Hello group
Let X(n) = Y for all n. X(n) is a stochastic process. Y is
random variable with uniform distribution in (0,1).
The autocorrelation of X, rxx(m) = E{X(n)X(n-m)} = E{Y.Y} =
E{Y^2}
1
E{Y^2} = integral y^2 pdf(y) dy = 1/3
0
rxx(m) = 1/3 which is a constant.
Matlab
N = 2500;
x = rand(1,N);
xc = xcorr(x,'unbiased');
The value of xc(1250) to corresponds rxx(0). Both are 1/3.
No problem so far.
My problem is this:
xc(x~=1250) = xc(x not equal 1250) is not 1/3. They are
all constant but not 1/3. What is the mistake I am making?
Thanks
A question on autocorrelation
Started by ●April 23, 2004
Reply by ●April 23, 20042004-04-23
innocent_802000@yahoo.com (Jay) writes:> Hello group > Let X(n) = Y for all n. X(n) is a stochastic process. Y is > random variable with uniform distribution in (0,1). > The autocorrelation of X, rxx(m) = E{X(n)X(n-m)} = E{Y.Y} = > E{Y^2}Nope! E{X(n)X(n-m)} = E{Y.Y} if m=0 0 if m != 0 This is because X(n) and X(n-m) are independent random variables.> 1 > E{Y^2} = integral y^2 pdf(y) dy = 1/3 > 0 > > rxx(m) = 1/3 which is a constant. > > Matlab > N = 2500; > x = rand(1,N); > xc = xcorr(x,'unbiased'); > The value of xc(1250) to corresponds rxx(0). Both are 1/3. > No problem so far. > > My problem is this: > xc(x~=1250) = xc(x not equal 1250) is not 1/3. They are > all constant but not 1/3. What is the mistake I am making?Ciao, Peter K. -- Peter J. Kootsookos "I will ignore all ideas for new works [..], the invention of which has reached its limits and for whose improvement I see no further hope." - Julius Frontinus, c. AD 84
Reply by ●April 23, 20042004-04-23
> > Nope! > > E{X(n)X(n-m)} = E{Y.Y} if m=0 > 0 if m != 0 > > This is because X(n) and X(n-m) are independent random variables. >Thanks for your tip on independence. 1 1 rxx(m) = E{X(n)X(n-m)} = integral integral x y joint pdf of (x,y) dx dy 0 0 1 1 = integral x pdf of x dx * integral y pdf of y dy 0 0 this is by independence = 1/2 * 1/2 = 1/4. Matlab also gave me the same answer 0.254 for m != 0.
Reply by ●April 23, 20042004-04-23
Peter J. Kootsookos wrote:> innocent_802000@yahoo.com (Jay) writes: > > >>Hello group >> Let X(n) = Y for all n. X(n) is a stochastic process. Y is >>random variable with uniform distribution in (0,1). >> The autocorrelation of X, rxx(m) = E{X(n)X(n-m)} = E{Y.Y} = >>E{Y^2} > > > Nope! > > E{X(n)X(n-m)} = E{Y.Y} if m=0 > 0 if m != 0 > > This is because X(n) and X(n-m) are independent random variables.Err, no they're not. For one instance of the random process, all the x[n] have the same value. Carlos --
Reply by ●April 23, 20042004-04-23
Carlos Moreno wrote:> Peter J. Kootsookos wrote: > >> innocent_802000@yahoo.com (Jay) writes: >> >> >>> Hello group >>> Let X(n) = Y for all n. X(n) is a stochastic process. Y is >>> random variable with uniform distribution in (0,1). >>> The autocorrelation of X, rxx(m) = E{X(n)X(n-m)} = E{Y.Y} = >>> E{Y^2} >> >> >> >> Nope! >> >> E{X(n)X(n-m)} = E{Y.Y} if m=0 0 if m != 0 >> >> This is because X(n) and X(n-m) are independent random variables. > > > Err, no they're not. For one instance of the random process, all the > x[n] have the same value. > > Carlos > --rannd(N,1) is not equal to c*ones(N,1)
Reply by ●April 23, 20042004-04-23
Stan Pawlukiewicz wrote:> Carlos Moreno wrote: > >> Peter J. Kootsookos wrote: >> >>> innocent_802000@yahoo.com (Jay) writes: >>> >>> >>>> Hello group >>>> Let X(n) = Y for all n. X(n) is a stochastic process. Y is >>>> random variable with uniform distribution in (0,1). >>>> The autocorrelation of X, rxx(m) = E{X(n)X(n-m)} = E{Y.Y} = >>>> E{Y^2} >>> >>> >>> >>> >>> Nope! >>> >>> E{X(n)X(n-m)} = E{Y.Y} if m=0 0 if m != 0 >>> >>> This is because X(n) and X(n-m) are independent random variables. >> >> >> >> Err, no they're not. For one instance of the random process, all the >> x[n] have the same value. >> >> Carlos >> -- > > rannd(N,1) is not equal to c*ones(N,1)Ok, but then the bug is in using rand(N,1) -- or in the OP's description of the problem. See above: "Let X(n) = Y for all n" That tells me that the random process is constant -- to obtain a realization of the random process, you draw a random variable Y with uniform distribution in (0,1), and the process will take that value for all time-indices. Carlos --
