From: <gyadav_at_ccilindia.co.in>

Date: Wed, 09 May 2007 14:28:53 +0530

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 09 May 2007 - 09:03:04 GMT

Date: Wed, 09 May 2007 14:28:53 +0530

Hi Horace and Mark

@@@ i myself know that this may be of little help but then also i am going
with this. Secondly the in the solution by Horace if corr(x, y) is beta
then

it implies that var(x) = var(y). Is that you want Mark. Well what i did i
am writing it down hereas under, may be its wrong.Please comment

var(y_t) = var(beta * x_t) + e_t)

=> var(y_t) = beta * var(x_t) + var(e_t) + cov(beta * x_t , e_t)

as cov(beta * x_t , e_t) = 0 hence

var(y_t) = beta * var(x_t) + var(e_t)............(i)

then

corr(x_t, y_t) = beta = cov(x_t, y_t)/ (sigma_x * sigma_y)

................ (ii)

further E[y_t] = beta * E[x_t] + E[e_t].........as E[e_t] = 0 hence

beta = E[y_t] / E[x_t]............(iii)

Now what to do ???

Mark, I suppose you make the usual assumptions, ie. E[x]=0, E[x*epsilon]=0, the correlation is just simply,

corr(x,y) = beta * ( var(x) / var(y) )

And you could get var(y) from var(x) and var(epsilon).

**HTH.
**
Horace

This is not an R question but if anyone can help me, it's much appreciated.

Suppose I have a series ( stationary ) y_t and a series x_t ( stationary )and x_t has variance sigma^2_x and epsilon is normal (0, sigma^2_epsilon )

and the two series have the relation

y_t = Beta*x_t + epsilon

My question is if there are particular values that sigma^2_x and sigma^2_epsilon have to take in order for corr(x_t,y_t) to equal Beta ?

I attempted to figure this out using two different methods and in one case I end up involving sigma^2_epsilon and in the other I don't and I'm not sure if either method is correct. I think I need to use results form the conditional bivariate normal but i'm really not sure. Also, it's not a homework problem because I am too old to have homework. Thanks for any insights/solutions.

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