91Ó°ÊÓ

Suppose a random variable X with probability density function \(f\left( {x|\theta } \right)\) is said to belong an exponential family if the pdf is of the form:

\(f\left( {x|\theta } \right) = a\left( \theta \right)b\left( x \right)\exp \left( {c\left( \theta \right)d\left( x \right)} \right)\).

Where,\(a\left( \theta \right)\)and\(c\left( \theta \right)\)are arbitrary function of\(\theta \)and a(x) and d(x) are arbitrary function of x.

(a)

Let, p.m.f. of Bernoulli distribution,

\(f\left( {x/p} \right) = {p^x}{\left( {1 - p} \right)^{1 - x}}\)

Therefore,

\(\begin{aligned}{}f\left( {x/p} \right) &= {p^x}{\left( {1 - p} \right)^{1 - x}}\\ &= \left( {1 - p} \right){\left( {\frac{p}{{1 - p}}} \right)^x}\end{aligned}\)

Therefore,

\(\begin{aligned}{}a\left( p \right) &= 1 - p,\\b\left( x \right) &= 1,\\c\left( p \right) &= \log \left( {\frac{p}{{1 - p}}} \right),\\d\left( x \right) &= x\end{aligned}\)

Hence, the family of Bernoulli distributions with an unknown value of the parameter pis an exponential family.

Let, the p.m.f. of Poisson distribution as

\(f\left( {x/\theta } \right) = \frac{{{e^{\left( { - \theta } \right)}}{\theta ^x}}}{{x!}}\)

Therefore,

\(\begin{aligned}{}a\left( \theta \right) &= {e^{\left( { - \theta } \right)}},\\b\left( x \right) = \frac{1}{{x!}},\\c\left( \theta \right) &= \log \theta ,\\d\left( x \right) &= x\end{aligned}\)

Hence, thefamily of Poisson distributions with an unknown mean is an exponential family.

c

Let, the p.m.f. of as

\(f\left( {x/p} \right) = \left( {\begin{aligned}{{}{}}{r + x - 1}\\x\end{aligned}} \right){p^r}{\left( {1 - p} \right)^x}\)

Therefore,

\(\begin{aligned}{}a\left( p \right) &= {p^r},\\b\left( x \right) &= \left( {\begin{aligned}{{}{}}{r + x - 1}\\x\end{aligned}} \right),\\c\left( p \right) &= \log \left( {1 - p} \right),\\d\left( x \right) &= x\end{aligned}\)

Hence, thefamily of negative binomial distribution for which the value of r is known and the value of p is unknown is an exponential family.

(d)

Let, p.d.f of normal distribution as,

\(f\left( {x/\mu } \right) = \frac{1}{{{{\left( {2\pi {\sigma ^2}} \right)}^{1/2}}}}\exp \left( { - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}} \right)\)

\( = \frac{1}{{{{\left( {2\pi {\sigma ^2}} \right)}^{1/2}}}}\exp \left( { - \frac{{{x^2}}}{{2{\sigma ^2}}}} \right)\exp \left( { - \frac{{{\mu ^2}}}{{2{\sigma ^2}}}} \right)\exp \left( {\frac{{\mu x}}{{{\sigma ^2}}}} \right)\)

Therefore,

\(\begin{aligned}{}a\left( \mu \right) &= \frac{1}{{{{\left( {2{\sigma ^2}} \right)}^{1/2}}}}\exp \left( { - \frac{{{\mu ^2}}}{{2{\sigma ^2}}}} \right),\\b\left( x \right) &= \exp \left( { - \frac{{{x^2}}}{{2{\sigma ^2}}}} \right),\\c\left( \mu \right) &= \frac{\mu }{{{\sigma ^2}}},\\d\left( x \right) = x\end{aligned}\)

Hence,the family of normal distributions with an unknown mean and a known variance is an exponential family.

(e)

Let, p.d.f of normal distribution as,

\(f\left( {x/{\sigma ^2}} \right) = \frac{1}{{{{\left( {2\pi {\sigma ^2}} \right)}^{1/2}}}}\exp \left( { - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}} \right)\)

Therefore,

\(\begin{aligned}{}a\left( {{\sigma ^2}} \right) &= \frac{1}{{{{\left( {2\pi {\sigma ^2}} \right)}^{1/2}}}},\\b\left( x \right) &= 1;\\c\left( {{\sigma ^2}} \right) &= - \frac{1}{{2{\sigma ^2}}},\\d\left( x \right) &= \log \;x\end{aligned}\)

Hence, the family normal distributions with an unknown variance and a known mean are an exponential family.

f

Let, p.d.f of gamma distribution as,

\(f\left( {x/a} \right) = \frac{{{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}}{x^{\alpha - 1}}\exp \left( { - \beta x} \right)\)

Therefore,

\(\begin{aligned}{}a\left( \alpha \right) = \frac{{{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}},\\b\left( x \right) = {x^{\alpha - 1}}\\c\left( \alpha \right) = 1\\d\left( x \right) = x\\\end{aligned}\)

Hence, the family of gamma distributions for which the value of α is unknown and the value of β is known is an exponential family.

(g)

Let, p.d.f as,

\(f\left( {x/a} \right) = \frac{{{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}}{x^{\alpha - 1}}\exp \left( { - \beta x} \right)\)

Therefore,

\(\begin{aligned}{}\alpha \left( \beta \right) &= {\beta ^\alpha },\\b\left( x \right) &= \frac{{{x^{\alpha - 1}}}}{{\Gamma \left( \alpha \right)}},\\c\left( \beta \right) &= - \beta ,\\d\left( x \right) &= x\end{aligned}\)

Hence, the family of gamma distributions for which the value of α is known and the value of β is unknown is an exponential family.

Let, p.d.f of beta distribution as,

\(f\left( {x|\beta } \right) = \frac{{\Gamma \left( {\alpha + \beta } \right)}}{{\Gamma \left( \alpha \right)\,\Gamma \left( \beta \right)}}{x^{\alpha - 1}}\,{\left( {1 - x} \right)^{\beta - 1}}\)

\(\begin{aligned}{}a\left( \alpha \right) &= \frac{{\Gamma \left( {\alpha + \beta } \right)}}{{\Gamma \left( \alpha \right)}},\\b\left( x \right) &= \frac{{{{\left( {1 - x} \right)}^{\beta - 1}}}}{{\Gamma \left( \beta \right)}},\\c\left( \alpha \right) &= \alpha - 1\end{aligned}\)

Hence, the beta distributions for which the value of α is unknown and the value of β is known is an exponential family.

i

Let, the p.d.f of beta distribution as,

\(f\left( {x|\beta } \right) = \frac{{\Gamma \left( {\alpha + \beta } \right)}}{{\Gamma \left( \alpha \right)\,\Gamma \left( \beta \right)}}{x^{\alpha - 1}}\,{\left( {1 - x} \right)^{\beta - 1}}\)

Therefore,

\(\begin{aligned}{}\alpha \left( \beta \right) &= \frac{{\Gamma \left( {\alpha + \beta } \right)}}{{\Gamma \left( \beta \right)}},\\b\left( x \right) &= \frac{{{x^{\alpha - 1}}}}{{\Gamma \left( \alpha \right)}},\\c\left( \beta \right) &= \beta - 1,\\d\left( x \right) &= \log \left( {1 - x} \right)\end{aligned}\)

Hence, the beta distributions for which the value of α is known and the value of β is unknown is an exponential family