Question

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

Answer 1

Hint:
Find E(x) and E(y) separately.
Then use E(X+Y)=E(X)+E(Y)

Answer 2

Well, I don't really understand what it means by the interval 0,1

Answer 3

Interval [0,1] is a set of all possible real values between 0 and 1, inclusively.
Since it says randomely chosen from the interval, so the random variable is a uniform distribution between 0 and 1. The probability distribution looks like the following:

The function of the distribution is therefore
\( P(x)=1\ if\ 0 \le x \le 1\ and \)
\(\ \ \ \ \ \ = 0\ elsewhere \)
The expected value of a probability distribution p(x) is
\( \int_{-\infty}^{\infty}xP(x)dx \\
\)
which for a uniform distribution [a,b] is simply the mean of a and b.

Hope this gets you started.

Answer 4

Yeah, I already got the question sorted out ;)

Answer 5

Thanks anyways though

Answer 6

yw!