Rajan_sust's blog

By Rajan_sust, history, 12 months ago, , Let, x1 <  = x2 <  = x3....... <  = xn

and

p1 + p2 + p3 + ....... + pn = 1

We all know that average of x1, x2, x3......., xn is in [x1,xn] and it is easy to understand.

In a contest, I assumed Expected value = p1 * x1 + p2 * x2 + p3 * x3 + ....... + 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 Comments (5)
 » 12 months ago, # | ← Rev. 5 →   The sum will be minimal if p1 = 1, because otherwise we can make p1 = 1 with moves which don't increase the value of the sum. Take some non-0 pi (i > 1) and add pi to p1 and make pi 0. This way the sum will change by  - pixi + pix1. Because x1 <  = xi this move will not increase the sum, the minimum is if p1 = 1, but then clearly the sum is at least x1. Same proof can be given for the fact that xn is the maximum.
•  » » Same thing, but a bit more intuitive for me so I'll share it:p1x1 + p2x2 + ... + pnxn ≥ p1x1 + p2x1 + ... + pnx1 = x1, since for each i, xi ≥ x1. Similarly, p1x1 + p2x2 + ... + pnxn ≤ p1xn + p2xn + ... + pnxn = xn, 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 [x1, xn]. Simply note the followingx1 = p1x1 + p2x1 + ... + pnx1 ≤ p1x1 + p2x2 + ... + pnxn ≤ p1xn + p2xn + ... + pnxn = xn.
•  » » Thanks for help.