I saw one question in which the sample mean was estimated as follows (I don't know why they divided by $n-1$ instead of $n$ here for estimation), $$ \widehat{\mu} = \frac{\sum_{i=1}^n x_i}{n-1} $$ where $n$ is the sample size.
Now if I take the expected value of sample mean, I get

$$ E[\widehat{\mu}] = \frac{\sum_{i=1}^n E[x_i]}{n-1} = \frac{n}{n-1}\mu $$ where $\mu$ is true mean of the population.
Now what the people around are saying is : estimated mean is not equal to true mean; hence the estimator is biased. But I have different views. Isn't the meaning of bias is additional term in estimation? For example, if I'm estimating $\widehat\mu = \mu + b $, then constant $b$ is called bias. Here, our estimator does not have such additional term, so I argue that this estimator is unbiased.

I know the definition that estimator is unbiased only if it is equal to true mean, which is not the case here, but they are proportional here, so, shouldn't the estimator be unbiased here? I mean, if we change the true mean, the sample mean will also change proportionally. There is no bias here. In short, the bias in my head means, the additional term (some people also call it error), in the estimation. I think that if estimated mean is proportional to true mean, then it should be unbiased estimator, but people around me are restricting this further, by saying that only if they both are exactly equal.

  • 2
    $\begingroup$ You can define bias as you like and call it the Mike Patel definition, but that only goes so far. Bias can be much more than just being off by an identifiable constant. The bias can vary with sample size, as here, and indeed with other controls. Equating the kind of bias in your example to lack of bias can't help any discussion or your own understanding of standard principles. $\endgroup$
    – Nick Cox
    Commented Mar 16, 2020 at 17:09

1 Answer 1


An estimator $\hat{\theta}$ of $\theta$ (based on any observed data $x$) is biased if its expected value is different from $\theta$.

In your example $E[\hat{\mu}] = \frac{n}{n-1}\mu = \mu + \frac{1}{n-1}\mu$, hence the bias is $\frac{1}{n-1}\mu$.


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