Hello,
This doesn't make sense, if you have only one vector you can estimate
its variance with
var(x)
but there is no covariance, the joint variance of two rv's. "co" or
joint with what if you have only x?
Note that the variance of x[1] or any other vector element is zero, it's
only one value therefore it does not vary. A similar reasonong can be
applied to cov(x[1], x[2]), etc.
Hope this helps,
Rui Barradas
Às 12:14 de 04/10/2024, Steven Yen escreveu:
Hello
I have a vector:
set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23017749
1.55870831 I like to create a matrix with elements containing variances
and covariances of x. That is var(x[1]) cov(x[1],x[2]) cov(x[1],x[3])
cov(x[2],x[1]) var(x[2]) cov(x[2],x[3]) cov(x[3],x[1]) cov(x[3],x[2])
var(x[3]) And I like to do it with "apply". Thanks.
On 10/4/2024 6:35 PM, Rui Barradas wrote:
Hello,
If you have a numeric matrix or data.frame, try something like
cov(mtcars)
Hope this helps,
Rui Barradas
Às 10:15 de 04/10/2024, Steven Yen escreveu:
On 10/4/2024 5:13 PM, Steven Yen wrote:
Pardon me!!!
What makes you think this is a homework question? You are not
obligated to respond if the question is not intelligent enough for you.
I did the following: two ways to calculate a covariance matrix but
wonder how I might replicate the results with "apply". I am not too
comfortable with the online documentation of "apply".
set.seed(122345671) > n<-3 > x<-rnorm(n); x [1] 0.92098449 0.80940115
0.60374785 > cov1<-outer(x-mean(x),x-mean(x))/(n-1); cov1 [,1] [,2]
[,3] [1,] 0.0102159207 0.00224105983 -0.0124569805 [2,] 0.0022410598
0.00049161983 -0.0027326797 [3,] -0.0124569805 -0.00273267965
0.0151896601 > cov2<-(x-mean(x))%*%t((x-mean(x)))/(n-1); cov2 [,1]
[,2] [,3] [1,] 0.0102159207 0.00224105983 -0.0124569805 [2,]
0.0022410598 0.00049161983 -0.0027326797 [3,] -0.0124569805
-0.00273267965 0.0151896601 >
On 10/4/2024 4:57 PM, Uwe Ligges wrote:
Homework questions are not answered on this list.
Best,
Uwe Ligges
On 04.10.2024 10:32, Steven Yen wrote:
The following line calculates standard deviations of a column vector:
se<-apply(dd,1,sd)
How can I calculate the covariance matrix using apply? Thanks.
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