I believe a continuous uniform distribution over an interval $[a,b]$ to be the probability distribution describing an equally likely chance for any outcome on $[a,b]$. Other than by mathematical definition, how can I see that the avg/mean/expected value is $\frac{a+b}{2}$? Isn't the mean an indicator of likelihood? Or is it just that 'expected value' is poor terminology for the uniform distribution?

For example: if I had a probability distribution shaped like an upward parabola (over a relatively large interval, where the vertex was a the center of the distribution, then the mean would be dead center of the interval but a very inaccurate indicator of what value you could expect to see during an experiment. Is this right?

I'm looking for some intuition here, because it doesn't make sense to me that some process can be completely random yet have an average value, especially when I think of an extremely large $|b-a|$.

  • 3
    $\begingroup$ mean is not likelihood. Think about a discrete distribution when $X$ is equally likely to take value of 0 and 1. The mean of $X$ is 0.5, even though $X$ can never be 0.5. $\endgroup$
    – Guangliang
    Mar 3 '17 at 22:01
  • $\begingroup$ The intuitive answer is that if the distribution is symmetric then the mean must be exactly in the middle. This does not have to be true for the mode $\endgroup$
    – Henry
    Mar 3 '17 at 22:54

Let's take the example of rolling a die. This has a uniform distribution over the discrete set $\{1,2,3,4,5,6\}$. The mean then is the average value (not the most frequent value) you expect to get over many rolls:

$$\frac{1+2+3+4+5+6}{6} = \frac{21}{6} = \frac{7}{2} = 3.5$$

You never roll a $3.5$, but it is the "expected value".

Moving to the analogous continuous case over the interval $[1,6]$, we have the mean as

$$\frac{1+6}{2} = \frac{7}{2} = 3.5$$

  • $\begingroup$ So, is "expected value" poor terminology? $\endgroup$
    – Zduff
    Mar 3 '17 at 22:05
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    $\begingroup$ @Zduff Perhaps, depending on what "expected" means to you. If I let you roll a die a million times and gave you a dollar for each dot that appeared on the top of the die after each roll, you would "expect" to earn around $3.5$ million dollars. This is the statistical sense of expected value. $\endgroup$ Mar 3 '17 at 22:11
  • $\begingroup$ Okay, I see. I was taking it to mean, "expected per individual event". It's more just telling you where the center of mass is of the distribution. Is that right? @Alexis Olson $\endgroup$
    – Zduff
    Mar 3 '17 at 22:16
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    $\begingroup$ @Zduff Yes. The most frequently occurring event has a different name. It is called the "mode". $\endgroup$ Mar 3 '17 at 22:17
  • $\begingroup$ But there is no mode for continuous distributions, right? And there are events under normal distributions that occur more frequently? Or is that just an error in thought I get by applying the smooth normal distribution to (in reality) a discrete population? $\endgroup$
    – Zduff
    Mar 3 '17 at 22:20

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