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June 29th, 2017, 04:52 PM  #1 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0  Compute E(XY) and E(XZ)
Let X be a random variable distributed uniformlyE(XY) 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(XY) Attempt: for this I tried to use the formula for E(XY) and got (Y^2/20)+1Y, 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? 
June 29th, 2017, 09:33 PM  #2 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 
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 $\displaystyle X_i$...

June 29th, 2017, 11:27 PM  #3  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,600 Thanks: 816  Quote:
$x \geq Y(x),~\forall x \in [0,20] \Rightarrow XY = XY$ $0 \leq Y(x) < 1$ If you look at $XY$ 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 $E[XY] = 0.5$  
June 30th, 2017, 12:41 AM  #4  
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  Quote:
 
June 30th, 2017, 12:49 AM  #5  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,600 Thanks: 816  Quote:
$x \pmod{1} = x  \left \lfloor x \right \rfloor$ i.e. the fractional part of $x$  
June 30th, 2017, 09:14 AM  #6 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,600 Thanks: 816  
July 2nd, 2017, 12:15 PM  #7 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 Thanks: 79 
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.

July 3rd, 2017, 11:42 AM  #8 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 
Hey Romsek, could u show the steps on how E(XY)=0.5?


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