p********0
发帖数: 186 | 1 HI, assume variable X, Y are independent variables, and Z is the predictor
Z ~ a X + b (a coefficient is positive)
Z ~ c Y + d( c coefficient is positive).
a new model Z ~ e X + f Y + g, will (e and f both be definitely postive???),
If X and Y are correlated, I can see the sign of e and f may not be both
positive. | h***i
发帖数: 3844 | 2 Yes
here, assume, a,c,e,f are true value instead of estimates.
),
【在 p********0 的大作中提到】
: HI, assume variable X, Y are independent variables, and Z is the predictor
: Z ~ a X + b (a coefficient is positive)
: Z ~ c Y + d( c coefficient is positive).
: a new model Z ~ e X + f Y + g, will (e and f both be definitely postive???),
: If X and Y are correlated, I can see the sign of e and f may not be both
: positive.
| p********0
发帖数: 186 | 3 Thanks, can you elaborate/proof it? | h***i
发帖数: 3844 | 4 I guess, you can prove,
b=d=g
a=e
c=f
by definition of conditional expectation and marginal expectation.
【在 p********0 的大作中提到】
: Thanks, can you elaborate/proof it?
| a*z
发帖数: 294 | 5 Recently I read something similar.
logit(Z) -- aX + e1,
logit(Z) -- bY + e2,
logit(Z) -- eX + fY + e3
a, b are positive
But e is negative, f is positive. All a, b, e, f are model estimates. | h***i
发帖数: 3844 | 6 if they are estimates,
1. usually, models are not written in this way.
2. if variance is > 0, anything could happen
【在 a*z 的大作中提到】
: Recently I read something similar.
: logit(Z) -- aX + e1,
: logit(Z) -- bY + e2,
: logit(Z) -- eX + fY + e3
: a, b are positive
: But e is negative, f is positive. All a, b, e, f are model estimates.
| a*z
发帖数: 294 | 7 Thank you, hezhi,
yes, the model in R is glm(Z ~ X, family=binomial), glm(Z ~ Y, family=
binomial) and glm(Z~ X + Y, family=binomial).
I do not understand your second comment. Model's estimates of variances of
e1, e2, and e3 should all be positive, am I right? | h***i
发帖数: 3844 | 8 i mean it is random thing
【在 a*z 的大作中提到】
: Thank you, hezhi,
: yes, the model in R is glm(Z ~ X, family=binomial), glm(Z ~ Y, family=
: binomial) and glm(Z~ X + Y, family=binomial).
: I do not understand your second comment. Model's estimates of variances of
: e1, e2, and e3 should all be positive, am I right?
| h***i
发帖数: 3844 | 9 i mean it is random thing
【在 a*z 的大作中提到】
: Thank you, hezhi,
: yes, the model in R is glm(Z ~ X, family=binomial), glm(Z ~ Y, family=
: binomial) and glm(Z~ X + Y, family=binomial).
: I do not understand your second comment. Model's estimates of variances of
: e1, e2, and e3 should all be positive, am I right?
| p********0
发帖数: 186 | 10 Thanks for the input,
in case of estimate
Z ~ a X + b + e1
Z ~ c Y + d + e2
If X Y are truly uncorrelated,
Z ~ e X + fY +g + e3, we can use matrix algebra to get new d and e, either X
or Y can be more dominant in decide the true Z value, then e may be
negative I guess,
matrix algebra can be used to decide e and f value, but it depends on the
true X, Y Z observed data then. | h******g
发帖数: 209 | 11 It seems that you do not understand the basic concepts of linear regression.
You do not understand the meaning of coefficients. In the model Z ~ a X +
b, a is the change of Z associated with one unit change of X. In the model,
Z ~ e X + f Y + g, e is the change of Z asscoiated with one unit change of
X holding others constant. In other words, f is fixed at a constant when you
discussing the meaning of e in the second model. Therefore, the meanings of
a and e are very different. a is positive does not assure that e should be
positive. For instance, say x is number of bedrooms, y is housing size, z is
housing price. a and c usually should be positive. However, given housing
size fixed, more bedrooms may lower housing price. |
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