Home page for accesible maths 3.4 Gamma Distribution:

\operatorname{\mathsf{Gam}}(\alpha,\beta)

3.4.1 Gamma function 3.5 The Beta Distribution:

\operatorname{\mathsf{Beta}}(\alpha_{1},\alpha_{2})

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3.4.2 Gamma distribution

A random variable $X$ has a gamma distribution with shape parameter $\alpha$ and rate parameter $\beta$ if its pdf is given by

\displaystyle f_{X}(x;\boldsymbol{\theta})=\left\{\begin{array}[]{ll}\frac{% \beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}\exp(-\beta x)&\quad x\geq 0\\ 0&\quad\text{otherwise}\end{array}\right.

(3.1)

with $\boldsymbol{\theta}=(\alpha,\beta)$ , where $\alpha>0$ and $\beta>0$ . We write $X\sim\operatorname{\mathsf{Gam}}(\alpha,\beta)$ .

Firstly, with $\alpha=1$ the density is $\beta\exp(-\beta x)$ for $x>0$ and so the Exponential distribution is a special case of the Gamma distribution.

We next check that the above is a density. Firstly, $f(x)\geq 0$ , and, secondly,

\displaystyle\int_{-\infty}^{\infty}f_{X}(s)\,\mathrm{d}s=\frac{\beta^{\alpha-% 1}}{\Gamma(\alpha)}\int_{0}^{\infty}s^{\alpha-1}\exp(-\beta s)\beta\,\mathrm{d% }s.

Substituting $\beta s=t$ , so that $\beta\,\mathrm{d}s=\,\mathrm{d}t$ gives,

\displaystyle\int_{-\infty}^{\infty}f_{X}(s)\,\mathrm{d}s=\frac{\beta^{\alpha-% 1}}{\Gamma(\alpha)}\int_{0}^{\infty}\frac{t^{\alpha-1}}{\beta^{\alpha-1}}\exp(% -t)\,\mathrm{d}t=1.

Figure 3.3 (First Link, Second Link) and Figure 3.4 (First Link, Second Link) show the pdf for four different sets of parameters.

Figure 3.3: First Link, Second Link, Caption: Pdfs for two gamma random variables

X

with different parameter values.

Figure 3.4: First Link, Second Link, Caption: Pdfs for two more gamma random variables

X

with different parameter values.

Convolution property: When $\alpha$ is an integer the Gamma $(\alpha,\beta)$ distribution is the distribution of the length of time you have to wait until a total of $\alpha$ events have occurred where the time between each event follows an Exponential $(\beta)$ distribution (see Chapter 8). More generally, the gamma distribution provides a flexible class of pdfs which may describe the distribution of a non-negative variable even when there is no strong probability based justification.

We cannot evaluate the cdf in closed form for a general (non-integer) value of $\alpha$ .

A short-cut method for evaluating the moments of the gamma distribution (as well as for the exponential and beta distributions, and others) is to use the unit integrability property of a suitable density; i.e. that $\int_{-\infty}^{\infty}f_{Y}(y)=1$ , when $f_{Y}(y)$ is a density.

To evaluate the $r$ -th moment of a $\operatorname{\mathsf{Gam}}(\alpha,\beta)$ distribution we use the fact that $f_{Y}(y)=\frac{\beta^{\alpha+r}}{\Gamma(\alpha+r)}y^{\alpha+r-1}e^{-\beta y}$ is a density (of a $\operatorname{\mathsf{Gam}}(\alpha+r,\beta)$ random variable).

	$\displaystyle\operatorname{\mathsf{E}}\left[{X^{r}}\right]$	$\displaystyle=\frac{\beta^{\alpha}}{\Gamma(\alpha)}\int_{0}^{\infty}x^{r}x^{% \alpha-1}\exp(-\beta x)\,\mathrm{d}x$
		$\displaystyle=\frac{\beta^{\alpha}\Gamma(\alpha+r)}{\beta^{\alpha+r}\Gamma(% \alpha)}\times\frac{\beta^{\alpha+r}}{\Gamma(\alpha+r)}\int_{0}^{\infty}x^{r+% \alpha-1}\exp(-\beta x)\,\mathrm{d}x$
		$\displaystyle=\frac{\Gamma(\alpha+r)}{\beta^{r}\Gamma(\alpha)}\times 1.$

$\operatorname{\mathsf{E}}\left[{X}\right]=\frac{\Gamma(\alpha+1)}{\beta\Gamma(% \alpha)}=\frac{\alpha}{\beta}$ ,
$\operatorname{\mathsf{E}}\left[{X^{2}}\right]=\frac{\Gamma(\alpha+2)}{\beta^{2% }\Gamma(\alpha)}=\frac{(\alpha+1)\alpha}{\beta^{2}}$ ,
${\operatorname{\mathsf{Var}}}\left[{X}\right]=\frac{(\alpha+1)\alpha}{\beta^{2% }}-\frac{\alpha^{2}}{\beta^{2}}=\frac{\alpha}{\beta^{2}}$ .

Example 3.4.1.

Lifetimes of batteries (in hours) are believed to independently follow an $\operatorname{Exponential}(1/10)$ distribution. You buy a pack of $4$ batteries and use them sequentially. Find:

(a)

The distribution of the lifetime of the pack.
(b)

The expected lifetime of the pack.
(c)

The probability that the lifetime of the pack exceeds $40$ hours.

Solution.

(a)

Let $X$ be the lifetime (in hours) of the pack. Then $X$ is the sum of the $4$ lifetimes of the batteries each of which has an Exponential $(1/10)$ distribution, thus

$\displaystyle{\color[rgb]{0.76,0.01,0}X\sim\operatorname{Gamma}(4,1/10).}$
(b)

The expected lifetime (in hours) is: $\operatorname{\mathsf{E}}\left[{X}\right]={\color[rgb]{0.76,0.01,0}\frac{4}{1/% 10}=40.}$
(c)

The probability that the lifetime exceeds 40 hours is $\operatorname{\mathsf{P}}\left({X>40}\right)=1-\operatorname{\mathsf{P}}\left(% {X\leq 40}\right),$ which can be evaluated in R:

⬇

> 1-pgamma(40,4,1/10)

[1] 0.4334701

Thus $\operatorname{\mathsf{P}}\left({X>40}\right)\approx 0.4335$ .