aldo_tapia - 25 days ago 3

R Question

I want to fit a linear model with no slope and extract information of it. My objective is to know which is the best y-intercept for an horizontal line in a data set and also evaluate from derived linear fit to identify if *y* has a particular behavior (*x* is date). I've using

`range`

Removing y-intercept:

`X <- 1:10`

Y <- 2:11

lm1 <- lm(Y~X + 0, data = data.frame(X=X,Y=Y)) # y-intercept remove opt 1

lm1 <- lm(Y~X - 1, data = data.frame(X=X,Y=Y)) # y-intercept remove opt 2

lm1 <- lm(Y~0 + X, data = data.frame(X=X,Y=Y)) # y-intercept remove opt 3

lm1$coefficients

X

1.142857

summary(lm1)$r.squared

[1] 0.9957567

All the

`lm`

`lm2 <- lm(Y~1, data = data.frame(X=X,Y=Y))`

lm2$coefficients

(Intercept)

6.5

summary(lm2)$r.squared

[1] 0

There is a way to calculate out of

`lm`

Answer

Let `lmObject`

be your linear model returned by `lm`

(called with `y = TRUE`

to return `y`

).

If your model has intercept, then R-squared is computed as

`with(lmObject, 1 - c(crossprod(residuals) / crossprod(y - mean(y))) )`

If your model does not have an intercept, then R-squared is computed as

`with(lmObject, 1 - c(crossprod(residuals) / crossprod(y)) )`

Note, if your model is only an intercept (so it is certainly from the 1st case above), you have

```
residuals = y - mean(y)
```

thus R-squared is always `1 - 1 = 0`

.

In regression analysis, it is always recommended to include intercept in the model to get unbiased estimate. A model with intercept only is the NULL model. Any other model is compared with this NULL model for further analysis of variance.

A note. The value / quantity you want has nothing to do with regression. You can simply compute it as

```
c(crossprod(Y - mean(Y)) / crossprod(Y)) ## `Y` is your data
#[1] 0.1633663
```

Alternatively, use

```
(length(Y) - 1) * var(Y) / c(crossprod(Y))
#[1] 0.1633663
```