
not consistent with the description where min <= x <= max
If the lower extreme is not considered because cdf(x) = p(X<x) ? {and
not p(X<=x)} the problem stands in p(X<11) which should be the sum of
everything. ( P(1) + P(2) + ... + P(10) )
What is happening here?
The uniform distribution is continuous. Your interval has length 9 (10-1 = 9), so the density 1/9. Multiplied by 10 it gives you your answer. Same for the cumulative probability distribution (punif) - it is zero at x=1 because that's where your interval starts. Peter