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June 29th, 2017, 03:52 PM   #1
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Compute E(|X-Y|) and E(|X-Z|)

Let X be a random variable distributed uniformlyE(|X-Y|) over [0, 20]. Define a new
random variable Y by Y = [X] (the greatest integer in X). Find the expected
value of Y. Compute E(|X-Y|)


Attempt: for this I tried to use the formula for E(|X-Y|) and got (Y^2/20)+1-Y, then
I plugged in 20 from the interval, but got 1. Then I attempted to take the integral $$\int_{0}^{20}([X])dx,$$
but got a number too high to be E(Y). How would u solve this?
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June 29th, 2017, 08:33 PM   #2
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Can you find any other definition of the $\displaystyle [X]$ notation in the text? Because right now I'm not sure how it's actually defined. If it means 'the greatest integer in the support of X', that's just 20 and certainly not random. If it means $\displaystyle \max_{X_i \in \mathbb{Z}} X_i$, that's a perfectly reasonable definition, but I don't see anywhere in the question where they mentioned $\displaystyle X_i$...
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June 29th, 2017, 10:27 PM   #3
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Quote:
Originally Posted by poopeyey2 View Post
Let X be a random variable distributed uniformlyE(|X-Y|) over [0, 20]. Define a new
random variable Y by Y = [X] (the greatest integer in X). Find the expected
value of Y. Compute E(|X-Y|)


Attempt: for this I tried to use the formula for E(|X-Y|) and got (Y^2/20)+1-Y, then
I plugged in 20 from the interval, but got 1. Then I attempted to take the integral $$\int_{0}^{20}([X])dx,$$
but got a number too high to be E(Y). How would u solve this?

$x \geq Y(x),~\forall x \in [0,20] \Rightarrow |X-Y| = X-Y$

$0 \leq Y(x) < 1$

If you look at $X-Y$ you'll see it's a sawtooth

$Y(x) = x \mod 1,~x \in [0, 20]$

So you have 20 copies of $U[0,1)$ each weighted by the original distribution of $\dfrac 1 {20}$ and thus this corresponds to

$Y$ being distributed as $U[0,1)$

and thus $E[|X-Y|] = 0.5$
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June 29th, 2017, 11:41 PM   #4
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Quote:
Originally Posted by romsek View Post
$x \geq Y(x),~\forall x \in [0,20] \Rightarrow |X-Y| = X-Y$

$0 \leq Y(x) < 1$

If you look at $X-Y$ you'll see it's a sawtooth

$Y(x) = x \mod 1,~x \in [0, 20]$

So you have 20 copies of $U[0,1)$ each weighted by the original distribution of $\dfrac 1 {20}$ and thus this corresponds to

$Y$ being distributed as $U[0,1)$

and thus $E[|X-Y|] = 0.5$
That was smart, I had no idea the square brackets were denoting equivalence classes! Could you clarify my confusion for me though? TBH I'm still a bit confused as to how $Y$ is defined - and in particular I don't understand the use of modular arithmetic with non-integral values...
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June 29th, 2017, 11:49 PM   #5
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Quote:
Originally Posted by 123qwerty View Post
That was smart, I had no idea the square brackets were denoting equivalence classes! Could you clarify my confusion for me though? TBH I'm still a bit confused as to how $Y$ is defined - and in particular I don't understand the use of modular arithmetic with non-integral values...
$Y$ is $\left \lfloor X \right \rfloor$

$x \pmod{1} = x - \left \lfloor x \right \rfloor$

i.e. the fractional part of $x$
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June 30th, 2017, 08:14 AM   #6
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Quote:
Originally Posted by romsek View Post
and thus this corresponds to

$Y$ being distributed as $U[0,1)$
This is a mistake, apologies.

I meant that $X-Y = |X-Y|$ is distributed as $U[0,1)$

and thus $E[|X-Y|] = 0.5$
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July 2nd, 2017, 11:15 AM   #7
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The mean of all the numbers from x to x + 1 is x + 0.5. This is true even if x isn't an integer. Therefore if y = the mean of all the numbers from x to x + 1, y - x = 0.5.
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July 3rd, 2017, 10:42 AM   #8
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Hey Romsek, could u show the steps on how E(|X-Y|)=0.5?
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