Lecture 10 Improving Estimators: Rao–Blackwell and Lehmann–Scheffe

In lecture 8 we have seen that a sufficient statistic $T$ can be used to summarise all the information in the data about the parameter. In this lecture we will use this idea to construct improved versions of estimators. The Rao–Blackwell theorem states that conditioning an unbiased estimator on the value of $T$ never increases its variance. The Lehmann–Scheffe theorem states that, if $T$ is a complete sufficient statistic, this procedure gives the unique best unbiased estimator. Finally, Basu’s theorem gives the independence of $\bar{X}$ and $S^{2}$ for normal samples and this result will be used in lectures 25 and 29. The proofs of these three theorems all use facts about conditional expectations, as shown in propositions A.9 and A.11 in appendix A.

10.1 The Rao–Blackwell Theorem

Assume that we want to estimate a function $g(\theta)$ of the parameter, e.g. we could be interested in the probability $\mathbb{P}_{\theta}(X_{1}=0)$ of zero counts. As in lecture 2, an estimator $U$ is unbiased for $g(\theta)$ , if $\mathbb{E}_{\theta}(U)=g(\theta)$ for all $\theta\in\Theta$ . While unbiased estimators are often easy to find, they often turn out to be wasteful, since they only use a small amount of the available sample. The Rao–Blackwell theorem uses a sufficient statistic $T$ to improve an existing estimator by replacing $U$ with the conditional expectation $V=\mathbb{E}_{\theta}(U\mskip 1.0mu|\mskip 1.0muT)$ , i.e. the average of $U$ for all samples which have the same value of $T$ as the observed sample.

Theorem 10.1 (Rao–Blackwell).

Let $T$ be a sufficient statistic for $\theta$ and let $U$ be an unbiased estimator for $g(\theta)$ with $\mathop{\mathrm{Var}}\nolimits_{\theta}(U)<\infty$ for all $\theta$ . Then $V=\mathbb{E}_{\theta}(U\mskip 1.0mu|\mskip 1.0muT)$ satisfies the following properties.

1.

The random variable $V$ does not depend on $\theta$ , but is a function of $T$ alone and thus a statistic.
2.

The statistic $V$ is an unbiased estimator for $g(\theta)$ .
3.

For all $\theta\in\Theta$ we have $\mathop{\mathrm{Var}}\nolimits_{\theta}(V)\leq\mathop{\mathrm{Var}}\nolimits_{% \theta}(U)$ , with equality if and only if $\mathbb{P}_{\theta}(U=V)=1$ .

Proof.

From definition 8.1 we know that the conditional distribution of the sample $(X_{1},\dots,X_{n})$ given $T$ does not depend on $\theta$ . Since $U$ is a function of the sample, the conditional distribution and thus the conditional expectation given $T$ do not depend on $\theta$ . Thus, $V$ is a function of $T$ and by definition 1.3 a statistic.

For the second statement, using the tower property from proposition A.9, we find

\mathbb{E}_{\theta}(V)=\mathbb{E}_{\theta}\bigl{(}\mathbb{E}_{\theta}(U\mskip 1% .0mu|\mskip 1.0muT)\bigr{)}=\mathbb{E}_{\theta}(U)=g(\theta)

for all $\theta$ . This shows that $V$ is unbiased.

Finally, for the third statement, we can use the law of total variance, proposition A.11, with $U$ instead of $X$ and $T$ instead of $Y$ . Since $\mathbb{E}_{\theta}(U\mskip 1.0mu|\mskip 1.0muT)=V$ we get

\mathop{\mathrm{Var}}\nolimits_{\theta}(U)=\mathbb{E}_{\theta}\bigl{(}\mathop{% \mathrm{Var}}\nolimits_{\theta}(U\mskip 1.0mu|\mskip 1.0muT)\bigr{)}+\mathop{% \mathrm{Var}}\nolimits_{\theta}(V).

