Moment Generating Function of Gamma Distribution, Harmonic Mean and Mode of Gamma Distribution
The mean and variance of a Gamma distribution is α. The mode of the gamma distribution is given by the solution of f' (x) = 0 and f '' (x) < 0. The mode of the gamma distribution is α - 1 for α > 1. It is noted that if α < 1, the mode is zero.The sum of independent gamma vareties is also a gamma varieties. That is, If X1, X2, ..., Xn are n independent gamma varieties with parametersα1,α2, . . .,αn respectively, then X1+ X2 + . . . + Xn. is also a gamma variate with parameterα1 +α2 + . . . +αn.
Summary
The mean and variance of a Gamma distribution is α. The mode of the gamma distribution is given by the solution of f' (x) = 0 and f '' (x) < 0. The mode of the gamma distribution is α - 1 for α > 1. It is noted that if α < 1, the mode is zero.The sum of independent gamma vareties is also a gamma varieties. That is, If X1, X2, ..., Xn are n independent gamma varieties with parametersα1,α2, . . .,αn respectively, then X1+ X2 + . . . + Xn. is also a gamma variate with parameterα1 +α2 + . . . +αn.
Things to Remember
- The mean and variance of a Gamma distribution is α
- The mode of the gamma distribution isα - 1 for α > 1. It is noted that if α < 1, the mode is zero.
- The sum of independent gamma varities is also a gamma varieties. That is, If X1, X2, ..., Xn are n independent gamma varieties with parametersα1,α2, . . .,αn respectively, then X1+ X2 + . . . + Xn. is also a gamma variate with parameterα1 +α2 + . . . +αn.
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Moment Generating Function of Gamma Distribution, Harmonic Mean and Mode of Gamma Distribution
Let X follow a gamma distribution with the parameterα i.e. X ~ G(α). The moment generating function of gamma distribution is given by
$$M_{x}(t) \ = \ E({e^t}^{x})$$
$$= \ \int_{0}^{\infty} \ {e^t}^{x} \ f(x) \ dx \ = \ \int_{0}^{\infty} \ {e^t}^{x} \ \frac{{e^-}^{x} \ {x^\alpha}^{-1}}{\Gamma\alpha} \ dx$$
$$= \ \frac{1}{\Gamma\alpha} \ \int_{0}^{\infty} \ {e^-}^{(1-t)x} \ {x^\alpha}^{x-1} \ dx$$
$$= \ \frac{1}{\Gamma\alpha} \ \int_{0}^{\infty} \ {e^-}^{y} \ { \left ( \ \frac{y}{1-t} \ \right ) ^\alpha}^{-1} \ \frac{dy}{1-t}$$
$$Where, \ y \ = \ (1-t) \ x \ \Rightarrow \ dy \ = \ (1-t) \ dx$$
$$= \ \frac{1}{\Gamma\alpha} \ \frac{1}{(1-t)^\alpha} \ \int_{o}^{\infty} \ {e^-}^{y} \ {y^\alpha}^{-1} \ dy$$
$$= \ \frac{1}{\Gamma\alpha} \ \frac{1}{(1-t)^\alpha} \ \Gamma\alpha$$
$$= \ \frac{1}{(1-t)^\alpha}$$
$$\Rightarrow M_X(t) \ = \ {(1-t)^-}^{\alpha} \ \ \ \ ; \ | t | \ < \ 1$$
Now, the moments of gamma distribution can be determined by expanding the moment generating function Mx(t).
ExpandingMx(t), we have
$$M_{x}(t) \ = \ 1 \ + \ \alpha \ + \\binom{\alpha + 1}{2} \ t^2 \ + \ . \ . \ . \ + \ \binom{\alpha + r - 1}{r} \ t^r \ + \ . \ . \ .$$
$$\therefore \ \mu_{r}^{'} \ = \ Coefficient \ of \ \frac{t^r}{r!}$$
$$= \ \frac{(\alpha \ + \ r \ - \ 1)!}{(\alpha \ - \ 1)!}$$
$$\mu_{r}^{'} \ = \ \frac{\Gamma(\alpha \ + \ r)}{\Gamma\alpha} \ \ \ \ ; \ r \ = \ 1, \ 2, \ 3, \ 4$$
Then putting r = 1, 2, 3, 4, the first four moments about origin can be determined and then central moments of the gamma distribution can be determined.
