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Topic: distribution of regression coefficients
Replies: 9   Last Post: Nov 12, 2010 11:58 AM

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 RGVickson@shaw.ca Posts: 1,600 Registered: 12/1/07
Re: distribution of regression coefficients
Posted: Nov 11, 2010 5:15 PM

On Nov 11, 12:12 pm, "Rod" <rodrodrod...@hotmail.com> wrote:
> "Paul" <paul_ru...@att.net> wrote in message
>
> On Nov 11, 5:45 am, "Rod" <rodrodrod...@hotmail.com> wrote:
>
>
>

> > "Ray Koopman" <koop...@sfu.ca> wrote in message
>
>
> > > On Nov 10, 4:07 am, "Rod" <rodrodrod...@hotmail.com> wrote:
> > >> On Nov 10, 2:20 am, Ray Koopman <koop...@sfu.ca> wrote:
> > >>> On Nov 10, 12:44 am, "Rod" <rodrodrod...@hotmail.com> wrote:
>
> > >>>> In regression y = a + b*x
>
> > >>>> I know how to compute the covariance matrix for a and b.
> > >>>> I also know that a and b are normally distributed,
> > >>>> but what is the joint distribution of a and b?
> > >>>> Its tempting to guess bivariate normal
> > >>>> but I don't see how to show that.

>
> > >>> y = X beta + e
>
> > >>> W = (X'X)^-1 X'
>
> > >>> b = Wy
> > >>> = beta + We

>
> > >>> If e is multivariate normal then so is We, and hence b.
>
> > >> ditto a, but what of the joint distribution given that a and b are
> > >> correlated?

>
> > > Sorry, I should have been more explicit (and used only low-ascii
> > > characters). In what I wrote, X is a given n by p matrix
> > > of predictors, where n is the # of cases and p is the # of
> > > predictors, beta is a p-vector of unknown coefficients, and e is
> > > a random n-vector. If one of the columns of X is a dummy predictor
> > > whose value is 1 for every case then the corresponding element in
> > > beta is the intercept (your a ). So the intercept is "just another
> > > coefficient".

>
> > > Whatever the distribution of e may be, if its mean vector and
> > > covariance matrix are m and S then the mean vector and covariance
> > > matrix of b are beta + Wm and WSW'. (Note: we usually assume
> > > m = [0,...,0]'.)

>
> > > If e is multivariate normal then b is also multivariate normal.
>
> > I rather hastily assumed my b was the same as yours, sorry.
> > OK I get it that if the e are normal then b is just a linear combination
> > and
> > hence also normal.
> > Either I am not understanding what you are saying (likely), or you haven't
> > yet answered my question fully.

>
> It's the former.
>

> > To keep it simple lets keep the e normal and independent from each other.
> > I am after the joint probability P(a,b) which because a and b are
> > correlated
> > is different to the product of the two distributions for a and b
> > separately.
> > I would put money on P being bivariate normal but for the life of me I
> > can't
> > see how to work that out.

>
> As Ray said, b = \beta + We, where W is computed from the X matrix.
> The theoretical variance-covariance matrix of b is E[(b-\beta)(b-
> \beta)'] = E[Wee'W']. Treat X, and therefore W, as constant with
> respect to the expectation. Since the e are assumed i.i.d. with zero
> mean and variance sigma^2, E[ee'] should be obvious (left to the
>
> /Paul
>
> Maybe I'm asking a silly question, or at least not communicating it
> correctly. But I am after the functional form of the joint probability of a
> and b or b_0 and b_k if you prefer.
>
> Thanks all for your contributions.
>
> Rod

You need to write out in detail those formulas that people have
already given to you in general terms.

Let the true model be yi = a + b*xi + ei, where the ei are iid with
distribution N(0,s^2); I really should use alpha and beta instead of a
and b, but that makes reading and writing harder. Instead, use small
letters for the unknown true values and capital letters for the
regression estimates.

The fitted equation is y = A + B*x, where A = [Sx^2 * Sy - Sx * Sxy]/D
and B = [n*Sxy - Sx * Sy]/D, with D = n* Sx^2 - (Sx)^2. Here, Sw means
the sum of w_i for i = 1,...,n, so Sx^2 = sum(xi^2), etc. Plugging in
the expression yi = a + b*xi + ei, i=1,...,n, we get: A = a + Ea, B =
b + Eb, where Ea = [Sx^2 * Se - Sx * Sxe]/D and Eb = [n * Sxe - Sx *
Se]/D. This exhibits A-a and B-b as explicit linear combinations of
e1,...,en (with constant coefficients), so A and B are automatically
bivariate normal. In fact, if we write Ea = sum(ui * ei,i=1..n) and Eb
= sum(vi * ei, i=1..n), then the bivariate generating function of
(A,B) is E exp(r*A + s*B) = E exp(sum(r*ui + s*vi)*ei,i=1..n)) =
product([E exp(r*ui+s*vi)*ei)],i=1..n) = product(exp((r*ui+s*vi)^2 *
sigma^2/2),i=1..n) = exp(sum(...),i=1..n). You can write the exponent
as (1/2)[U*r^2 + 2*W*r*s + V*s^2], so Var(A) = U, Var(B) = V and
cov(A,B) = W. (A,B) had bivariate normal distribution with mean =
(a,b) and variance-covariance matrix [[U,W],[W,V]], where this means
matrix = [row1, row2]. You have everything you need to write out U, V
and W in detail.

R.G. Vickson

Date Subject Author
11/10/10 Rod
11/10/10 Ray Koopman
11/10/10 Ray Koopman
11/10/10 Rod
11/10/10 Ray Koopman
11/11/10 Rod
11/11/10 Paul
11/11/10 Rod
11/11/10 RGVickson@shaw.ca
11/12/10 Rod