Since the conditional variance $\mathop{\mathrm{Var}}\nolimits_{\theta}(U\mskip 1.0mu|\mskip 1.0muT)$ is non-negative, we find $\mathop{\mathrm{Var}}\nolimits_{\theta}(V)\leq\mathop{\mathrm{Var}}\nolimits_{% \theta}(U)$ . Equality holds, if and only if $\mathbb{E}_{\theta}\bigl{(}\mathop{\mathrm{Var}}\nolimits_{\theta}(U\mskip 1.0% mu|\mskip 1.0muT)\bigr{)}=0$ and since a non-negative random variable with expectation zero is zero with probability one, this is the case if and only if $\mathop{\mathrm{Var}}\nolimits_{\theta}(U\mskip 1.0mu|\mskip 1.0muT)=0$ with probability one, i.e. if and only if $U$ coincides with its conditional expectation $V$ with probability one. This completes the proof. ∎

Since both estimators are unbiased, the mean squared errors equal the variances, as given in theorem 2.7, and thus $V$ is at least as good as $U$ , and strictly better unless $U$ was already a function of $T$ . The procedure of passing from $U$ to $V$ is called Rao–Blackwellisation. The necessity of sufficiency is highlighted by the fact that conditioning on a statistic which is not sufficient in general will result in a quantity which depends on $\theta$ but is not an estimator.

Example 10.2.

Let $X_{1},\dots,X_{n}$ be i.i.d. Poisson with parameter $\theta$ and assume that we want to estimate $g(\theta)=\mathbb{P}_{\theta}(X_{1}=0)=e^{-\theta}$ . The indicator $U=\mathbf{1}_{\{X_{1}=0\}}$ is unbiased for $g(\theta)$ , since $\mathbb{E}_{\theta}(U)=\mathbb{P}_{\theta}(X_{1}=0)$ . By corollary 8.3 the sum $T=\sum_{i}X_{i}$ is sufficient. Given $T=t$ , the first observation is distributed as $\text{Binomial}(t,1/n)$ and thus we have

V=\mathbb{E}_{\theta}(U\mskip 1.0mu|\mskip 1.0muT)=\mathbb{P}_{\theta}(X_{1}=0% \mskip 1.0mu|\mskip 1.0muT)=\Bigl{(}1-\frac{1}{n}\Bigr{)}^{T}.

The Rao–Blackwellised estimator now uses all $n$ observations. Exercises 10.1 and 10.2 show how to find the conditional distribution and how to apply the same procedure to a continuous model.

10.2 Completeness and the Lehmann–Scheffe Theorem

The Rao–Blackwell theorem improves an existing estimator, but two different unbiased estimators could in principle be improved to two different functions of $T$ . The concept of completeness, definition 8.8, solves this problem: if $T$ is complete, there can only be one unbiased estimator for $g(\theta)$ among the functions of $T$ and thus all Rao–Blackwellisations will converge to this estimator. We start our discussion by naming the target.

Definition 10.3.

An unbiased estimator $V$ for $g(\theta)$ with $\mathop{\mathrm{Var}}\nolimits_{\theta}(V)<\infty$ for all $\theta\in\Theta$ is a uniformly minimum variance unbiased estimator (UMVUE) for $g(\theta)$ , if $\mathop{\mathrm{Var}}\nolimits_{\theta}(V)\leq\mathop{\mathrm{Var}}\nolimits_{% \theta}(W)$ for all $\theta\in\Theta$ and for every unbiased estimator $W$ for $g(\theta)$ with finite variance.

The word “uniformly” refers to the parameter: the inequality has to hold for every $\theta$ at once, and there is no reason a priori why such an estimator should exist. The following theorem shows that, whenever a complete sufficient statistic exists, a UMVUE exists as well and is easy to identify.

Theorem 10.4 (Lehmann–Scheffe).

Let $T$ be a complete sufficient statistic for $\theta$ and let $V=\varphi(T)$ be a function of $T$ , an unbiased estimator for $g(\theta)$ with $\mathop{\mathrm{Var}}\nolimits_{\theta}(V)<\infty$ for all $\theta$ . Then $V$ is a UMVUE for $g(\theta)$ . Furthermore, $V$ is the unique UMVUE: any other unbiased estimator $W$ for $g(\theta)$ with $\mathop{\mathrm{Var}}\nolimits_{\theta}(W)=\mathop{\mathrm{Var}}\nolimits_{% \theta}(V)$ for all $\theta$ satisfies $\mathbb{P}_{\theta}(W=V)=1$ for all $\theta$ .

Proof.

