R help - May 2011 - Convolution confusion:

Hi,

I'm new to R, and I'm a bit confused with the "convolve()"
function.
If I do:
x<-c(1, 2, 3)
convolve(x, rev(x), TRUE, "open")
= 9 12 10 4 1

But I expected: 3 8 14 8 3 (like in Octave/MATLAB - conv(x, reverse(x)) )

3 2 1 x 1 2 3
= 3 2 1
     0 6 4 2
     0 0 9 6 3
= 3 8 14 8 3

The thing is, that "convolve(x, x, TRUE, "open")" works.
For me it feels very confusing, that convolution does the reverse itself 
but the help suggest to reverse it again.

The help file says: "Note that the usual definition of convolution of 
two sequences x and y is given by convolve(x, rev(y), type =
"o")."

Thanks for your help,

Alex

On May 18, 2011, at 15:47 , Alex Hofmann wrote:
> Hi,
> 
> I'm new to R, and I'm a bit confused with the
"convolve()" function.
> If I do:
> x<-c(1, 2, 3)
> convolve(x, rev(x), TRUE, "open")
> = 9 12 10 4 1
> 
> But I expected: 3 8 14 8 3 (like in Octave/MATLAB - conv(x, reverse(x)) )
> 
> 3 2 1 x 1 2 3
> = 3 2 1
>    0 6 4 2
>    0 0 9 6 3
> = 3 8 14 8 3
> 
> The thing is, that "convolve(x, x, TRUE, "open")"
works.
> For me it feels very confusing, that convolution does the reverse itself
but the help suggest to reverse it again.
> 
> The help file says: "Note that the usual definition of convolution of
two sequences x and y is given by convolve(x, rev(y), type =
"o")."
> 
> Thanks for your help,
This confuses me every time as well. One way of putting it is that R's
"convolve" is really what others call "correlate": the
product-sum between x and y shifted by k, for k=(1-n):(n-1) (adding appropriate
padding):
> z <- 1:3
> crossprod(z,z)
     [,1]
[1,]   14
> crossprod(c(z,0),c(0,z))
     [,1]
[1,]    8
> crossprod(c(z,0,0),c(0,0,z))
     [,1]
[1,]    3

Notice that this always comes out symmetric if x==y. 

However in convolution you want sum(x_j, y_(k-j)) so y is used in reverse order.

One way of spotting the issue is that if x represents the distribution of a
binary random variable X, then the convolution of x with itself should be the
distribution of the sum of two independent such variables.
> x
[1] 0.05 0.95
> convolve(x,x,type="o")
[1] 0.0475 0.9050 0.0475
> convolve(x,rev(x),type="o")
[1] 0.0025 0.0950 0.9025

... and it is pretty obviously not the case that the sum of two highly skewed
distributions is symmetric, so the 2nd line is right.
  
> dbinom(0:2,p=.95,size=2)
[1] 0.0025 0.0950 0.9025

-- 
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com