How do I find the bias of an estimator?
My notes say
$$B(\hat\theta) = E(\hat\theta) - \theta $$
And I understand that the bias is the difference between a parameter and the expectation of its estimator. What I don't understand is how to calulate the bias given only an estimator? My notes lack ANY examples of calculating the bias, so even if anyone could please give me an example I could understand it better!
$\endgroup$ 42 Answers
$\begingroup$The concept of bias is related to sampling distribution of the statistic. Consider, for example, a random sample $X_{1},X_{2},\cdots X_{n}$ from $N(\mu, \sigma^{2})$. Then, it is easy to observe that, the sampling distribution of the sample mean $\bar{X}$ is $N(\mu,\frac{1}{n}\sigma^{2})$. we note that, $E(\bar{X})=\mu$. That is, the center of the sampling distribution of $\bar{X}$ is also $\mu$. Now consider, the statistics, \begin{equation*} S_{1}^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2},\qquad\qquad S_{2}^{2}=\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2} \end{equation*} as estimators of the parameter $\sigma^{2}$. It can be shown that \begin{equation*} E(S_{1}^{2})=\sigma^{2} \mbox{ and } E(S_{2}^{2})=\frac{n-1}{n} \sigma^{2} \end{equation*} The sampling distribution of $S_{1}^{2}$ is centered at $\sigma^{2}$, where as that of $S_{2}^{2}$ is not. We say that, the estimator $S_{2}^{2}$ is a biased estimator for $\sigma^{2}$. Now using the definition of bias, we get the amount of bias in $S_{2}^{2}$ in estimating $\sigma^{2}$.
$\endgroup$ 1 $\begingroup$Roughly speaking there are two favorable attributes for an estimator $T$ of a parameter $\tau$, accuracy and precision. Accuracy is lack of bias and precision is small variance. If an estimator is unbiased, then we just look at its variance. If it is biased we sometimes look at 'mean squared error', which is $$MSE_\tau = E[(T - \tau)^2] = B^2(T) + Var(T).$$
As an example, consider data $X_1, X_2, \dots, X_n \stackrel{iid}{\sim} UNIF(0, \tau).$ The estimator $T_1 = 2\bar X$ is unbiased, and the estimator $T_2 = X_{(n)} = \max(X_i)$ is biased because $E(T_2) = \frac{n}{n+1}\tau.$
As a substitute for a (fairly easy) analytical proof, here is a simulation to show that $T_2$ is 'better' in the sense that its MSE is smaller. We look at a million samples of size $n = 5$ from $UNIF(0, \tau = 1).$
m = 10^6; n = 5; tau = 1
x = runif(m*n, 0, 1)
DTA = matrix(x, nrow=m) # each row a sample of n
t1 = 2*rowMeans(DTA); t2 = apply(DTA, 1, max)
mean(t1); mean(t2)
## 0.9997444 # aprx E(T1) = 1 unbiased
## 0.8332033 # aprx E(T2) = 5/6 biased
n/(n+1)
## 0.8333333
var(t1); var(t2)
## 0.06665655 # aprx Var(T1)
## 0.01983109 # aprx Var(T2) < Var(T1)
mse.t1 = mean((t1-tau)^2); mse.t2 = mean((t2-tau)^2)
mse.t1; mse.t2
## 0.06665655 # aprx MSE(T1)
## 0.04765219 # aprx MSE(T2) < MSE(T1)We see that the smaller variance of $T_2$ is enough to overcome its bias to give it the smaller MSE. It is possible to 'unbias' $T_2$ by multiplying by $(n+1)/n$ to get $T_3 = \frac{6}{5}T_2,$ which is unbiased and still has smaller variance than $T_1:$ $Var(T_3) \approx 0.029 < Var(T_1) \approx 0.067.$ The simulated distributions of the three estimators are shown in the figure below.
