0
$\begingroup$

Given an existing regression curve, how do I properly account for the known variance of the dependent variable when back-calculating for the (nominally) independent variable? If I had an observation $Y_{new}=1.09$ with a variance $\sigma_Y^2=0.012$ then how should I incorporate that information into my final answer for $X_{new}$ ?

I can build a (very simple) regression model with two vectors. Using R notation:

x <- c(8, 10, 50, 200, 350, 500, 1000, 2000)
y <- c(0.012, 0.016, 0.078, 0.333, 0.583, 0.799, 1.643, 3.002)
simple.lm <- lm(y ~ x,data=data.frame(x,y))

And back-calculation of any new value of X given a value of Y can be found by inverting the linear equation. Easy enough.

A <- coef(simple.lm)[1]
B <- coef(simple.lm)[2]
predict_X <- (y - A) / B

Y_new <- 1.09
X_new <- (Y_new - A) / B

But that doesn't deliver the prediction interval for the $X_{new}$ from that regression curve. It also assumes I either don't know, or don't care to include, the variability attached to my observation of Y. When I have $\sigma_Y^2$, I would like to carry that forward into the reported variance of X, from which I can calculate the prediction interval for X. This is basically the inverse of my prior question but I wanted to lay it out, and answer it, because this is the problem more likely to be searched for.

$\endgroup$

1 Answer 1

1
$\begingroup$

Consider three possible measures of variance in a new $y_0$ where the first variance value is treated as "no (known) variance," the second a low variance roughly on par with the experimentally expected values for the described problem, the third a high variance proposed in the question.

Rearranging the regression equation to solve for x, we can calculate the matrix and transposed matrix $X_p$ of that equation. The covariance matrix can be taken from the regression solution, which leads to both $\sigma_\text{regression}^2$ and $\sigma_\text{residuals}^2$. The prediction interval of a back-calculated concentration is then

$$ PI_\text{multiplier} \sqrt{\sigma_\text{regression}^2 + \sigma_\text{residuals}^2} $$

The known variance of that new $y_0$ is applied as the coefficient of variation and added to the overall standard error

$$ \sigma_x^2 = \sigma_\text{regression}^2 + \sigma_\text{residuals}^2 \\ CV_y = \frac{\sqrt{\sigma_y^2}}{\bar{y}} \\ se_y = \sigma_{x(\text{vary})} = \sigma_x + CV_y \\ x_\text{predict} = PI_\text{multiplier} \times se_y $$

Y_new <- 1.09 
variance_Y <- c(0,7.68E-8,1.2E-2) 

lvl <- 0.95
deg_freedom <- 6
pi_multiplier_knownDF <- qt((1 - lvl) / 2, deg_freedom, lower.tail = FALSE)
Xp <- matrix(c(-1/B,(A-Y_new)/(B^2)),ncol=2)
V <- vcov(simple.lm)
variance_regression <- diag(Xp %*% V %*% t(Xp)) 
variance_residuals <- sum((x-predict_X)^2) / deg_freedom
cv_Y <- sqrt(variance_Y) / mean(y)
se_var <- sqrt(variance_regression + variance_residuals) + cv_Y 
x_PI_with_yvar <- pi_multiplier_knownDF * se_var

This is comparable to the calculations done in an errors-in-variables model, where the reliability ratio $\lambda$ is applied as a divisor to the slope.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.