It is noted that the moments of the gamma distribution can also be found by using the derivatives of Mx(t) at t = 0. i.e.
$$\mu_{r}^{'} \ = \ \frac{d^r}{dt^r} \ M_{x}(t) \mid_{t=0}$$
$$= \ \frac{d^r}{dt^r} \ {(1-t)^-}^\alpha \ \mid_{t=0}$$
$$\therefore \ \mu_{r}^{'} \ = \ \frac{d}{dt} \ {(1-t)^-}^\alpha \ \mid_{t=0}$$
$$= \ - \ \alpha \ {(1-t)^-}^\alpha \ (-1) \ \mid_{t=0}$$
$$ = \ \alpha$$
$$\Rightarrow \ Mean \ = \ \alpha.$$
$$\mu_{2}^{'} \ = \ \frac{d^2}{dt^2} \ {(1-t)^-}^\alpha \ \mid_{t=0}$$
$$= \ \alpha \ (- \ \alpha \ - \ 1) \ {(1 \ - \ t)^-}^{\alpha \ - \ 2} \ (- \ 1) \ \mid_{t=0}$$
$$= \ \alpha \ (\alpha \ + \ 1)$$
$$\mu_{2} \ = \ \mu_{2}^{'} \ - \ (\mu_{1}^{'})^2 \ = \ \alpha \ (\alpha \ + \ 1) \ - \ {\alpha}^2 \ = \ \alpha.$$
$$\Rightarrow \ Variance \ = \ \alpha.$$
Harmonic mean and mode of Gamma distribution
We have gamma distribution G(α) has the probability density function (pdf)
$$f(x) \ = \ \frac{{e^-}^{x} \ {x^\alpha}^{-1}}{\Gamma\alpha} \ \ \ \ ; \ 0 \ < x \ < \ \infty$$
The harmonic mean H of the gamma distribution is given by
$$\frac{1}{H} = \ E \ \left ( \frac{1}{X} \right )$$
$$= \ \int_{0}^{\infty} \ \frac{1}{x} \ f(x) \ dx$$
$$= \ \frac{1}{\Gamma\alpha} \ \int_{0}^{\infty} \ \frac{1}{x} \ {e^-}^{-x}. \ {x^\alpha}^{-1} \ dx$$
$$= \ \frac{1}{\Gamma\alpha} \ \int_{0}^{\infty} \ {e^-}^{x}. \ {x^\alpha}^{-2} \ dx$$
$$= \ \frac{\Gamma \ (\alpha \ - \ 1)}{\Gamma\alpha} \ = \ \frac{(\alpha - 2)!}{(\alpha - 1)!} \ = \ \frac{1}{\alpha - 1}$$
$$\therefore \ H \ = \ \alpha \ - \ 1$$
The mode of the gamma distribution is given by the solution of f' (x) = 0 and f '' (x) < 0.
Thus,
$$f^{' }(x) \ = \ 0$$
$$or, \ \ \frac{d}{dx} \ \left [ \frac{{e^-}^{x} \ {x^\alpha}^{-1}}{\Gamma\alpha} \ \right ] \ = \ 0$$
$$or, \ \ \frac{1}{\Gamma\alpha} \ [ \ {e^-}^{x} \ {x^\alpha}^{-1} \ + \ (\alpha \ - \ 1) \ {e^-}^{x} \ {x^\alpha}^{-2} \ ] \ = \ 0$$
$$0r, \ \ \frac{{e^-}^{x} \ {x^\alpha}^{-1}}{\Gamma\alpha} \ \left [ \ 1 \ - \ \frac{(\alpha \ - \ 1 \ )}{x} \ \right ] \ = \ 0$$
$$or, \ \ \ f(x) \ \left [ \frac{x - (\alpha - 1)}{x}\ \right ] \ = \ 0$$
$$or, \ \ \ \frac{x-(\alpha - 1)}{x} \ = \ 0$$
$$\Rightarrow \ x \ = \ \alpha \ - \ 1$$
Also, it can be proved that f '' (x) < 0 at x = α - 1. Hence, the mode of the gamma distribution isα - 1 forα > 1. It is noted that ifα < 1, the mode is zero.
Additive Property of Gamma Distribution
The sum of independent gamma varities is also a gamma varietes. That is, If X1, X2, ..., Xn are n independent gamma varieties with parametersα1,α2, . . .,αn respectively, then X1+ X2 + . . . + Xn. is also a gamma variate with parameterα1 +α2 + . . . +αn.
Bibliography
Sukubhattu N.P. (2013). Probability & Inference - II. Asmita Books Publishers & Distributors (P) Ltd., Kathmandu.
Larson H.J. Introduction to Probability Theory and Statistical Inference. WileyInternational, New York.
Lesson
continuious probablity distributions
Subject
Statistics
Grade
Bachelor of Science
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