To show the uniqueness of $V$ as an unbiased estimator for $g(\theta)$ which is a function of $T$ , let $\psi(T)$ be another such function. Then we have $\mathbb{E}_{\theta}\bigl{(}\psi(T)\bigr{)}=g(\theta)=\mathbb{E}_{\theta}\bigl{% (}\varphi(T)\bigr{)}$ for all $\theta$ . The function $h=\psi-\varphi$ satisfies $\mathbb{E}_{\theta}\bigl{(}h(T)\bigr{)}=0$ for all $\theta$ and, since $T$ is complete, by definition 8.8, we have $\mathbb{P}_{\theta}\bigl{(}h(T)=0\bigr{)}=1$ for all $\theta$ , i.e. $\psi(T)=\varphi(T)$ with probability one.

Now let $W$ be any unbiased estimator for $g(\theta)$ with finite variance. Then, by the Rao–Blackwell theorem, theorem 10.1, the random variable $\mathbb{E}_{\theta}(W\mskip 1.0mu|\mskip 1.0muT)$ is a function of $T$ which is unbiased for $g(\theta)$ . Thus, by the first step of the proof, it equals $V$ with probability one. By the Rao–Blackwell theorem we have then

\mathop{\mathrm{Var}}\nolimits_{\theta}(V)=\mathop{\mathrm{Var}}\nolimits_{% \theta}\bigl{(}\mathbb{E}_{\theta}(W\mskip 1.0mu|\mskip 1.0muT)\bigr{)}\leq% \mathop{\mathrm{Var}}\nolimits_{\theta}(W)

for all $\theta$ , i.e. $V$ is a UMVUE. If $\mathop{\mathrm{Var}}\nolimits_{\theta}(W)=\mathop{\mathrm{Var}}\nolimits_{% \theta}(V)$ for all $\theta$ , then the equality case of the Rao–Blackwell theorem implies $W=\mathbb{E}_{\theta}(W\mskip 1.0mu|\mskip 1.0muT)=V$ with probability one. This completes the proof. ∎

The theorem turns the search for a best unbiased estimator into a two-step recipe: find a complete sufficient statistic $T$ , then find any function of $T$ which is unbiased for $g(\theta)$ , by adjusting a constant or by Rao–Blackwellising a crude unbiased estimator. For the models of this module the following two facts settle the completeness.

Proposition 10.5.

1.

Let $X_{1},\dots,X_{n}$ be an i.i.d. sample from an exponential family of full rank, as given in definitions 7.1 and 7.10. Then the corresponding natural statistic $T_{n}=\sum_{i=1}^{n}T(X_{i})$ is complete.
2.

Let $X_{1},\dots,X_{n}$ be i.i.d. and uniformly distributed on the set $(0,\theta)$ where $\theta>0$ . Then the sample maximum $\max_{i}X_{i}$ is complete.

This proposition is given without proof. The first statement rests on the uniqueness theorem for Laplace transforms. The second statement rests on taking derivatives of the relation $\int_{0}^{\theta}h(t)\,t^{n-1}\,\mathrm{d}t=0$ with respect to $\theta$ . Both statements are allowed to be quoted. Together with corollary 8.3 and exercise 8.2 they form a complete sufficient statistic for all families of the form given in tables 1.1 and 1.2 (with $m$ known for the binomial). If we use a one-to-one function of a complete sufficient statistic, the new function is again complete and sufficient. Thus we can use the function $\bar{X}$ instead of $\sum_{i}X_{i}$ if required.

Example 10.6.

For the Poisson sample from example 10.2, the natural parameter $\eta=\log\theta$ takes values on all of $\mathbb{R}$ and thus the sum $T=\sum_{i}X_{i}$ is a complete and sufficient statistic. Since $\bar{X}=T/n$ is a function of $T$ and is unbiased for $\theta$ , it is the UMVUE for $\theta$ by theorem 10.4. Similarly, the Rao–Blackwellised indicator $V=(1-1/n)^{T}$ is the UMVUE for $e^{-\theta}$ .

Example 10.7.

Let $X_{1},\dots,X_{n}$ be i.i.d. $N(\mu,\sigma^{2})$ with both parameters unknown. This is a two-parameter exponential family of full rank with natural statistic $T(x)=(x,x^{2})$ , and thus the pair $(\bar{X},S^{2})$ is complete and sufficient; exercise 10.4 carries out the verification. Since $\bar{X}$ and $S^{2}$ are unbiased for $\mu$ and $\sigma^{2}$ by lemmas 2.2 and 2.3, they are the UMVUEs for these parameters.

