MATLAB Central Newsreader - random matrix generation

random matrix generation

arun — Sat, 07 Nov 2009 22:18:40 -0500

hi,

how can i generate a m*n matrix with the m-rows
1) dependent on each other
2) independent of each other

also, is it possible to create the same random matrix with just 1's
and 0's under the same 2 criterias?

thank you,
best, arun.

Re: random matrix generation

ImageAnalyst — Sat, 07 Nov 2009 23:49:38 -0500

On Nov 7, 5:18 pm, arun <aragorn1...@gmail.com> wrote:
> hi,
>
> how can i generate a m*n matrix with the m-rows
> 1) dependent on each other
> 2) independent of each other
>
> also, is it possible to create the same random matrix with just 1's
> and 0's under the same 2 criterias?
>
> thank you,
> best, arun.

--------------------------------------------------------------------
1) depends on what you mean by dependent.
2) randomArray = rand(100, 50);

binaryRandomArray = randomArray > 0.5;

Re: random matrix generation

arun — Sun, 08 Nov 2009 14:42:34 -0500

On Nov 8, 12:49 am, ImageAnalyst <imageanal...@mailinator.com> wrote:
> On Nov 7, 5:18 pm, arun <aragorn1...@gmail.com> wrote:
>
> > hi,
>
> > how can i generate a m*nmatrixwith the m-rows
> > 1) dependent on each other
> > 2) independent of each other
>
> > also, is it possible to create the samerandommatrixwith just 1's
> > and 0's under the same 2 criterias?
>
> > thank you,
> > best, arun.
>
> --------------------------------------------------------------------
> 1) depends on what you mean by dependent.
> 2) randomArray = rand(100, 50);
>
> binaryRandomArray = randomArray > 0.5;

what i mean by independent is that, if I take any 2 rows and compute a
measure of dependence (say, corr), and then permute one of the rows
and compute *corr* a 1000 times and obtain its distribution, then they
should be that of a uniform distribution's pdf (equally likely). Or,
equivalently, the cdf of p-values obtained directly from corr should
be that of a uniform distribution's pdf.

if its not, they are dependent in some way.
best, arun.

Re: random matrix generation

arun — Sun, 08 Nov 2009 14:43:49 -0500

On Nov 8, 3:42 pm, arun <aragorn1...@gmail.com> wrote:
> On Nov 8, 12:49 am, ImageAnalyst <imageanal...@mailinator.com> wrote:
>
>
>
> > On Nov 7, 5:18 pm, arun <aragorn1...@gmail.com> wrote:
>
> > > hi,
>
> > > how can i generate a m*nmatrixwith the m-rows
> > > 1) dependent on each other
> > > 2) independent of each other
>
> > > also, is it possible to create the samerandommatrixwith just 1's
> > > and 0's under the same 2 criterias?
>
> > > thank you,
> > > best, arun.
>
> > --------------------------------------------------------------------
> > 1) depends on what you mean by dependent.
> > 2) randomArray = rand(100, 50);
>
> > binaryRandomArray = randomArray > 0.5;
>
> what i mean by independent is that, if I take any 2 rows and compute a
> measure of dependence (say, corr), and then permute one of the rows
> and compute *corr* a 1000 times and obtain its distribution, then they
> should be that of a uniform distribution's pdf (equally likely). Or,
> equivalently, the cdf of p-values obtained directly from corr should
> be that of a uniform distribution's pdf.
>
> if its not, they are dependent in some way.
> best, arun.

**** Or, equivalently, the cdf of p-values obtained directly from corr
should
be that of a uniform distribution's pdf.

This line should be:
Or,
equivalently, the cdf of p-values obtained directly from corr should
be that of a uniform distribution's *CDF*.

Re: random matrix generation

arun — Sun, 08 Nov 2009 14:43:56 -0500

Re: random matrix generation

arun — Sun, 08 Nov 2009 14:45:47 -0500

On Nov 8, 3:43 pm, arun <aragorn1...@gmail.com> wrote:
> On Nov 8, 3:42 pm, arun <aragorn1...@gmail.com> wrote:
>
>
>
> > On Nov 8, 12:49 am, ImageAnalyst <imageanal...@mailinator.com> wrote:
>
> > > On Nov 7, 5:18 pm, arun <aragorn1...@gmail.com> wrote:
>
> > > > hi,
>
> > > > how can i generate a m*nmatrixwith the m-rows
> > > > 1) dependent on each other
> > > > 2) independent of each other
>
> > > > also, is it possible to create the samerandommatrixwith just 1's
> > > > and 0's under the same 2 criterias?
>
> > > > thank you,
> > > > best, arun.
>
> > > --------------------------------------------------------------------
> > > 1) depends on what you mean by dependent.
> > > 2) randomArray = rand(100, 50);
>
> > > binaryRandomArray = randomArray > 0.5;
>
> > what i mean by independent is that, if I take any 2 rows and compute a
> > measure of dependence (say, corr), and then permute one of the rows
> > and compute *corr* a 1000 times and obtain its distribution, then they
> > should be that of a uniform distribution's pdf (equally likely). Or,
> > equivalently, the cdf of p-values obtained directly from corr should
> > be that of a uniform distribution's pdf.
>
> > if its not, they are dependent in some way.
> > best, arun.
>
> **** Or, equivalently, the cdf of p-values obtained directly from corr
> should
> be that of a uniform distribution's pdf.
>
> This line should be:
> Or,
> equivalently, the cdf of p-values obtained directly from corr should
> be that of a uniform distribution's *CDF*.

