## Example Problems on Estimators

ST702 Homework 5 on Estimators

# 7.45

Let $X_1, X_2, \dots X_n$ be iid from a distribution with mean $\mu$ and variance $\sigma^2$, and let $S^2$ be the usual unbiased estimator of $\sigma^2$. In Example 7.3.4 we saw that, under normality, the MLE has smaller MSE than $S^2$. In this exercise we will explore variance estimators some more.

## a

Show that, for any estimator of the form $aS^2$, where $a$ is some constant,

$MSE(aS^2) = E[aS^2 - \sigma^2]^2 = a^2 Var(S^2) + (a - 1)^2 \sigma^4$ \begin{align} MSE(aS^2) & = E[W - \theta]^2 \\ & = E[aS^2 - \sigma^2]^2 \\ & = Var(aS^2) - (Bias(aS^2))^2 \\ & = a^2 Var(S^2) - (E(aS^2) - \sigma^2)^2 \\ & = a^2 Var(S^2) - (a^2 E(S^2)^2 - 2a E(S^2)\sigma^2 + \sigma^4)\\ & = a^2 Var(S^2) - (a^2 (\sigma^2)^2 - 2a \sigma^2 \sigma^2 + \sigma^4) & \text{unbiased}\\ & = a^2 Var(S^2) + (a-1)^2 \sigma^4 \end{align}

## b

Show that

$Var(S^2) = \frac{ 1 }{ n } \Big( \kappa - \frac{ n-3 }{ n-1 } \Big)\sigma^4$

where $\kappa = E[X-\mu]^4 / \sigma^4$ is the kurtosis.

\begin{align} V(S^2) & = V\Big( \frac{ 1 }{ n-1 } \sum (X_i - \overline{ X })^2 \Big) \\ & = \frac{ 1 }{ (n-1)^2 } \sum V( (X_i - \overline{ X })^2) \\ & = \frac{ 1 }{ (n-1)^2 } \sum \Big[ E( ((X_i - \overline{ X })^2)^2 ) - E((X_i - \overline{ X })^2)^2 \Big] \\ & = \frac{ 1 }{ (n-1)^2 } \sum \Big[ E( (X_i - \overline{ X })^4 ) - E((X_i - \overline{ X })^2)^2 \Big] \\ & = \frac{ 1 }{ (n-1)^2 } \sum \kappa \sigma^4 - (\sigma^2)^2 \\ & = \frac{ 1 }{ (n-1)^2 } ( n \kappa \sigma^4 - n \sigma^4) \end{align}

## c

Show that under normality, the kurtosis is $3$ and establish that, in this case, the estimator of the form $aS^2$ with the minimum MSE is $\frac{ n-1 }{ n+1 }S^2$. (Lemma 3.6.5 may be helpful)

We can bring the $\sigma^4$ inside the fraction to get

$\kappa = E\Big[ \Big( \frac{ X-\mu }{ \sigma } \Big)^4 \Big].$

Notice that the inside function now follows a $N(0,1)$ distribution. We can also make use of Lemma 3.6.5 which says

$E\Big[ g(x) (x - \mu) \Big] = \sigma^2 E(g'(x)).$

Now we can take $g(Z) = Z^3$. Then,

\begin{align} \kappa & = E[Z^4] \\ & = E[g(Z) (Z - 0)] \\ & = \sigma^2 E[g'(Z)] \\ & = \sigma^2 E[3 Z^2] \\ & = 3 \sigma^2 E[Z^2] \\ & = 3 \sigma^2 (V(Z) + E(Z)^2) \\ & = 3 \cdot 1 ( 1 + 0) \\ & = 3. \end{align}

Now we can use our MSE equation from (a) and our variance equation from (b). We are trying to optimize with respect to $a$.

\begin{align} MSE & = a^2 \frac{ 1 }{ n } \Big( \kappa - \frac{ n-3 }{ n-1 } \Big)\sigma^4 + (a-1)^2 \sigma^4 \\ \frac{ \partial MSE }{\partial a} & = 2 a \frac{ 1 }{ n } \Big( \kappa - \frac{ n-3 }{ n-1 } \Big)\sigma^4 + 2(a-1) \sigma^4 = 0 \\ a & = \frac{ \sigma^4 }{ \frac{ 1 }{ n } \Big( \kappa - \frac{ n-3 }{ n-1 } \Big)\sigma^4 + \sigma^4 } \\ & = \frac{ \sigma^4 }{ \frac{ 1 }{ n } \Big( 3 - \frac{ n-3 }{ n-1 } \Big)\sigma^4 + \sigma^4 } \\ & = \frac{ 1 }{ \frac{ 1 }{ n } \Big( 3 - \frac{ n-3 }{ n-1 } \Big) + 1 } \\ & = \frac{ n-1 }{ n+1 } \end{align}

## d

If normality is not assumed, show that $MSE(aS^2)$ is minimized at

$a = \frac{ n-1 }{ n+1 + \frac{ (\kappa - 3) (n-1) }{ n } }$

which is useless as is depends on a parameter.