The second example puts section 2.3 into perspective: there the biased estimator $\frac{1}{n+1}\sum_{i}(X_{i}-\bar{X})^{2}$ had smaller mean squared error than $S^{2}$ . No contradiction, since the Lehmann–Scheffe theorem compares $S^{2}$ to unbiased estimators only.

10.3 Ancillary Statistics and Basu’s Theorem

A sufficient statistic contains all the information about $\theta$ . At the opposite extreme are statistics which contain none.

Definition 10.8.

A statistic $A=A(X_{1},\dots,X_{n})$ is ancillary for $\theta$ , if the distribution of $A$ under $\mathbb{P}_{\theta}$ does not depend on $\theta\in\Theta$ .

The running example is a normal sample $X_{1},\dots,X_{n}\sim N(\mu,\sigma^{2})$ with known variance $\sigma^{2}$ and unknown mean $\mu$ . Writing $Z_{i}=X_{i}-\mu$ we have $X_{i}-\bar{X}=Z_{i}-\bar{Z}$ and the $Z_{i}$ are i.i.d. $N(0,\sigma^{2})$ , for every value of $\mu$ . Thus, the vector of residuals $(X_{1}-\bar{X},\dots,X_{n}-\bar{X})$ is ancillary for $\mu$ and every function of the residuals is ancillary, e.g. the sample variance $S^{2}$ and the range $\max_{i}X_{i}-\min_{i}X_{i}$ ; we will consider the range in exercise 10.5.

One might expect an ancillary statistic to have nothing to do with a sufficient one. Under completeness this becomes a theorem, and it is a surprisingly effective way of proving independence.

Theorem 10.9 (Basu).

Let $T$ be a complete sufficient statistic for $\theta$ and let $A$ be an ancillary statistic. Then $T$ and $A$ are independent of each other under $\mathbb{P}_{\theta}$ for all $\theta\in\Theta$ .

Proof.

Let $B$ be a set of possible values of $A$ . Since $A$ is ancillary, the probability $p_{B}=\mathbb{P}_{\theta}(A\in B)$ does not depend on $\theta$ . We compare this probability to the conditional probability

h(T)=\mathbb{P}_{\theta}(A\in B\mskip 1.0mu|\mskip 1.0muT)=\mathbb{E}_{\theta}% \bigl{(}\mathbf{1}_{\{A\in B\}}\!\mathrel{\big{|}}\!T\bigr{)}.

Since $A$ is a function of the sample and since $T$ is sufficient, the conditional distribution of $A$ given $T$ does not depend on $\theta$ . Thus, $h(T)$ is a function of $T$ alone. Using the tower property from proposition A.9 we find

\mathbb{E}_{\theta}\bigl{(}h(T)\bigr{)}=\mathbb{P}_{\theta}(A\in B)=p_{B}

for all $\theta$ . Thus the function $h-p_{B}$ satisfies $\mathbb{E}_{\theta}\bigl{(}h(T)-p_{B}\bigr{)}=0$ for all $\theta$ . Since $T$ is complete, we have $h(T)=p_{B}$ with probability one for every $\mathbb{P}_{\theta}$ .

It remains to show that this implies independence. For any set $C$ of possible values of $T$ , the indicator function $\mathbf{1}_{\{T\in C\}}$ is a function of $T$ and can be taken out of the conditional expectation given $T$ (see proposition A.10 in appendix A). Using the tower property we find

\displaystyle\mathbb{P}_{\theta}(T\in C,\ A\in B)

\displaystyle=\mathbb{E}_{\theta}\bigl{(}\mathbf{1}_{\{T\in C\}}\mathbf{1}_{\{% A\in B\}}\bigr{)}=\mathbb{E}_{\theta}\Bigl{(}\mathbf{1}_{\{T\in C\}}\,\mathbb{% E}_{\theta}\bigl{(}\mathbf{1}_{\{A\in B\}}\!\mathrel{\big{|}}\!T\bigr{)}\Bigr{)}

\displaystyle=\mathbb{E}_{\theta}\bigl{(}\mathbf{1}_{\{T\in C\}}\,p_{B}\bigr{)% }=\mathbb{P}_{\theta}(T\in C)\,\mathbb{P}_{\theta}(A\in B).

