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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

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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