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I had seen two types of bias function:

$$BIAS = \frac{y-y'}{n}$$

$$BIAS = \frac{y'-\bar{y}}{n}$$

  • $y$ = actual
  • $y'$ = predicted
  • $n$ = no. of data

Which one is the correct way to calculate BIAS? Can anyone help explain to me?

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  • $\begingroup$ What is "$\bar y$"? $\endgroup$ – whuber Jan 18 '17 at 14:24
  • $\begingroup$ @whuber mean of actual $\endgroup$ – bbadyalina Jan 18 '17 at 14:32
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In general, bias is (or more precisely is defined by me as) the dfference

observed value $-$ true value.

I won't define "observed", but often the words "measured" or "estimated" would fit in the same place as well or better.

Here "true" could mean known in some absolute sense. For example, we "know" the constant $\pi = 3.14159...$ to many, many decimal places, but there are contexts in which it might amuse us to estimate it from simulations, so several observed or estimated values of $\pi$ and one true value are easy to imagine.

Perhaps more frequently, "true" could just be a polite or honorific name for our best measurement or estimate of something. Some fields talk about "gold standard" measurements, which here has nothing to do with economics or finance but just means using the best available technology, usually of some physical or chemical or biological quantity, such as the concentration of an element in a substance.

There are many different ways to make that broad idea more concrete. That definition allows bias to be simply the difference between two numbers. If someone said that my age is 42, I can tell you that answer is biased, and the bias is then just the difference between 42 and my real age.

In many more serious contexts, estimating or calculating bias includes taking an average over either observed or "true" values.

Both those definitions could make sense, except that I'd expect averaging to imply summation first! Without context, we can't say which might be right for any particular problem.

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  • $\begingroup$ Just asking, IF RMSE is low and the bias is high what is that mean? $\endgroup$ – bbadyalina Jan 18 '17 at 14:34
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    $\begingroup$ It means what it says: I know that will seem facetious but I don't know what else to say. It is like saying if height is short and weight is large, what does that mean. A short heavy person is the answer. $\endgroup$ – Nick Cox Jan 18 '17 at 14:45
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A more formal answer is that the bias is the difference between the expectation value of the estimator and the true value, for example: $$b = E[\hat{M}] - \mu$$

where the estimator $\hat{M}$ is a random variable (a function of the sample) and $\mu$ is the true value. One normally chooses this function so that its variance and bias is small (sometimes there is a trade-off between the two). As a simple example, if $\mu$ represents the population mean, and $\hat{M}$ is chosen to be the sample mean, then the bias will be zero.

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  • $\begingroup$ Is it different for bias in prediction? where bias in prediction using the predicted minus the observed data? $\endgroup$ – bbadyalina Jan 18 '17 at 15:06
  • $\begingroup$ The second form of "bias" in the question is not of this sort, because it does not involve any parameters at all: it's a linear combination of two statistics. Thus, this reply doesn't seem to answer the question. $\endgroup$ – whuber Jan 18 '17 at 15:20
  • $\begingroup$ Contrary to the other answer, bias is NOT "observed - true". Bias is a property of the estimator, not a property of the sample. As an example: you use a mis-calibrated thermometer to estimate the temperature, on average the thermometer reports a value that is 2 degrees higher than the correct temperature. There are other factors outside of your control that cause the temperature reading to not be identical even though the temperature has not changed (you might call that noise). Suppose the true temp is 40 degrees, and your thermometer reports 41. The bias is 2 degrees! $\endgroup$ – Dean Jan 19 '17 at 22:02
  • $\begingroup$ Because the question concerns differences between statistics (and does not involve either parameters or expectations), the standard definition of "bias" of an estimator that you use here--although correct--is not applicable. $\endgroup$ – whuber Jan 21 '17 at 22:52
  • $\begingroup$ You dispute my answer here but not there. As my answer indicates, it's consistent with averaging over any concrete or imaginary set of replications. I think it's dogmatic to imply, as I think you are doing, that you can't talk about a sample's bias if a sample differs from a true value. The idea of bias is older and bigger than any formalisation in statistical inference in terms of properties of estimators; otherwise we could hardly talk about biased measurements. $\endgroup$ – Nick Cox Jan 22 '17 at 16:22

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