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July 4th, 2017, 04:21 PM  #1 
Member Joined: Apr 2017 From: PA Posts: 34 Thanks: 0  Compute $E(XZ)$.
Let $X$ be a random variable distributed uniformly on [0,20]. Define a new random variable $Z$ by $Z$ = [X+.5 ] (the greatest integer in X). Find the expected value of $Z$. Compute $E(XZ)$. <br><br> **Attempt:** I attempted to set up the integral $$\int_{0}^{20}\frac 1 {20}\,dx,$$ since I found out that it was uniformly distributed on $[0,1]$. Also, I thought the distribution fuction woud be $\dfrac 1 {20}$. Then, I used the formula $E(XZ)$=E(X)E(Z), I found out that $E(X)$=10, after doing the calculation, then I took the integral $$\int_{0}^{20}(X^2+.5X),dx,$$, but was unable to find the right answer. 
July 5th, 2017, 05:15 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,256 Thanks: 507 
Z=0 for X < .5, XZ=X Z=1 for .5<X<1.5, XZ is uniform in the interval (0,.5) Z=2, etc. leads to XZ uniform in the interval (0,.5) You should be able to do the rest. 
July 6th, 2017, 10:01 AM  #3 
Member Joined: Apr 2017 From: PA Posts: 34 Thanks: 0 
So, we should stop at Z=10? Where 8.5<X<9.5?

July 6th, 2017, 01:17 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,256 Thanks: 507 
Z=10 for (9.5<X<10.5). Your original statement has X up to 20.

July 6th, 2017, 02:38 PM  #5 
Member Joined: Apr 2017 From: PA Posts: 34 Thanks: 0 
So, we need to find out how much Z=? When 19.5<X <20.5

July 7th, 2017, 01:14 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,256 Thanks: 507  
July 10th, 2017, 12:12 PM  #7 
Member Joined: Apr 2017 From: PA Posts: 34 Thanks: 0 
So, we should stop at Z=19.5, where 19.5<X <20?


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