Since $B$ and $C$ were arbitrary, we have shown the independence of $T$ and $A$ and the proof is complete. ∎

The proof uses each hypothesis exactly once: ancillarity makes the unconditional probability constant in $\theta$ , sufficiency makes the conditional probability a statistic, and completeness forces these two to be the same. The most important application of this result in this module is the following.

Corollary 10.10.

Let $X_{1},\dots,X_{n}$ be i.i.d. $N(\mu,\sigma^{2})$ . Then $\bar{X}$ and $S^{2}$ are independent.

Proof.

Fixing the value of $\sigma^{2}$ , we can consider $\mu\in\mathbb{R}$ to be the only parameter. The density of a single observation is

f(x;\mu)=\frac{1}{\sqrt{2\pi\sigma^{2}}}\,e^{-x^{2}/(2\sigma^{2})}\exp\Bigl{(}% \frac{\mu}{\sigma^{2}}\,x-\frac{\mu^{2}}{2\sigma^{2}}\Bigr{)}.

This is of the form given in definition 7.1 with natural statistic $T(x)=x$ and natural parameter $\eta(\mu)=\mu/\sigma^{2}$ . As $\mu$ ranges over $\mathbb{R}$ , so does $\eta(\mu)=\mu/\sigma^{2}$ , that is, the natural parameter takes all values in $\mathbb{R}$ , and thus the family is of full rank. By corollary 8.3 and proposition 10.5, the sum $\sum_{i}X_{i}$ , and thus $\bar{X}$ , is a complete and sufficient statistic for $\mu$ . Furthermore, $S^{2}$ is a function of the residuals $X_{i}-\bar{X}$ and thus, as shown above, is ancillary for $\mu$ . By Basu’s theorem, theorem 10.9, the statistics $\bar{X}$ and $S^{2}$ are independent under $\mathbb{P}_{\mu,\sigma^{2}}$ for every $\mu$ and, since $\sigma^{2}$ was arbitrary, the claim is proved. ∎

This completes the proof of the independence statement of proposition A.3. The result from lectures 25 and 29 on the ratio of $\bar{X}-\mu$ to $S$ uses this proposition. The proof contains a useful trick: when both parameters are unknown, neither $\bar{X}$ nor $S^{2}$ can be ancillary and we cannot apply Basu’s theorem directly. The solution to this problem is to fix the value of $\sigma^{2}$ , i.e. to consider the independence statement for one distribution at a time.

In lecture 13 we will see a second way to derive optimality, using the Cramer–Rao bound, which does not require conditioning.

Summary.

•

Rao–Blackwell: If an unbiased estimator is conditioned on a sufficient statistic $T$ , the conditioned estimator is unbiased and has no larger variance.
•

A UMVUE is an unbiased estimator which has minimal variance among all unbiased estimators, for all $\theta$ .
•

Lehmann–Scheffe: If $T$ is complete and sufficient, then the only function of $T$ which is unbiased for $g(\theta)$ is the UMVUE for $g(\theta)$ .
•

The natural statistic for a full-rank exponential family and the maximum of a uniform sample are complete, but without proof.
•

Basu: A complete sufficient statistic is independent of all ancillary statistics. Using this result, one can show that $\bar{X}$ and $S^{2}$ of a normal sample are independent.

Exercise 10.1.

Let $X_{1},\dots,X_{n}$ be i.i.d. Poisson with parameter $\theta>0$ , let $T=\sum_{i=1}^{n}X_{i}$ and let $U=\mathbf{1}_{\{X_{1}=0\}}$ , as in example 10.2. You may use the fact that a sum of independent Poisson random variables is again Poisson, with the parameters added.

1.

Show that $U$ is an unbiased estimator for $e^{-\theta}$ and determine the variance of this estimator.
2.

Show that for $t\in\{0,1,2,\dots\}$ and $x\in\{0,1,\dots,t\}$ we have

$\mathbb{P}_{\theta}(X_{1}=x\mskip 1.0mu|\mskip 1.0muT=t)=\binom{t}{x}\Bigl{(}% \frac{1}{n}\Bigr{)}^{x}\Bigl{(}1-\frac{1}{n}\Bigr{)}^{t-x},$

i.e. that the conditional distribution of $X_{1}$ given $T=t$ is $\text{Binomial}(t,1/n)$ and that this distribution does not depend on $\theta$ .
3.

Deduce that $V=\mathbb{E}_{\theta}(U\mskip 1.0mu|\mskip 1.0muT)=(1-1/n)^{T}$ .
4.