The problem when I generate using X = rand(50,100); is that, the
columns also become independent. But I don't (necessarily) want this.

thank you.

Re: random matrix generation

dpb — Sun, 08 Nov 2009 15:36:24 -0500

arun wrote:
> hi,
>
> how can i generate a m*n matrix with the m-rows
> 1) dependent on each other
> 2) independent of each other
>
> also, is it possible to create the same random matrix with just 1's
> and 0's under the same 2 criterias?

Well, you have to define what dependent means, specifically, in order to
have a generating function even theoretically. Realizing it may be
something else, but you have to start w/ something.

Also, as written, 1) and 2) are contradictory if somehow this is to be a
single matrix satisfying both.

You need a far more precise problem definition to get anywhere methinks.

--

Re: random matrix generation

arun — Sun, 08 Nov 2009 15:52:48 -0500

On Nov 8, 4:36 pm, dpb <n...@non.net> wrote:
> arun wrote:
> > hi,
>
> > how can i generate a m*n matrix with the m-rows
> > 1) dependent on each other
> > 2) independent of each other
>
> > also, is it possible to create the same random matrix with just 1's
> > and 0's under the same 2 criterias?
>
> Well, you have to define what dependent means, specifically, in order to
> have a generating function even theoretically. Realizing it may be
> something else, but you have to start w/ something.
>
> Also, as written, 1) and 2) are contradictory if somehow this is to be a
> single matrix satisfying both.
>
> You need a far more precise problem definition to get anywhere methinks.
>
> --

Hi dpb,
I have to create 2 separate matrices (not every other row being
dependent and independent of each other), one satisfying the first
criteria and the other the second.
Regarding the problem definition, did you read my reply to
imageAnalyst? I have given the requirement there.

Re: random matrix generation

arun — Sun, 08 Nov 2009 15:53:08 -0500

Re: random matrix generation

dpb — Sun, 08 Nov 2009 16:13:33 -0500

arun wrote:
...

> Regarding the problem definition, did you read my reply to
> imageAnalyst? I have given the requirement there.

How is one to derive a generating function from that "requirement", pray
tell?

I can make a second row of a random matrix be dependent upon the first by

x=rand(1,10);
x=[x;x];

Is that ok?

It meets the definition stated.

--

Re: random matrix generation

Bruno Luong — Mon, 09 Nov 2009 07:31:03 -0500

arun <aragorn168b@gmail.com> wrote in message <8a05888a-f76b-4565-992d-802d92ce953e@v30g2000yqm.googlegroups.com>...
> On Nov 8, 12:49?am, ImageAnalyst <imageanal...@mailinator.com> wrote:
> > On Nov 7, 5:18?pm, arun <aragorn1...@gmail.com> wrote:
> >
> > > hi,
> >
> > > how can i generate a m*nmatrixwith the m-rows
> > > 1) dependent on each other
> > > 2) independent of each other
> >
> > > also, is it possible to create the samerandommatrixwith just 1's
> > > and 0's under the same 2 criterias?
> >
> > > thank you,
> > > best, arun.
> >
> > --------------------------------------------------------------------
> > 1) depends on what you mean by dependent.
> > 2) ?randomArray = rand(100, 50);
> >
> > binaryRandomArray = randomArray > 0.5;
>
> what i mean by independent is that, if I take any 2 rows and compute a
> measure of dependence (say, corr), and then permute one of the rows
> and compute *corr* a 1000 times and obtain its distribution, then they
> should be that of a uniform distribution's pdf (equally likely). Or,
> equivalently, the cdf of p-values obtained directly from corr should
> be that of a uniform distribution's pdf.
>
> if its not, they are dependent in some way.
> best, arun.

1. Dependent
epsilon = 0.1
A = ones(100,50)+epsilon*randn(100,50)

2. Independent
A = randn(100,50)

Bruno

Re: random matrix generation

ImageAnalyst — Mon, 09 Nov 2009 11:27:03 -0500

I think dpb and Bruno give excellent examples. For one more, you
could get vertical correlation betrween rows by doing a convolution in
the vertical direction with a rectangular window (i.e. just average
some number of rows in a given column). I believe that according to
the central limit theorem, this will tend toward giving a probability
density function that is a Gaussian. Again, this will satisfy your
vague, overly broad definition.

You can generate random numbers according to any given pdf if you know
the pdf that you want. You sort of generate the cdf and invert the
equation - standard stuff out of many textbooks. But the key word is
"given." If you don't give anything then you need to take what we
give you.