From above

\begin{align} a & = \frac{ \sigma^4 }{ \frac{ 1 }{ n } \Big( \kappa - \frac{ n-3 }{ n-1 } \Big)\sigma^4 + \sigma^4 } \\ & = \frac{ 1 }{ \frac{ 1 }{ n } \Big( \kappa - \frac{ n-3 }{ n-1 } \Big) + 1 }\\ & = \frac{ n-1 }{ n+1 + \frac{ (\kappa - 3) (n-1) }{ n } } \end{align}

## e

Show that

### i

for distributions with $\kappa > 3$, the optimal $a$ will satisfy $a < \frac{ n-1 }{ n+1 }$.

Taking $\kappa > 3$ will make the second term in our denominator positive, which will make the denominator larger than $n+1$. So our overall expression will be less than $\frac{ n-1 }{ n+1 }$.

### ii

for distributions with $\kappa < 3$, the optimal $a$ will satisfy $\frac{ n-1 }{ n+1 } < a < 1$.

Similarly, making $\kappa < 3$ will make the second term in our denominator negative, which will make the denominator smaller than $n+1$. So our overall expression will be greater than $\frac{ n-1 }{ n+1 }$.

# 7.48

Suppose that $X_i, \ i = 1, \dots , n$ are iid $Bernoulli(p)$.

## a

Show that the variance of the MLE of $p$ attains the Cramér-Rao Lower Bound.

The MLE is,

\begin{align} L(p | x) & = \prod f(x_i | p) \\ & = \prod p^{x_i} (1-p)^{1-x_i}\\ & = p^{\sum x_i} (1-p)^{1-\sum x_i} \\ \\ \ell(p|x) & = \sum x_i \log(p) + (n - \sum x_i) \log(1-p) \\ \\ \frac{ \partial \ell }{\partial p } & = \sum x_i \frac{ 1 }{ p } + -(n - \sum x_i) \frac{ 1 }{ 1-p } \\ 0 & = \sum x_i \frac{ 1 }{ p } + -(n - \sum x_i) \frac{ 1 }{ 1-p } \\ \\ \sum x_i \frac{ 1 }{ p } & = (n - \sum x_i) \frac{ 1 }{ 1-p } \\ \frac{ 1-p }{ p } & = \frac{ n - \sum x_i }{ \sum x_i } \\ \widehat p & = \overline{ X }. \end{align}

The Cramér-Rao lower bound is,

\begin{align} V_\theta ( W(X)) & \geq \frac{ \Big( \frac{ \partial }{\partial \theta} E_\theta(W(X)) \Big)^2 }{ -E_\theta \Big( \frac{ \partial^2 }{\partial \theta^2} \log(f(x|\theta)) \Big) } \\ -E_\theta \Big( \frac{ \partial^2 }{\partial \theta^2} \log(f(x|\theta)) \Big) & = -E_\theta \Big( \frac{ \partial }{\partial \theta} \sum \frac{ x_i }{ p } - (n - \sum x_i) \frac{ 1 }{ 1-p } \Big) \\ & = -E_\theta \Big( - \frac{ \sum x_i }{ p^2 } + \frac{ -(n-\sum x_i) }{ (1-p)^2 } \Big) \\ & = \frac{ 1 }{ p^2 } \sum E(x_i) + E\Big( \frac{ n = \sum x_i }{ (1-p)^2 } \Big) \\ & = \frac{ n }{ p^2 } p + \frac{ n-np }{ (1-p)^2 } \\ & = \frac{ n }{ p } + \frac{ n(1-p) }{ (1-p)^2 } \\ \\ CRLB & = \frac{ 1 }{ \frac{ n }{ p } + \frac{ n(1-p) }{ (1-p)^2 } } \\ & = \frac{ p(1-p) }{ n }. \end{align}

Now we can find the variance of our estimator.

\begin{align} V(\widehat p) & = V \Big( \frac{ \sum x_i }{ n } \Big) \\ & = \frac{ 1 }{ n^2 } \sum V(x_i) \\ & = \frac{ 1 }{ n^2 } n p (1-p) \\ & = \frac{ p(1-p) }{ n }\\ & = CRLB \end{align}

## b

For $n \geq 4$, show that the product $X_1 X_2 X_3 X_4$ is an unbiased estimator of $p^4$. and use this fact to find the best unbiased estimator of $p^4$.

\begin{align} E\Big[ X_1 X_2 X_3 X_4 \Big] & = E(X_1)E(X_2)E(X_3)E(X_4) & \text{independence} \\ & = p^4. \end{align}

Also $\sum_{i=1}^{n} x_i$ is sufficient for $p$ by the Factorization Theorem.

\begin{align} f(X | p) & = \prod p^{x_i} (1-p)^{1-x_i} \\ & = p^{\sum x_i} (1-p)^{1-\sum x_i} \end{align}

We can use the Rao-Blackwell theorem to find a UMVUE.

