Question

Question:Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from a distribution for which the p.d.f. or the p.f. is f (x|θ ), where the value of the parameter θ is unknown. Let\({\bf{X = }}\left( {{{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}} \right)\)and let T be a statistic. Assuming that δ(X) is an unbiased estimator of θ, it does not depend on θ. (If T is a sufficient statistic defined in Sec. 7.7, then this will be true for every estimator δ. The condition also holds in other examples.) Let\({{\bf{\delta }}_{\bf{0}}}\left( {\bf{T}} \right)\)denote the conditional mean of δ(X) given T.

a. Show that\({{\bf{\delta }}_{\bf{0}}}\left( {\bf{T}} \right)\)is also an unbiased estimator of θ.

b. Show that\({\bf{Va}}{{\bf{r}}_{\bf{\theta }}}\left( {{{\bf{\delta }}_{\bf{0}}}} \right) \le {\bf{Va}}{{\bf{r}}_{\bf{\theta }}}\left( {\bf{\delta }} \right)\)for every possible value of θ. Hint: Use the result of Exercise 11 in Sec. 4.7.

Short answer

a. Proved.\({\delta _0}\left( T \right)\)is an unbiased estimator of\(\theta \)

b. Proved. \(Va{r_\theta }\left( {\delta \left( X \right)} \right) \ge Va{r_\theta }\left( {{\delta _0}\left( X \right)} \right)\)

Step-by-step solution

Given information

\({X_1},...,{X_n}\)random sample

\(X = \left( {{X_1},...,{X_n}} \right)\) and T be a statistic

(a) Finding the unbiased estimator

By the law of total probability for expectation,

Let X and Y be random variables such that,

\(E\left[ {E\left( {Y\left| X \right.} \right)} \right] = E\left( Y \right)\)

Similarly,

\(\begin{aligned}{}E\left( \delta \right) &= E\left( {E\left( {\delta \left| T \right.} \right)} \right)\\ &= E\left( {{\delta _0}} \right)\end{aligned}\)

Therefore,\(\delta \,\,and\,\,{\delta _0}\)have the same expectation.

Since\(\delta \)is unbiased,\(E\left( {{\delta _0}} \right) = \theta \)also, in other words,\({\delta _0}\)is also unbiased.

Therefore, \({\delta _0}\left( T \right)\) is an unbiased estimator of \(\theta \)

(b) Finding the variance

By the law of total probability for variances,

Let X and Y be arbitrary random variables for which the necessary expectations and variances exist, then,

\(Var\left( Y \right) = E\left( {Var\left( {Y\left| X \right.} \right)} \right) + Var\left( {E\left( {Y\left| X \right.} \right)} \right)\)

Similarly,

Let,\(Y = \delta \left( X \right)\,\,\,and\,\,\,X = T\)

\(Va{r_\theta }\left( {\delta \left( X \right)} \right) = Va{r_\theta }\left( {{\delta _0}\left( X \right)} \right) + {E_\theta }Var\left( {\delta \left( X \right)\left| T \right.} \right)\)

Since,

\(Var\left( {\delta \left( X \right)\left| T \right.} \right) \ge 0\)

So, too is\({E_\theta }Var\left( {\delta \left( X \right)\left| T \right.} \right)\)

So, \(Va{r_\theta }\left( {\delta \left( X \right)} \right) \ge Va{r_\theta }\left( {{\delta _0}\left( X \right)} \right)\)[Proved]