Properties of Estimators

Let $\hat \theta$ be an estimator of $\theta$ :

1. $Bias(\hat \theta) = E(\hat \theta)-\theta$

• We say $\hat \theta$ is unbiased for estimation $\theta$ if the bias is 0
2. $Var(\hat \theta) = E[(\hat \theta-E(\hat \theta))^2]$

3. Mean-Squared Error

• $MSE(\hat \theta) = E[(\hat \theta - \theta)^2]$

• Note: $MSE(\hat \theta) = Var(\hat \theta)\Leftrightarrow Bias(\hat \theta)=0$

$$\begin{split} MSE(\hat \theta)&=E[(\hat \theta - \theta)^2] \\ &=E[(\hat \theta - E(\hat \theta)+E(\hat \theta)-\theta)^2]\\ &=E[(\hat \theta-E(\hat \theta))^2]+E[2(\hat \theta-E(\hat \theta))\underbrace{(E(\hat \theta)-\hat \theta)}_{\text{constant}}]+E[(E(\hat \theta)-\hat \theta)^2] \\ & = E[(\hat \theta-E(\hat \theta))^2]+2(E(\hat \theta)-\hat \theta))\underbrace{E(\hat \theta-E(\hat \theta))}_{\underbrace{E(\hat \theta)-E(E(\hat \theta))}_{\text{zero}}}]+\underbrace{E[(E(\hat \theta)-\hat \theta)^2]}_{Bias(\hat \theta)}\\ &=Var(\hat \theta) + Bias(\hat \theta)^2 \end{split}$$

1. Consistency

• Let $\hat \theta_n$ be estimator based on sample size $n$, we say that estimator is consistent if $\lim_{n \to \infty}Bias(\hat \theta_n)=0$ and $\lim_{n \to \infty}Var(\hat \theta_n)=0$.

Ex 1. $x_1, x_2, \dots, x_n \overset{\text{i.i.d}}\sim N(\mu, \sigma^2)$.
recall $MLE$ : $\hat \mu = \bar x$, $\hat {\sigma^2} = \frac{n-1}{n}S^2$

To get $\hat \mu$:
$$E(\hat\mu)=E(\bar x)=\mu$$
so $\hat \mu$ is an unbiased estimator of $\mu$.
$$Var(\hat \mu)= Var(\bar x)=\frac{\sigma^2}{n}$$
so $$MSE(\hat \mu)=Var(\hat \mu)=\frac{\sigma^2}{n}$$
and it is consistent since $Bias=0$, $\lim_{n \to \infty}\frac{\sigma^2}{n}=0$

To get $\hat \sigma^2$:

$$\begin{split} E(\hat \sigma^2)&=E(\frac{n-1}{n}S^2) \\ &=\frac{n-1}{n}E(S^2)\\ &=\frac{n-1}{n}\sigma^2 \end{split}$$

where $E(S^2) = \sigma^2$ in general.

$$\begin{split} Bias(\hat \sigma^2)&=E(\hat \sigma^2)-\sigma^2\\ &=-\frac{\sigma^2}{n} \end{split}$$

$$\begin{split} Var(\hat \sigma^2)&=Var(\frac{n-1}{n}S^2) \\ &=Var(\frac{\sigma^2}{n}\cdot \frac{n-1}{\sigma^2}S^2) \\ &=\frac{\sigma^4}{n^2}\cdot 2(n-1) \end{split}$$

where $\frac{(n-1)S^2}{\sigma^2}\sim \chi_{n-1}^2$ and $Var(\chi_{n-1}^2)=2(n-1)$

$$MSE(\hat \sigma^2)=\frac{\sigma^4}{n^2}\cdot 2(n-1) + \underbrace{\frac{\sigma^4}{n^2}}_{Bias^2}$$

$\Rightarrow \hat \sigma^2$ is a constant estimator of $\sigma^2$ since $\lim_{n \to \infty}-\frac{\sigma^2}{n}=0$ and $\lim_{n \to \infty}\frac{2\sigma^4(n-1)}{n^2}=0$.

We can construct a function to eliminate the Bias of MLE, e.g. the estimator $\hat \sigma_{\text{unbiased}}^2=S^2$ would be unbiased for estimating $\sigma^2$, however, if we reduce the bias, the variance will increase and vice versa.