Reply by ●April 23, 20042004-04-23
Carlos Moreno wrote:> Stan Pawlukiewicz wrote: > >> Carlos Moreno wrote: >> >>> Peter J. Kootsookos wrote: >>> >>>> innocent_802000@yahoo.com (Jay) writes: >>>> >>>> >>>>> Hello group >>>>> Let X(n) = Y for all n. X(n) is a stochastic process. Y is >>>>> random variable with uniform distribution in (0,1). >>>>> The autocorrelation of X, rxx(m) = E{X(n)X(n-m)} = E{Y.Y} = >>>>> E{Y^2} >>>> >>>> >>>> >>>> >>>> >>>> Nope! >>>> >>>> E{X(n)X(n-m)} = E{Y.Y} if m=0 0 if m != 0 >>>> >>>> This is because X(n) and X(n-m) are independent random variables. >>> >>> >>> >>> >>> Err, no they're not. For one instance of the random process, all the >>> x[n] have the same value. >>> >>> Carlos >>> -- >> >> >> rannd(N,1) is not equal to c*ones(N,1) > > > Ok, but then the bug is in using rand(N,1) -- or in the OP's > description of the problem. See above: "Let X(n) = Y for all n" > > That tells me that the random process is constant -- to obtain > a realization of the random process, you draw a random variable > Y with uniform distribution in (0,1), and the process will take > that value for all time-indices. > > Carlos > --Most books don't agree on simple notation. I don't expect to see any better on a newsgroup particularly when the person posting is also trying to get the basic concepts. It's really up to the original poster to elaborate on his or her meaning.
Reply by ●April 23, 20042004-04-23
"Jay" <innocent_802000@yahoo.com> wrote in message news:7f6c0a9a.0404230334.338af3f1@posting.google.com... ......> this is by independence > = 1/2 * 1/2 = 1/4. > Matlab also gave me the same answer 0.254 for m != 0.The difference is due to the fact that MATLAB, a US product, does not use the metric system in its internal calculations, but does use it in the external display because "scientists" use the metric system. Note that one inch is defined to be EXACTLY 2.54 cm in the US (something slightly different in the rest of the world). Thus, the answer that MATLAB computed is actually 0.1 (inch), and it converted it to cm to give 0.254 as the answer. What is amazing, of course, that Jay's back-of-the-enevelope calculation gives 1/4 = 0.25 which is so close to the correct answer. \include{Smileys all around to circumvent expostulations by the humor-impaired}
Reply by ●April 23, 20042004-04-23
Carlos Moreno <moreno_at_mochima_dot_com@xx.xxx> wrote in message news:<uw9ic.58817$mK3.695597@weber.videotron.net>...> Peter J. Kootsookos wrote: > > > innocent_802000@yahoo.com (Jay) writes: > > > > > >>Hello group > >> Let X(n) = Y for all n. X(n) is a stochastic process. Y is > >>random variable with uniform distribution in (0,1). > >> The autocorrelation of X, rxx(m) = E{X(n)X(n-m)} = E{Y.Y} = > >>E{Y^2} > > > > > > Nope! > > > > E{X(n)X(n-m)} = E{Y.Y} if m=0 > > 0 if m != 0 > > > > This is because X(n) and X(n-m) are independent random variables. > > Err, no they're not. For one instance of the random process, all the > x[n] have the same value. > > Carlos > --No, Peter is correct. What we have to understand is that at each time instant in the random process, the value of the random process is uniformly distributed. However, for the next time instant, the value of the random process will be uniformly distributed and independent of the previous time instant. This process is a white random process with uniformly distributed samples. Susheem
Reply by ●April 23, 20042004-04-23
Thanks everyone. What I meant is the following: The value X(n) takes is drawn from a uniform random distribution. X(n) and X(m) are not the same random variables if n and m are different. X(n) is drawn from uniform distribution. X(m) is again drawn from the same distribution. More formally, the pdf of X(n) is uniform in (0,1). The pdf of X(m) is again uniform in (0,1). X(n) and X(m) need not be neccessarily equal if n~=m. They are equal if n = m.