Show directly that $V$ is an unbiased estimator for $e^{-\theta}$ , using the probability generating function $\mathbb{E}_{\theta}(s^{T})=\exp\bigl{(}n\theta(s-1)\bigr{)}$ of the Poisson distribution with parameter $n\theta$ .
5.

Explain why $V$ is the UMVUE for $e^{-\theta}$ and discuss the differences between this UMVUE and the MLE $e^{-\bar{X}}$ , introduced in exercise 5.7.

Exercise 10.2.

Let $n\geq 2$ and let $X_{1},\dots,X_{n}$ be i.i.d. exponential with rate $\theta>0$ , i.e. $f(x;\theta)=\theta e^{-\theta x}$ for $x\geq 0$ and let $T=\sum_{i=1}^{n}X_{i}$ . You may use the fact that the sum of $k$ independent random variables of this type has density $f_{k}(t)=\theta^{k}t^{k-1}e^{-\theta t}/(k-1)!$ for $t\geq 0$ .

1.

Explain why $X_{1}$ is an unbiased estimator for the mean $1/\theta$ and why $T$ is sufficient for $\theta$ .
2.

Let $R=\sum_{i=2}^{n}X_{i}$ . Use the independence of $X_{1}$ and $R$ to find the joint density of $(X_{1},T)$ . Deduce the conditional density of $X_{1}$ given $T=s$ . The result is

$f(x\mskip 1.0mu|\mskip 1.0mus)=\frac{(n-1)(s-x)^{n-2}}{s^{n-1}}\qquad\text{for% }0<x<s,$

and zero elsewhere. Check that this density does not depend on $\theta$ .
3.

Show that $\mathbb{E}_{\theta}(X_{1}\mskip 1.0mu|\mskip 1.0muT=s)=s/n$ and conclude that Rao–Blackwellising $X_{1}$ leads to the sample mean $\bar{X}$ , again unbiased for $1/\theta$ .
4.

Find a second argument for $\mathbb{E}_{\theta}(X_{1}\mskip 1.0mu|\mskip 1.0muT)=T/n$ , based on the symmetry of the sample, without any integration.

Exercise 10.3.

Let $X_{1},\dots,X_{n}$ be i.i.d. uniformly distributed on the set $(0,\theta)$ where $\theta>0$ , and let $M=\max_{i}X_{i}$ .

1.

Using the mean of $M$ from exercise 2.2, show that $\hat{\theta}=(n+1)M/n$ is an unbiased estimator for $\theta$ .
2.

Show that $\hat{\theta}$ is the UMVUE for $\theta$ .
3.

The estimator $\tilde{\theta}=2\bar{X}$ is also unbiased for $\theta$ . Determine the variances of $\hat{\theta}$ and of $\tilde{\theta}$ and verify that they are ordered as in theorem 10.4.
4.

Without computing any conditional expectations, find $\mathbb{E}_{\theta}(\tilde{\theta}\mskip 1.0mu|\mskip 1.0muM)$ .

Exercise 10.4.

Let $X_{1},\dots,X_{n}$ be i.i.d. $N(\mu,\sigma^{2})$ , where both parameters are unknown. This is the situation of example 10.7.

1.

Write the density of an observation in the form from definition 7.8 and identify the natural statistic $T(x)=(x,x^{2})$ and the natural parameter $\eta(\mu,\sigma^{2})$ .
2.

Show that the family of densities has full rank, in the sense of definition 7.10, by determining the range of $\eta$ .
3.

Deduce that $\bigl{(}\sum_{i}X_{i},\sum_{i}X_{i}^{2}\bigr{)}$ is complete and sufficient and argue that the same holds for the pair $(\bar{X},S^{2})$ .

Exercise 10.5.

Let $X_{1},\dots,X_{n}$ be i.i.d. $N(\mu,\sigma^{2})$ where $\sigma^{2}$ is known and $\mu\in\mathbb{R}$ is unknown. Furthermore, let $R=\max_{i}X_{i}-\min_{i}X_{i}$ be the sample range.

1.

Show that $R$ is ancillary for $\mu$ .
2.

Show that $\bar{X}$ and $R$ are independent.
3.

Now assume that $\mu$ is known and $\sigma^{2}>0$ is unknown. Is $R$ ancillary for $\sigma^{2}$ ? Is $R/\sigma$ a statistic?