Density, distribution function, and random generation for the exponential growth distribution.
Usage
dexpgrowth(x, min = 0, max = 1, r, log = FALSE)
pexpgrowth(q, min = 0, max = 1, r, lower.tail = TRUE, log.p = FALSE)
rexpgrowth(n, min = 0, max = 1, r)Arguments
- x, q
Vector of quantiles.
- min
Minimum value of the distribution range. Default is 0.
- max
Maximum value of the distribution range. Default is 1.
- r
Rate parameter for the exponential growth.
- log, log.p
Logical; if TRUE, probabilities p are given as log(p).
- lower.tail
Logical; if TRUE (default), probabilities are P[X ≤ x], otherwise, P[X > x].
- n
Number of observations. If
length(n) > 1, the length is taken to be the number required.
Value
dexpgrowth gives the density, pexpgrowth gives the distribution
function, and rexpgrowth generates random deviates.
The length of the result is determined by n for rexpgrowth, and is the
maximum of the lengths of the numerical arguments for the other functions.
Details
The exponential growth distribution is defined on the interval [min, max] with rate parameter (r). Its probability density function (PDF) is:
$$f(x) = \frac{r \cdot \exp(r \cdot (x - min))}{\exp(r \cdot max) - \exp(r \cdot min)}$$
The cumulative distribution function (CDF) is:
$$F(x) = \frac{\exp(r \cdot (x - min)) - \exp(r \cdot min)}{ \exp(r \cdot max) - \exp(r \cdot min)}$$
For random number generation, we use the inverse transform sampling method:
Generate \(u \sim \text{Uniform}(0,1)\)
Set \(F(x) = u\) and solve for \(x\): $$ x = min + \frac{1}{r} \cdot \log(u \cdot (\exp(r \cdot max) - \exp(r \cdot min)) + \exp(r \cdot min)) $$
This method works because of the probability integral transform theorem, which states that if \(X\) is a continuous random variable with CDF \(F(x)\), then \(Y = F(X)\) follows a \(\text{Uniform}(0,1)\) distribution. Conversely, if \(U\) is a \(\text{Uniform}(0,1)\) random variable, then \(F^{-1}(U)\) has the same distribution as \(X\), where \(F^{-1}\) is the inverse of the CDF.
In our case, we generate \(u\) from \(\text{Uniform}(0,1)\), then solve \(F(x) = u\) for \(x\) to get a sample from our exponential growth distribution. The formula for \(x\) is derived by algebraically solving the equation:
$$ u = \frac{\exp(r \cdot (x - min)) - \exp(r \cdot min)}{\exp(r \cdot max) - \exp(r \cdot min)} $$
When \(r\) is very close to 0 (\(|r| < 1e-10\)), the distribution approximates a uniform distribution on [min, max], and we use a simpler method to generate samples directly from this uniform distribution.
Examples
x <- seq(0, 1, by = 0.1)
probs <- dexpgrowth(x, r = 0.2)
cumprobs <- pexpgrowth(x, r = 0.2)
samples <- rexpgrowth(100, r = 0.2)