Question

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

Answer 1

Without doing any math, what do you expect the value to be? Just like, make a guess.

Answer 2

just what i was going to say
before you do any math, make a hypothesis
there is only one answer that should make any sense

Answer 3

1/2

Answer 4

i know its about 1/2 but i don't know the exact answer

Answer 5

What is the maximum and minimum value of (x+y) ?

Answer 6

it is one half for one of the random variables, not for the sum of the two

Answer 7

is it 0.75

Answer 8

What is the maximum and minimum value of (x+y) ?

Answer 9

1,0

Answer 10

Can't we pick x=1 and y=1 or x=0 and y=0?

This would give us (1+1) and (0+0) for our max and min, meaning we have 2 and 0 as the range of possible values.

Answer 11

@Kainui maybe it is time to "do the math"

Answer 12

Ok, I'll let you do the math if that's what you want @satellite73 ;P

Answer 13

@satellite73

Answer 14

no i don't want to really
but i thought the answer would be obvious

Answer 15

\[E(x+y)=E(x)+E(y)\] here and since these are identical random variables this is the same as \(2E(x)\)

Answer 16

\[E(x)=\int_{-\infty}^{\infty}xP(x)\] in this case
\[P(x)= \left\{\begin{array}{rcc}
1 & \text{if} &0<x<1 \\
0& \text{otherwise}
\end{array}
\right.
\]

Answer 17

so
\[E(x)=\int _0^1xdx=\frac{1}{2}\]

Answer 18

so 1/2

Answer 19

its wrong

Answer 20

@satellite73

Answer 21

\(E(x)=\frac{1}{2}. 2E(x)=1\)

Answer 22

really it should be way more obvious than all this silly math
they are uniform
the smallest is zero the largest is two and the average is one

Answer 23

Let's pretend you have two friends. Each of them are either going to give you $0 or $1. So what amount of money could you possibly get? Well they could either both give you 1 dollar, or both 0 dollars, or they could give you either 1 or 0 dollars.

1+1=2
0+0=0
1+0=1
0+1=1

It looks like out of all 4 possibilities one of these is more likely than the others because it appears twice. Why is that?

Answer 24

Since we can add expected values of independent events, the expected value of the sum is the sum of the expected values of each. Since we are just as likely to pick x as 1-x, the expected value of x (and likewise y) is equal to \frac12. Therefore the expected value of the sum is \frac12+\frac12=\boxed1.

Answer 25

wait i just figure it out

Answer 26

these are dependent events. Sry for not realizing fast enough

Answer 27

Wait... They are independent according to what you told us in the question. Whatever x is has no bearing on what y is and vice versa.

Answer 28

0<=x+y<=2

Answer 29

sry typo