Ex 2. $x_1, x_2, \dots, x_n \overset{\text{i.i,d}}\sim f(x|\theta)=\frac{1}{\theta},\ 0<x<\theta$
Recall MLE: $\hat \theta = max\lbrace x_i \rbrace$
a) Derive CDF/PDF of $U = max\lbrace x_i \rbrace$
Bounds: $0<\mu<\theta$

$$\begin{split} p(U\leq \mu)&=p(max\lbrace x_i \rbrace \leq \mu) \\ &=p(x_1 \leq \mu, x_2 \leq \mu,\dots, x_n \leq \mu) \\ &=p(x_1\leq \mu)p(x_2 \leq \mu)\dots p(x_n\leq \mu) \\ &=[p(x_i \leq \mu)]^n \\ &=[\int_0^\mu\frac{1}{\theta}dx]^n \\ &=(\frac{\mu}{\theta})^n \end{split}$$
So $U$ has PDF, $f(U)=\frac{d}{d\mu}[(\frac{\mu}{\theta})^n]= \frac{n}{\theta^n}\mu^{n-1}$, where $0 < \mu< 0$

b) Bias, Variance, Consistency of $\hat \theta$
$$\begin{split} E(\hat \theta) = E(U) & = \int_0^\theta \mu \frac{n}{\theta^n}\mu^{n-1}d\mu \\ &=\frac{n}{\theta^n}\cdot \frac{\mu^{n+1}}{n+1}|_0^\theta \\ &=\frac{n}{n+1}\theta \end{split}$$

$$\begin{split} Bias(\hat \theta)&=\frac{n}{n+1}\theta-\theta \\ &= -\frac{\theta}{n+1} \end{split}$$

$$\begin{split} E(U^2)& = \int_0^\theta \mu^2 \frac{n}{\theta^n}\mu^{n-1}d\mu\\ &=\frac{n}{\theta^n}\frac{\mu^{n+1}}{n+2}|_0^\theta \\ &=\frac{n}{n+2}\theta^2 \end{split}$$

$$\begin{split} Var(\hat \theta)& = E(U^2)-E(U)^2 \\ &=\frac{n}{n+2}\theta^2 - (\frac{n}{n+2}\theta)^2 \\ &=\theta^2\frac{[n(n+1)^2-n^2(n+2)]}{(n+2)(n+1)^2} \\ &=\frac{n}{(n+2)(n+1)^2}\theta^2 \end{split}$$, where $n(n+1)^2-n^2(n+2) = n^3+2n^2+n-(n^3+2n^2)$.

$$MSE(\hat \theta) = Var(\hat \theta)+[Bias(\hat \theta)]^2$$
$\hat \theta$ is consistent since $\lim_{n \to \infty}-\frac{\theta}{n+1}=0$, $\lim_{n \to \infty}-\frac{n}{(n+1)^2(n+2)}\theta^2=0$

• We can construct function to eliminate bias, $\hat \theta_1=\frac{n+1}{n}max\lbrace x_i \rbrace$ is unbiased
• For two unbiased estimators, it makes sense to compare their variances for the sample size $n$ (which estimator goves more precise results for the same amount of data?)

Definition: The relative efficiency of two unbiased estimators $\hat \theta_1$, and $\hat \theta_2$, is $\frac{\text{variance}(\hat \theta_2)}{\text{variance}(\hat \theta_1)}$.

• $\hat \theta_1$ is more efficient than $\hat \theta_2$ if $Var(\hat \theta_1)< Var(\hat \theta_2)$
• $\hat \theta_2$ is more efficient than $\hat \theta_1$ if $Var(\hat \theta_2)< Var(\hat \theta_1)$

Ex. $x_1, x_2, \dots, x_n \overset{\text{i.i.d}}\sim \text{uniform}(0, \theta)$, i.e. $f(x|\theta)=\frac{1}{\theta}$ for $0<x<\theta$
$\hat \theta_1 = \frac{n+1}{n}max \lbrace x_i \rbrace$ is unbiased estimator of $\theta$.
$\hat \theta_2 = 2 \bar x$ is also unbiased since $E(x_i)=\frac{\theta}{2}$, $Var(x_i)=\frac{\theta^2}{12}$.
probability of uniform distribution:
$E(\bar x)=\frac{\theta}{2}, Var(\bar x)=\frac{\theta^2}{12}$
And so $E(\hat \theta_2)=E(2\bar x)=2E(\bar x)=2\cdot \frac{\theta}{2}=\theta$, which is unbiased.
if $\hat \theta_1$ or $\hat \theta_2$ more efficient?

$$\begin{split} Var(\hat \theta_1) &= (\frac{n+1}{n})^2 Var(max \lbrace x_i \rbrace) \\ &=(\frac{n+1}{n})^2\frac{n}{(n+1)^2(n+2)}\theta^2 \\ &=\frac{1}{n(n+2)}\theta^2 \end{split}$$

$$\begin{split} Var(\hat \theta_2) &= 2^2 Var(\bar x) \\ &=4 \frac{\theta^2}{12n} \\ &=\frac{\theta^2}{3n} \end{split}$$

$$\begin{split} \frac{Var(\hat \theta_2)}{Var(\hat \theta_1)} &= \frac{\frac{\theta^2}{3n}}{\frac{\theta^2}{n(n+2)}} \\ & = \frac{n+2}{3} \end{split}$$

• if $n>1$, then $\hat \theta_1$ is more efficient than $\hat \theta_2$
• if $n=1$, then equal efficiency
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