Betting on a sample from a known distribution This was an interview question. Given a known distribution, sample a value from it with replacement for many times. Two people A and B bet on the sample with their own guesses, and the one closer to the actual sample value wins. What's your betting strategy?
The interviewer claimed the median value is better than mean. I didn't get it. Can someone explain? Thanks.
 A: The mean is affected by outliers in the distribution, so does not inform us about the mid-point of the distribution. The median is not affected by outliers.
Consider three example distributions and see how the presence of one outlier affects the mean but not the median:




Example
Distribution
Median
Mean




1
[1, 2, 3, 4, 5]
3
3


2
[1, 2, 3, 4, 10]
3
4


3
[1, 2, 3, 4, 20]
3
6




So relating these examples back to your interview question: assume A picks the median, so 50% of the time the sample will be less than A's guess and 50% of the time greater. Therefore, if B picks a different value from A, A's guess must be closer to the sample value at least 50% of the time.
Now assume B picks the mean. In the first case, they pick the same number, so the outcome is a draw. In the second case, A wins 60% of the time, B wins 20% of the time and it's a draw 20% of the time. In the third case, A wins 80% of the time and B wins 20% of the time.
A: The median minimizes expected absolute distance between observations sampled from the distribution and a single value. That is:
$$
\underset{m}{\text{min}}\text{ }\mathbb E\left[
\left\vert X - m\right\vert
\right] = \text{median}
$$
This is in contrast to the mean, which minimizes expected squared distance.
$$
\underset{m}{\text{min}}\text{ }\mathbb E\left[
\left( X - m\right)^2
\right] = \text{mean}
$$
I think this is what the interviewer wanted to test if you knew.
