The process of drawing a sample from a population and to gather information about the parameter through a reasonably close function is known as estimation. This is done in order to find out the unknown parameters which we encounter in a population and which hinders our estimation and conclusion about a population. Thus this process is used to find out the parameters. The value obtained from such a process is known as estimated value and the function is called estimator.

An estimator is supposed to possess the qualities stated below to perform well. They are:

**Un biasedness**

A statistic t is an unbiased estimator of a parameter , if E[t] =

If not, then the estimator is biased.

There are quite a few theorems stated below which proves this.

**Theorem 1:**

Prove that the sample mean x is an unbiased estimator of the population mean µ

**Proof: **

Let x₁, x₂, …, , be a simple random sample with replacement from a finite population of size N, say, X₁, X₂,… ,

µ=

Prove E(x) =µ

While drawing xi, it can be one of the population members i.e., the probability distribution of Xi can be taken as follows:

X₁ | X₂ | … | |

probability | 1/N | 1/N | 1/N |

Therefore,

E ( = X₁* + X₂* +….+ *

=

= µ, i= 1, 2,…., n

E ( = E [( ]

= [E (

If the population is finite or the sampling is done without replacement, the same result will be obtained.

**Theorem 2:**

The sample variance S²= is a biased estimator of variance σ².

**Proof:**

Let x₁, x₂, …, be a random sample from an infinite population with mean σ and variance σ².

Then, E (x) = µ, Variance (x_{i}) = E (xi – µ)² = σ², where i= 1, 2, … ,n.

S²=

=

= where xi – µ and standard deviation is unaffected by change in origin.

= – µ)²

E (s²) = – µ)²

= )= σ²-

Thus, s² is a biased estimator of σ².

Also, Let S²=

E (s²) =

= σ²

= σ²

Therefore, s² is a biased estimator of σ².

**Example: **A population consists of 4 values 3, 7, 11, 15. Draw all possible sample of size two with replacement. Verify that the sample mean is an unbiased estimator of the population mean.

**Solution: **

No. of samples = 42 = 16, which are as below:

(3, 3), (7, 3), (11, 3), (15, 3)

(11 , 7), (15, 7), (11 , 11), (15, 11)

(11, 15), (15, 15), (3, 7), (7, 7),

(3, 11), (7, 11), (3, 15), (7, 15)

Population mean µ= = = 9

Sampling distribution of mean

Sample mean | Frequency f | .f |

3 | 1 | 3 |

5 | 2 | 10 |

7 | 3 | 21 |

9 | 4 | 36 |

11 | 3 | 33 |

13 | 2 | 26 |

15 | 1 | 15 |

Total | 16 | 144 |

Mean of sample = = 9

Since E ()= µ

Therefore, sample mean is an unbiased estimator of population mean.

**Consistency**

A statistic obtained from a random sample of size n is said to be a consistent estimator of a parameter if it converges in probability to θ as n tends to infinity.

Alternatively, If E [ ] θ and Var [ ] 0 as n ∞, then the statistic is said to be consistent estimator of θ.

**Example:**

When sampling from a population N,

E )= µ and )= → 0 as n→

Therefore, sample mean is a consistent estimator of population mean.

**Efficiency**

A parameter might comprise of more than one consistent estimator. Let T_{1} and T_{2} be two consistent estimators of a parameter θ. If Var (T₁) < Var (T₂) for all n, then T₁ is said to be more efficient than T₂ for all size.

**Sufficiency**

Let x₁, x ₂, … , be a random sample from a population whose pmfor pdfisf (x, 8). Then T is said to be a sufficient estimator of e if f (x₁, ). f(x₂, )…..f( , ) = g_{1}(T , ). g₂( x_{1},x_{2},…,

Where g_{1}(T , ) is the sampling distribution and g₂( x_{1},x_{2},…, is independent of .

Even though sufficient estimators exists only in certain cases, but when random sampling for a normal population, the sampling mean x is a sufficient estimator of µ.