For a continuous random variable x with CDF F(x), the expected value is x0, can I say F(x0) = 0.5?
If not, does the value x1 that makes CDF(x1) = 0.5 mean anything special?
1
-
1$\begingroup$ It's called the median, and median need not be the mean!! $\endgroup$– P.WindridgeMay 3, 2017 at 8:39
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
does the value x1 that makes CDF(x1) = 0.5 mean anything special?
Any value for which it's true is a median; for distributions that aren't continuous (such as discrete distributions) there may be no value at which you get exact equality. Consider the uppermost face on a toss of a fair six-sided die with faces numbered 1,2,3,1,2,3 $-$ the cdf of that variable is never exactly $\frac12$. More generally if there's no value at which it is $\frac12$ you find the first value for which the cdf exceeds $\frac12$.
See wikipedia on the median
For a continuous random variable x with CDF F(x), the expected value is x0, can I say F(x0) = 0.5?
Not in general. It's sometimes true that the expected value and the median are the same but typically it's not the case.
A sufficient condition is that the distribution is symmetric and the mean exists. Symmetry is not a necessary condition though.
