Expected Value...

Правка en1, от Rajan_sust, 2018-10-22 15:57:20

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

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en1 Английский Rajan_sust 2018-10-22 15:57:20 466 Initial revision (published)