# Use of mean over median [duplicate]

Although this topic has been discussed on the thread numerous times, I have yet to read a convincing argument for why the mean is favoured over the median as a measure of central tendency. This is particularly prevalent in financial economics and for decision making under uncertainty, where the expected value and the expectations operator are widely used.

Is it due to efficiency? Does the mean have favourable properties? Or is it perhaps just used for computational convenience?

I may be mixing several things up, in asking this question, so if you could clarify any misunderstandings, I would be very grateful.

• I posted one likely duplicate. See also this question and this question and this question and other questions tagged both "mean" and "median". Jul 26, 2017 at 9:28
• @DJohnson: (1) Makes no sense to talk of a statistic's being sufficient in general - it's sufficient (or not) for inference about an unknown parameter indexing a family of distributions. The second example of a sufficient statistic in Cox & Hinkley, Ch. 2.2, p19, is one where the sample maximum is sufficient, not the mean. (2) I don't know what's meant by using "all of the information available from the data" to calculate the mean, but I'm quite sure you can't calculate the median from just the sample minimum & maximum - consider the samples $\{1,2,4\}$ & $\{1,3,4\}$. Oct 8, 2017 at 21:38
• (3) Interestingly, it seems the median is never a sufficient statistic - see When if ever is a median statistic a sufficient statistic?. That's not the point though, its utility lies in its robustness to contamination or mis-specification of the distribution (C.& H., Ch. 9.4). Oct 8, 2017 at 21:40
• @DJohnson: C. &H. do discuss sufficiency of the mean - their first example is one where the mean is sufficient. (Similarly the web page you link to gives two examples, one where the mean is sufficient & one where it isn't.) Sure, when you're confident that a specified family of distributions indexed by the parameter $\theta$ is an adequate model for your data (so robustness isn't an issue), and a sufficient estimator for $\theta$ exists, and that sufficient estimator happens to be the mean; then it's hard to think of any reason to use the median to estimate $\theta$ - but ... Oct 19, 2017 at 21:42
• ... in that case no-one argues you should. Oct 19, 2017 at 21:42