91Ó°ÊÓ

Let \({X_1},...{X_n}\) be the random sample of size n from the specified distribution with density function \(f\left( {x,\theta } \right)\) where parameter \(\theta \).is unknown.

An estimator \(T = T\left( {{X_1},...{X_n}} \right)\) is said to be the sufficient estimator of \(\theta \) If conditional joint distribution\({X_1},...{X_n}\)given any value t of estimator is independent of \(\theta \)

Let \({X_1},...{X_n}\) be the random sample of size n from the specified distribution with density function\(f\left( {x,\theta } \right)\), here \(\theta \) is unknown, an estimator \(T = T\left( {{X_1},...{X_n}} \right)\) is said to be the sufficient estimator of \(\theta \) if \(L\left( {x,\theta } \right) = g\left( {t,\theta } \right)h\left( x \right)\)

Here\(L\left( {x,\theta } \right)\)is the likelihood function of\({X_1},...{X_n}\).

\(g\left( {t,\theta } \right)\)is the function of\({X_1},...{X_n}\)which depends on\(\theta \)

\(h\left( x \right)\)is the s function which is independent on\(\theta \).

Let \({x_1},...{x_n}\) be the random sample from normal distribution with mean \(\mu \) and standard deviation of \(\sigma \).

A pdf of a normal distribution is given as:

\(f\left( {x;\mu ,{\sigma ^2}} \right) = \frac{1}{{\sqrt {2\pi \sigma } }}{e^{ - \frac{{\left( {x - \mu } \right)}}{{2{\sigma ^2}}}}}\)

Using factorization theorem,

\(\begin{align}L\left( {\mu ,{\sigma ^2}} \right) &= f\left( {{x_1};\mu ,{\sigma ^2}} \right) \times f\left( {{x_2};\mu ,{\sigma ^2}} \right)... \times f\left( {{x_n};\mu ,{\sigma ^2}} \right)\\ &= \frac{1}{{\sqrt {2\pi \sigma } }}{e^{ - \frac{{\left( {{x_1} - \mu } \right)}}{{2{\sigma ^2}}}}} \times \frac{1}{{\sqrt {2\pi \sigma } }}{e^{ - \frac{{\left( {{x_2} - \mu } \right)}}{{2{\sigma ^2}}}}}... \times \frac{1}{{\sqrt {2\pi \sigma } }}{e^{ - \frac{{\left( {{x_n} - \mu } \right)}}{{2{\sigma ^2}}}}}\\ &= {\left( {\frac{1}{{\sqrt {2\pi \sigma } }}} \right)^n}{e^{ - \frac{{{{\left( {{x_1} - \mu } \right)}^2}}}{{2{\sigma ^2}}} + \frac{{{{\left( {{x_2} - \mu } \right)}^2}}}{{2{\sigma ^2}}} + ...\frac{{{{\left( {{x_n} - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}\end{align}\)

\(\begin{align}L\left( {\mu ,{\sigma ^2}} \right) &= {\left( {\frac{1}{{\sqrt {2\pi \sigma } }}} \right)^n}{e^{ - \frac{1}{2}\sum\limits_{i = 1}^n {\frac{{{{\left( {{x_i} - \mu } \right)}^2}}}{{{\sigma ^2}}}} }}\\ &= {\left( {\frac{1}{{\sqrt {2\pi \sigma } }}} \right)^n}{e^{ - \frac{1}{{2{\sigma ^2}}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}\end{align}\)

We can say that the 1^st bracket is independent of \({\sigma ^2}\) but the second term is dependent, therefore by factorization theorem \(T = \sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} \) is sufficient statistic. for \({\sigma ^2}\).