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Gamma
Probability density function
Gamma distribution pdf
Cumulative distribution function
Gamma distribution cdf
Parameters $ k > 0\, $ shape (real)
$ \theta > 0\, $ scale (real)
Support $ x \in [0; \infty)\! $
pdf $ x^{k-1} \frac{\exp\left(-x/\theta\right)}{\Gamma(k)\,\theta^k} $
cdf $ \frac{\gamma(k, x/\theta)}{\Gamma(k)} $
Mean $ k \theta\, $
Median
Mode $ (k-1) \theta\, $ for $ k \geq 1\, $
Variance $ k \theta^2\, $
Skewness $ \frac{2}{\sqrt{k}} $
Kurtosis $ \frac{6}{k} $
Entropy $ k\theta+(1-k)\ln(\theta)+\ln(\Gamma(k))\, $
$ +(1-k)\psi(k)\, $
mgf $ (1 - \theta\,t)^{-k} $ for $ t < 1/\theta $
Char. func. $ (1 - \theta\,i\,t)^{-k} $

In probability theory and statistics, the gamma distribution is a continuous probability distribution. For integer values of the parameter k it is also known as the Erlang distribution.

Probability density functionEdit

The probability density function of the gamma distribution can be expressed in terms of the gamma function:

$ f(x;k,\theta) = x^{k-1} \frac{e^{-x/\theta}}{\theta^k \, \Gamma(k)} \ \mathrm{for}\ x > 0 \,\! $

where $ k > 0 $ is the shape parameter and $ \theta > 0 $ is the scale parameter of the gamma distribution. (NOTE: this parameterization is what is used in the infobox and the plots.)

Alternatively, the gamma distribution can be parameterized in terms of a shape parameter $ \alpha = k $ and an inverse scale parameter $ \beta = 1/\theta $, called a rate parameter:

$ g(x;\alpha,\beta) = x^{\alpha-1} \frac{\beta^{\alpha} \, e^{-\beta\,x} }{\Gamma(\alpha)} \ \mathrm{for}\ x > 0 \,\! $

Both parameterizations are common because they are convenient to use in certain situations and fields.

PropertiesEdit

The cumulative distribution function can be expressed in terms of the incomplete gamma function,

$ F(x;k,\theta) = \int_0^x f(u;k,\theta)\,du = \frac{\gamma(k, x/\theta)}{\Gamma(k)} \,\! $

The information entropy is given by:

$ S=k\theta+(1-k)\ln(\theta)+\ln(\Gamma(k))+(1-k)\psi(k)\, $

where $ \psi(k) $ is the polygamma function.

If $ X_i \sim \mathrm{Gamma}(\alpha_i, \beta) $ for $ i=1, 2, \cdots, N $ and $ \bar{\alpha} = \sum_{k=1}^N \alpha_i $ then

$ \left[ Y = \sum_{i=1}^N X_i \right] \sim \mathrm{Gamma} \left( \bar{\alpha}, \beta \right) $

provided all $ X_i $ are independent. The gamma distribution exhibits infinite divisibility.

If $ X \sim \operatorname{Gamma}(k, \theta) $, then $ \frac X \theta \sim \operatorname{Gamma}(k, 1) $. Or, more generally, for any $ t > 0 $ it holds that $ tX \sim \operatorname{Gamma} (k, t \theta) $. That is the meaning of θ (or β) being the scale parameter.

Parameter estimation Edit

The likelihood function is

$ L=\prod_{i=1}^N f(x_i;k,\theta) $

from which we calculate the log-likelihood function

$ \ell=(k-1)\sum_{i=1}^N\ln(x_i)-\sum x_i/\theta-Nk\ln(\theta)-N\ln\Gamma(k) $

Finding the maximum with respect to $ \theta $ by taking the derivative and setting it equal to zero yields the maximum likelihood estimate of the $ \theta $ parameter:

$ \theta=\frac{1}{kN}\sum_{i=1}^N x_i $

Substituting this into the log-likelihood function gives:

$ \ell=(k-1)\sum_{i=1}^N\ln(x_i)-Nk-Nk\ln\left(\frac{\sum x_i}{kN}\right)-N\ln\Gamma(k) $

Finding the maximum with respect to $ k $ by taking the derivative and setting it equal to zero yields:

$ \ln(k)-\psi(k)=\ln\left(\frac{1}{N}\sum_{i=1}^N x_i\right)-\frac{1}{N}\sum_{i=1}^N\ln(x_i) $

where $ \psi(k)=\frac{\Gamma'(k)}{\Gamma(k)} $ is the digamma function.

There is no closed-form solution for $ k $. The function is numerically very well behaved, so if a numerical solution is desired, it can be found using Newton's method. An initial value of $ k $ can be found either using the method of moments, or using the approximation:

$ \ln(k)-\psi(k) \approx \frac{1}{k}\left(\frac{1}{2} + \frac{1}{12k+2}\right) $

If we let $ s = \ln\left(\frac{1}{N}\sum_{i=1}^N x_i\right)-\frac{1}{N}\sum_{i=1}^N\ln(x_i), $ then $ k $ is approximately

$ k \approx \frac{3-s+\sqrt{(s-3)^2 + 24s}}{12s} $

which is within 1.5% of the correct value.