\begin{align} \phi(T) & = E\Big( X_1 X_2 X_3 X_4 | \sum_{i=1}^{n} X_i = t \Big) \\ & = \frac{ P(X_1, X_2, X_3, X_4 = 1 \land \sum_{i=1}^{n} X_i = t) }{ P(\sum_{i=1}^{n} X_i = t) } \\ & = \frac{ P(X_1, X_2, X_3, X_4 = 1 \land \sum_{i=5}^{n} X_i = t-4) }{ P(\sum_{i=1}^{n} X_i = t) } \\ & = \frac{ P(X_1, X_2, X_3, X_4 = 1) P(\sum_{i=5}^{n} X_i = t-4) }{ P(\sum_{i=1}^{n} X_i = t) } & \text{independence} \\ & = \frac{ p^4 \cdot {n-4 \choose t-4} p^{t-4} (1-p)^{n-4-(t-4)} }{ {n \choose t} p^t (1-p)^{n-t} } \\ & = \frac{ {n-4 \choose t-4} }{ {n \choose t} } \end{align}

# 7.49

Let $X_1, \dots , X_n$ be iid $exponential(\lambda)$.

## a

Find an unbiased estimator of $\lambda$ based only on $Y = \min {X_1, \dots , X_n }$.

\begin{align} F_Y & = P(X < X_i) \\ & = [1 - F(X)]^n \\ \\ f_y & = n f(x) (1- F(X))^{n-1} \\ & = n \frac{ 1 }{ \lambda } e^{-\frac{ 1 }{ \lambda } x} \Big( 1 - (1 - e^{\frac{ -1 }{ \lambda } x}) \Big)^{n-1} \\ & = n \frac{ 1 }{ \lambda } e^{- n \frac{ 1 }{ \lambda } x}. \end{align}

So $Y\sim Exp(\lambda / b)$. So $nY$ is unbiased.

$E(nY) = n E(Y) = n \lambda/n = \lambda$

## b

Find a better estimator than the one in part (a). Prove that it is better.

Since it’s an exponential family, $\sum x_i$ is sufficient and complete. Notice that $\overline{ X }$ is a function of $\sum x_i$ and is unbiased.

$E(\overline{ X }) = \frac{ 1 }{ n } \sum E(x_i) = \frac{ 1 }{ n } n \lambda = \lambda$

By Lehmann-Scheffe, $\overline{ X }$ is UMVUE, so it is better than the estimator in (a).

## c

The following data are high-stress failure times (in hours) of Kevlar/epoxy spherical vessels used in a sustained pressure environment on the space shuttle:

$50.1, 70.1, 137.0, 166.9, 170.5, 152.8, 80.5, 123.5, 112.6, 148.5, 160.0, 125.4.$

Failure times are often modeled with the exponential distribution. Estimate the mean failure time using the estimators from parts (a) and (b).

From (a) $\widehat \lambda = 50.1 \cdot 12 = 601.2$. For (b) we can take the average $\widehat \lambda = 124.825$.

# 4

Consider a random sample $X_1, X_2, \dots X_n$ from the density $f(x| \theta) = \exp\Big[ -(x-\theta) \Big]$, $\theta < x < \infty$, and 0 elsewhere. Show that the first order statistic $X_{(1)} = \min X_i$ is a complete sufficient statistic for $\theta$, and find the UMVUE of $\theta$.

\begin{align} f(X | \theta) & = \prod e^{-(x-\theta)} I[\theta < x< \infty] \\ & = e^{-\sum x_i - n \theta} I[\theta < X_{(1)}] \\ & = e^{- \sum x_i} e^{n \theta} I[\theta < X_{(1)}] \end{align}

By the factorization theorem $X_{(1)}$ is sufficient. Now we can show that it is complete. First, we need its distribution.

\begin{align} f_{X_{(1)}} & = n \cdot f_x ( 1- F_X)^{n-1} \\ & = n e^{-(x-\theta)} \Big[ 1 - \int_{\theta}^{x} e^{-(t - \theta)} dt \Big]^{n-1} \\ & = n e^{-(x-\theta)} \Big[ 1 - (1 - e^{-x + \theta}) \Big]^{n-1} \\ & = n e^{-(x-\theta)} \Big[ e^{-x + \theta} \Big]^{n-1} \\ & = n e^{n(-x + \theta)} \end{align}

Now we can show completeness.

\begin{align} \int_{\theta}^{\infty} g(x) n e^{n(\theta - t)}dx & = n e^{n \theta} \int_{\theta}^{\infty} g(x) e^{- nt}dx \end{align}

This is only 0 if $g(x)$ is 0. Now we can look at the expectation of the minimum order statistic.

\begin{align} E(X_{(1)}) & = \int_{\theta}^{\infty} x n e^{(-x + \theta)n} dx \\ & = \frac{ n (1+n\theta) }{ n^2 } \\ & = \frac{ 1 }{ n } + \theta \end{align}

So $X_{(1)} - \frac{ 1 }{ n }$ in unbiased.