The process to generate the two RVs is as follows. We first draw T from U n i f o r m ( 0 , 1 ) . If T ≤ 0.5 we take X = T and draw Y from U n i f o r m ( 0 , 1 ) . Otherwise if T &gt; 0.5 , we take Y = T and draw X from U n i f o r m ( 0 , 1 ) . Running a simulation it seems like X and Y are positively correlated, though intuitively it seems like they should have no effect on each other. What is the explanation?

Question

The process to generate the two RVs is as follows. We first draw T from   U  n  i  f  o  r  m  (  0  ,  1  ) . If   T  ≤  0.5 we take   X  =  T and draw Y from   U  n  i  f  o  r  m  (  0  ,  1  ) . Otherwise if   T  &amp;gt;  0.5 , we take   Y  =  T and draw X from   U  n  i  f  o  r  m  (  0  ,  1  ) . Running a simulation it seems like X and Y are positively correlated, though intuitively it seems like they should have no effect on each other. What is the explanation?

minotaurafe · Accepted Answer

Step 1Let&#039;s compute   E  [  X  Y  ]  −  E  [  X  ]  E  [  Y  ] . We have   E  [  Y  ]  =      1    2    (  E  [  Y  ∣  T  ≤  0.5  ]  +  E  [  Y  ∣  T  &amp;gt;  0.5  ]  )  =            0.5      +      0.75        2   . Similarly, we have   E  [  X  ]  =      1    2    (  E  [  X  ∣  T  ≤  0.5  ]  +  E  [  X  ∣  T  &amp;gt;  0.5  ]  )  =            0.25      +      0.5        2   .Then, we have   E  [  X  Y  ]  =      1    2    (  E  [  X  Y  ∣  T  ≤  0.5  ]  +  E  [  X  Y  ∣  T  &amp;gt;  0.5  ]  ) . If   T  ≤  0.5 , Y is independent from X, so   E  [  X  Y  ∣  T  ≤  0.5  ]  =  0.25  ⋅  0.5 . Similarly, when   T  &amp;gt;  0.5 , Y is indepdent from X, so   E  [  X  Y  ∣  T  &amp;gt;  0.5  ]  =  0.5  ⋅  0.75 . Thus, we have   cov  ⁡  (  X  ,  Y  )  =      1    2    (  0.125  +  0.375  )  −            1.25      ⋅      0.75        4    ≈  0.0156 , indicating a positive correlation.

Darian Hubbard · Accepted Answer

Step 1Of course they are correlated. For one thing they are not independent (as X can only be bigger than 0.5 if   Y  &amp;gt;  0.5 and conversely Y can only be smaller than 0.5 if   X  ≤  0.5 ).But in fact they are positively correlated, as the cases with   X  &amp;gt;  0.5 ,   Y  &amp;lt;  0.5 cannot happen.Let’s calculate this:      E    (  X  )  =      ∫    0          0.5        x    d  x  +  0.5      ∫    0    1    x    d  x  =  (      0.5    2    +  0.5  )      /    2  =  0.75      /    2  =  0.375and      E    (  Y  )  =  0.5      ∫    0    1    x    d  x  +      ∫          0.5              1        x    d  x  =  (  0.5  +  (  1  −      0.5    2    )  )      /    2  =  0.625Then      E    (  (  X  −  0.375  )  (  Y  −  0.625  )  )  =      ∫    0          0.5        (  x  −  0.375  )      ∫    0    1    (  y  −  0.625  )    d  y    d  x  +      ∫    0    1    (  x  −  0.375  )      ∫          0.5        1    (  y  −  0.625  )    d  y    d  xwhich amounts to   1      /    64 for the covariance. So the correlation is positive. We’d still need to calculate the variance of X, Y (which is the same) for the correlation:      E    (      X    2    )  =      ∫    0          0.5            x    2      d  x  +  0.5      ∫    0    1        x    2      d  x  =  (      0.5    3    +  0.5  )      /    3  =  (  5      /    8  )      /    3  =  5      /    24so      V    a    r    X  =  5      /    24  −  (  3      /    8      )    2    =  13      /    192so we get a correlation of  (  1      /    64  )      /    (  13      /    192  )  =  3      /    13  ≈  0.23

The process to generate the two RVs is as follows. We first draw T from Uniform(0,1).

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