Generating Gamma random variables Edit

Given the scaling property above, it is enough to generate Gamma variables with $ \beta = 1 $ as we can later convert to any value of β with simple division.

Using the fact that if $ X \sim \operatorname{Gamma}(1, 1) $, then also $ X \sim \operatorname {Exp} (1) $, and the method of generating exponential variables, we conclude that if U is uniformly distributed on (0, 1], then $ -\ln U \sim \operatorname{Gamma} (1, 1) $. Now, using the "α-addition" property of Gamma distribution, we expand this result:

$ \sum _{k=1} ^n {-\ln U_k} \sim \operatorname{Gamma} (n, 1) $,

where $ U_k $ are all uniformly distributed on (0, 1 ] and independent.

All that is left now is to generate a variable distributed as $ \operatorname{Gamma} (\delta, 1) $ for $ 0 < \delta < 1 $ and apply the "α-addition" property once more. This is the most difficult part, however.

We provide an algorithm without proof. It is an instance of the acceptance-rejection method:

  1. Let m be 1.
  2. Generate $ V_{2m - 1} $ and $ V_{2m} $ — independent uniformly distributed on (0, 1] variables.
  3. If $ V_{2m - 1} \le v_0 $, where $ v_0 = \frac e {e + \delta} $, then go to step 4, else go to step 5.
  4. Let $ \xi_m = \left( \frac {V_{2m - 1}} {v_0} \right) ^{\frac 1 \delta}, \ \eta_m = V_{2m} \xi _m^ {\delta - 1} $. Go to step 6.
  5. Let $ \xi_m = 1 - \ln {\frac {V_{2m - 1} - v_0} {1 - v_0}}, \ \eta_m = V_{2m} e^{-\xi_m} $.
  6. If $ \eta_m > \xi_m^{\delta - 1} e^{-\xi_m} $, then increment m and go to step 2.
  7. Assume $ \xi = \xi_m $ to be the realization of $ \operatorname {Gamma} (\delta, 1) $.

Now, to summarize,

$ \frac 1 \beta \left( \xi - \sum _{k=1} ^{[\alpha]} {\ln U_k} \right) \sim \operatorname{Gamma}(\alpha, \beta) $ ,

where $ [\alpha] $ is the integral part of α, ξ has been generating using the algorithm above with $ \delta = \{\alpha\} $ (the fractional part of α), $ U_k $ and $ V_l $ are distributed as explained above and are all independent.

Related distributionsEdit

  • $ X \sim \mathrm{Exponential}(\theta) $ is an exponential distribution if $ X \sim \mathrm{Gamma}(1, \theta) $.
  • $ cX \sim \mathrm{Gamma}(k, c\theta) $ if $ X \sim \mathrm{Gamma}(k, \theta) $ for any c > 0 .
  • $ Y \sim \mathrm{Gamma}(N, \theta) $ is a gamma distribution if $ Y = X_1 + \cdots + X_N $ and if the $ X_i \sim \mathrm{Exponential}(\theta) $ are all independent and share the same parameter $ \theta $.
  • $ X \sim \chi^2(\nu) $ is a chi-square distribution if $ X \sim \mathrm{Gamma}(k=\nu/2, \theta = 2) $.
  • If $ k $ is an integer, the gamma distribution is an Erlang distribution (so named in honor of A. K. Erlang) and is the probability distribution of the waiting time until the $ k $-th "arrival" in a one-dimensional Poisson process with intensity $ 1/\theta $.
  • $ X \sim \mathrm{Gamma}(k, \theta) $ then $ Y \sim \mathrm{InvGamma}(k, \theta^{-1}) $ if $ Y = 1/X $, where $ \mathrm{InvGamma} $ is the inverse-gamma distribution.
  • $ Y = X_1/(X_1+X_2) \sim \mathrm{Beta} $ is a beta distribution if $ X_1 \sim \mathrm{Gamma} $< and $ X_2 \sim \mathrm{Gamma} $ and are also independent.
  • $ Y \sim \mathrm{Maxwell}(\beta) $ is a Maxwell-Boltzmann distribution if $ X \sim \mathrm{Gamma}(\alpha = 3/2, \beta) $.
  • $ Y \sim N(\mu = \alpha \beta, \sigma^2 = \alpha \beta^2) $ is a normal distribution as $ Y = \lim_{\alpha \to \infty} X $ where $ X \sim \mathrm{Gamma}(\alpha, \beta) $.
  • The real vector $ (X_1/S,\ldots,X_n/S)\sim \operatorname{Dirichlet}(\alpha_1,\ldots,\alpha_n) $ follows a Dirichlet distribution if $ X_i\sim\operatorname{Gamma}(\alpha_i,1) $ are independent, and $ S=X_1+\cdots+X_n $.

References Edit

  • R. V. Hogg and A. T. Craig. Introduction to Mathematical Statistics, 4th edition. New York: Macmillan, 1978. (See Section 3.3.)

See also Edit

es:Distribución gammafi:Gamma-jakauma sv:Gammafördelning

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