Jtdylktuy

I have the model

$$
y=x^a times z^b + e
$$

where $y$ is the dependent variable, $x$ and $z$ are explanatory variables, $a$ and $b$ are the parameters and $e$ is an error term. I have parameter estimates of $a$ and $b$ and a covariance matrix of these estimates. How do I test if $a$ and $b$ are significantly different?

Assessing the hypothesis that $a$ and $b$ are different is equivalent to testing the null hypothesis $a - b = 0$ (against the alternative that $a-bne 0$).

The following analysis presumes it is reasonable for you to estimate $a-b$ as $$U = hat a - hat b.$$ It also accepts your model formulation (which often is a reasonable one), which--because the errors are additive (and could even produce negative observed values of $y$)--does not permit us to linearize it by taking logarithms of both sides.

The variance of $U$ can be expressed in terms of the covariance matrix $(c_{ij})$ of $(hat a, hat b)$ as

$$operatorname{Var}(U) = operatorname{Var}(hat a - hat b) = operatorname{Var}(hat a) + operatorname{Var}(hat b) - 2 operatorname{Cov}(hat a, hat b) = c_{11} + c_{22} - 2c_{12}^2.$$

When $(hat a, hat b)$ is estimated with least squares, one usually uses a "t test;" that is, the distribution of $$t = U / sqrt{operatorname{Var(U)}}$$ is approximated by a Student t distribution with $n-2$ degrees of freedom (where $n$ is the data count and $2$ counts the number of coefficients). Regardless, $t$ usually is the basis of any test. You may perform a Z test (when $n$ is large or when fitting with Maximum Likelihood) or bootstrap it, for instance.

To be specific, the p-value of the t test is given by

$$p = 2t_{n-2}(-|t|)$$

where $t_{n-2}$ is the Student t (cumulative) distribution function. It is one expression for the "tail area:" the chance that a Student t variable (of $n-2$ degrees of freedom) equals or exceeds the size of the test statistic, $|t|.$

More generally, for numbers $c_1,$ $c_2,$ and $mu$ you can use exactly the same approach to test any hypothesis

$$H_0: c_1 a + c_2 b = mu$$

against the two-sided alternative. (This encompasses the special but widespread case of a "contrast".) Use the estimated variance-covariance matrix $(c_{ij})$ to estimate the variance of $U = c_1 a + c_2 b$ and form the statistic

$$t = (c_1 hat a + c_2 hat b - mu) / sqrt{operatorname{Var}(U)}.$$

The foregoing is the case $(c_1,c_2) = (1,-1)$ and $mu=0.$

To check that this advice is correct, I ran the following R code to create data according to this model (with Normally distributed errors e), fit them, and compute the values of $t$ many times. The check is that the probability plot of $t$ (based on the assumed Student t distribution) closely follows the diagonal. Here is that plot in a simulation of size $500$ where $n=5$ (a very small dataset, chosen because the $t$ distribution is far from Normal) and $a=b=-1/2.$

In this example, at least, the procedure works beautifully. Consider re-running the simulation using parameters $a,$ $b,$ $sigma$ (the error standard deviation), and $n$ that reflect your situation.

Here is the code.

#

# Specify the true parameters.

#

set.seed(17)

a <- -1/2

b <- -1/2

sigma <- 0.25 # Variance of the errors

n <- 5        # Sample size

n.sim <- 500  # Simulation size

#

# Specify the hypothesis.

#

H.0 <- c(1, -1) # Coefficients of `a` and `b`.

mu <- 0 

#

# Provide x and z values in terms of their logarithms.

#

log.x <- log(rexp(n))

log.z <- log(rexp(n))

#

# Compute y without error.

#

y.0 <- exp(a * log.x + b * log.z)

#

# Conduct a simulation to estimate the sampling distribution of the t statistic.

#

sim <- replicate(n.sim, {

  #

  # Add the errors.

  #

  e <- rnorm(n, 0, sigma)

  df <- data.frame(log.x=log.x, log.z=log.z, y.0, y=y.0 + e)

  #

  # Guess the solution.

  #

  fit.ols <- lm(log(y) ~ log.x + log.z - 1, subset(df, y > 0))

  start <- coefficients(fit.ols) # Initial values of (a.hat, b.hat)

  #

  # Polish it using nonlinear least squares.

  #

  fit <- nls(y ~ exp(a * log.x + b * log.z), df, list(a=start[1], b=start[2]))

  #

  # Test a hypothesis.

  #

  cc <- vcov(fit)

  s <- sqrt((H.0 %*% cc %*% H.0))

  (crossprod(H.0, coef(fit)) - mu) / s

})

#

# Display the simulation results.

#

summary(lm(sort(sim) ~ 0 + ppoints(length(sim))))

qqplot(qt(ppoints(length(sim)), df=n-2), sim, 

       pch=21, bg="#00000010", col="#00000040",

       xlab="Student t reference value", 

       ylab="Test statistic")

abline(0:1, col="Red", lwd=2)