I need to check if an estimator $\hat\theta$ for the parameter $\theta$ is biased. Theory says I should compare the expected value of $\hat\theta$ versus the expected value of $\theta$.

I assume the expected value of an estimator is the "weighted average" of the estimator and its distribution: $E[\hat\theta] = \int_0^\inf \hat\theta f(\hat\theta) d\hat\theta$*. If I'm right, to compute $E[\hat\theta]$ I need to know how $\hat\theta$ is distributed.

For example:

$X$ is a random var with support $0 <= X <= \theta$ and pdf $f(x;\theta) = 3x^2 / \theta^3$. Check if $\delta(x)=(4\bar X)/3$ is biased.

*is it $E[\hat\theta] = \int_0^\inf \hat\theta f(\hat\theta) d\hat\theta$ or $E[\hat\theta] = \int_0^\inf \hat\theta f(x) dX$?

  • 1
    $\begingroup$ Something strange happens at the beginning of the second paragraph: whereas the first paragraph refers to the expected value of an estimator $\hat\theta,$ the second paragraph refers to the expected value of a parameter (which in the theory of unbiased estimators makes no sense: a parameter is a constant). That may be why you're stuck. As far as the "just in case" part goes, what is $\mu$? $\endgroup$
    – whuber
    Nov 6, 2013 at 4:07
  • $\begingroup$ You appear to be working in a frequentist framework, in which case parameters are constants. It's the distribution of the random variable that you have to worry about in order to compute the bias, and your example specifies that. You compute $E(\hat \theta)$ ($\hat \theta$ is a function of data) and compare that with $\theta$. Where did $\mu$ come from? the question is about estimating $\theta$. $\endgroup$
    – Glen_b
    Nov 6, 2013 at 5:03
  • $\begingroup$ Since this appears to be standard bookwork, you should take a read through the self-study tag wiki and add the tag. $\endgroup$
    – Glen_b
    Nov 6, 2013 at 5:05
  • $\begingroup$ @whuber seems I'm confusing $\theta$ and $\hat\theta$ =/ Just to be clear, I need to compute both $E[\theta]$ and $E[\hat\theta]$: a) the first one is a constant $E[\theta] = \theta$; b) the second one is the "weighted average" $E[\hat\theta] = \int_0^\hat\theta \hat\theta (4\bar X)/3 \space dX$. Confirmed? $\endgroup$
    – mugen
    Nov 6, 2013 at 13:53
  • $\begingroup$ @Glen_b please, see the comment above (I can't tag more than one person per comment), thank you :) $\endgroup$
    – mugen
    Nov 6, 2013 at 13:54

1 Answer 1


You seem to have some conceptual issues.

In the classical non-bayesian context (the fact that your are learning about bias, and your working example, suggest that this is your context) the parameter $\theta$ is ... a parameter, a number; which is perhaps unknown to us but which takes nonetheless some determined fixed value. In short: $\theta$ is not a random variable.

The estimator, instead, is (in general) a random variable. Because $\hat{\theta}=g(\{X\})$ where $g(\cdot)$ is some function and $\{X\}$ is a list of realizations ($X_1,X_2 \cdots.. X_n$) of a random variable. (Think for example, of the sample average $(X_1+X_2+\cdots + X_n)/n$) This is to say: in different "experiments" (trials) we'll get different values of the estimator $\hat{\theta}$ . But in all experiments the parameter $\theta$ will be the same.

That's why it makes sense to ask if $E(\hat{\theta})=\theta$ (because the left side is the expectation of a random variable, the right side is a constant). And, if the equation is valid (it might or not be, according to the estimator) the estimator is unbiased.

In your example, you're using $\hat{\theta} = \frac{X_1+X_2+ \cdots + X_n}{n}\frac{4}{3}$. The expectation of this is $E(\hat{\theta} )= \frac{n E(X)}{n} \frac{4}{3}$ Now, we need to compute $E(X)$ (it will be a function of $\theta$) and check if that gives $\theta$.

  • $\begingroup$ Thank you! You showed I should operate algebraically from $ E(\hat\theta)$ until I get $ E(X) $, and then compute $ E(X) $ which will eventually be a function of $ \theta $. Once I understood that, I could move on with these exercises. $\endgroup$
    – mugen
    Nov 23, 2013 at 20:41

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