Computes the mean and variance of common discrete distributions given their parameters.
Usage
mbinom(size, prob)
mpois(lambda)
mgeom(prob)
mnbinom(size, prob)
mhyper(m, n, k)
mbenford(nDigits = 1)Arguments
- size
number of trials (binomial, negative binomial).
- prob
probability of success on each trial (binomial, geometric, negative binomial).
- lambda
mean (Poisson).
- m
number of white balls in the urn (hypergeometric).
- n
number of black balls in the urn (hypergeometric).
- k
number of balls drawn (hypergeometric).
- nDigits
number of leading digits for Benford's distribution, either
1(default, support {1,...,9}) or2(support {10,...,99}).
Details
Distribution | Mean | Variance |
| Binomial | \(np\) | \(np(1-p)\) |
| Poisson | \(\lambda\) | \(\lambda\) |
| Geometric | \(\frac{1-p}{p}\) | \(\frac{1-p}{p^2}\) |
| Negative binomial | \(\frac{r(1-p)}{p}\) | \(\frac{r(1-p)}{p^2}\) |
| Hypergeometric | \(\frac{km}{N}\) | \(\frac{km}{N}\frac{n}{N}\frac{N-k}{N-1}\) |
| Benford | \(\sum_d d\log_{10}\left(1 + \frac{1}{d}\right)\) | \(\sum_d d^2\log_{10}\left(1 + \frac{1}{d}\right) - \mu^2\) |
For the binomial distribution, \(n\) = size; for the negative
binomial distribution, \(r\) = size; and for the hypergeometric
distribution, \(N = m + n\). For Benford's distribution, the sum runs
over \(d \in \{1,\ldots,9\}\) for nDigits = 1 and
\(d \in \{10,\ldots,99\}\) for nDigits = 2. As there is no
closed-form solution, the moments are computed numerically.
References
Forbes, C., Evans, M., Hastings, N. and Peacock, B. (2011) Statistical Distributions. Fourth Edition. Wiley.
Johnson, N. L., Kotz, S. and Balakrishnan, N. (1995) Continuous Univariate Distributions, Vol. 2. Wiley.
Examples
mbinom(size = 10, prob = 0.5)
#> mean variance
#> 5.0 2.5
mpois(lambda = 3)
#> mean variance
#> 3 3
mgeom(prob = 0.3)
#> mean variance
#> 2.333333 7.777778
mnbinom(size = 5, prob = 0.3)
#> mean variance
#> 11.66667 38.88889
mhyper(m = 10, n = 5, k = 4)
#> mean variance
#> 2.6666667 0.6984127
mbenford(nDigits = 1)
#> mean variance
#> 3.440237 6.056513
mbenford(nDigits = 2)
#> mean variance
#> 38.58976 621.83174