Let, *x*1 < = *x*2 < = *x*3....... < = *xn*

and

*p*1 + *p*2 + *p*3 + ....... + *pn* = 1

We all know that average of *x*1, *x*2, *x*3......., *xn* is in [x1,xn] and it is easy to understand.

In a contest, I assumed Expected value = *p*1 * *x*1 + *p*2 * *x*2 + *p*3 * *x*3 + ....... + *pn* * *xn* is in [x1,xn] regardless how probability is distributed that means the sum of probability can be 1 in many different ways.

My assumption was right and got ac. I'm interested to know the proof.

**TIA**

The sum will be minimal if

p_{1}= 1, because otherwise we can makep_{1}= 1 with moves which don't increase the value of the sum. Take some non-0p_{i}(i> 1) and addp_{i}top_{1}and makep_{i}0. This way the sum will change by -p_{i}x_{i}+p_{i}x_{1}. Becausex_{1}< =x_{i}this move will not increase the sum, the minimum is ifp_{1}= 1, but then clearly the sum is at leastx_{1}. Same proof can be given for the fact thatx_{n}is the maximum.Same thing, but a bit more intuitive for me so I'll share it:

p_{1}x_{1}+p_{2}x_{2}+ ... +p_{n}x_{n}≥p_{1}x_{1}+p_{2}x_{1}+ ... +p_{n}x_{1}=x_{1}, since for eachi,x_{i}≥x_{1}. Similarly,p_{1}x_{1}+p_{2}x_{2}+ ... +p_{n}x_{n}≤p_{1}x_{n}+p_{2}x_{n}+ ... +p_{n}x_{n}=x_{n}, so you have both bounds.Many thanks.Got it.

A small thing about your notation, what you are writing doesn't really make sense, but I think I understand what you are asking about.

There is a simple argument why the expected value is in [

x_{1},x_{n}]. Simply note the followingx_{1}=p_{1}x_{1}+p_{2}x_{1}+ ... +p_{n}x_{1}≤p_{1}x_{1}+p_{2}x_{2}+ ... +p_{n}x_{n}≤p_{1}x_{n}+p_{2}x_{n}+ ... +p_{n}x_{n}=x_{n}.Thanks for help.