Question

Two numbers, x and y, are randomly chosen from the interval [0,1]. What is the expected value of (x+y)?

Answer 1

what is the expected value of \(x\) ?

Answer 2

it is clear that the expected value of \(x\) is \(\frac{1}{2}\) ?

Answer 3

not really?

Answer 4

well, one thing we could say, what else would it be?

Answer 5

you pick a number at random in the interval, why should any number be favored? have of the length of the interval is one half
do you have a definition for the expected value of a continuous variable?

Answer 6

not really no

Answer 7

hmm you have to have something right?

Answer 8

oh wait. I see. since they are all equally chosen, then there is 1(the distance) divided by 2 (the # of variables). Yep! makes sense

Answer 9

...right?

Answer 10

wait. then y and x would both be 1/2 so 1/2+1/2 = 1

Ooh! the answer is 1! Thanks!

Answer 11

hold on for a second, i don't want to steer you wrong

Answer 12

darn

Answer 13

ok i just had to check
you are right it is \(\frac{1}{2}+\frac{1}{2}=1\) for your answer

Answer 14

yay! is my reasoning right or...?

Answer 15

there is an actual definition in terms of integrals, but i have a feeling you have not covered it
yes, reasoning is right

Answer 16

Thanks so much :D

Answer 17

well actually it has nothing to do with the number of variables
the expected value of \(x\) is \(\frac{1}{2}\) as is the expected value of \(y\)
so since they are both one half, the expected value of the sum is one

Answer 18

oh. ok.

Answer 19

the reason the expected value is \(\frac{1}{2}\) is because
\[\int_0^1xdx=\frac{1}{2}\] but i sort of doubt you have this in your class
i could be wrong

Answer 20

yah. youre right. I have no idea what you mean. still a newb XD

Answer 21

don't worry about it