mysesshou
  • mysesshou
Need help :) Q in following post
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mysesshou
  • mysesshou
Any help or hints are appreciated :) Lots!
PaxPolaris
  • PaxPolaris
Explain E(X|Y)... expected value of X given Y is what? If the question is as is, does this mean the answer has a variable?
mysesshou
  • mysesshou
sloooow computer

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PaxPolaris
  • PaxPolaris
I'm just guessing here... when Y=0, E(X) = 0 when Y=1, E(X) = 1 so, E(X|Y)=Y
mysesshou
  • mysesshou
Ah, your prev post makes sense i think.
PaxPolaris
  • PaxPolaris
as for Covariance cov(X,Y) = E[XY] - E[X]*E[Y] E[X] & E[Y] should be 0.5 ..... not sure how you calculate E[XY]
PaxPolaris
  • PaxPolaris
You agree that E[X] = E[Y] = 0.5?
mysesshou
  • mysesshou
boo. the Y=0 didn't show here
PaxPolaris
  • PaxPolaris
I thought Cov(X,Y) should be numerical value... there is a 50% chance E(X)=0, there is a 50% chance E(X)=1,\[E(X)= 0.5*0\ +\ 0.5*1=0.5\]
mysesshou
  • mysesshou
You know, that makes sense. Ok. that works .
PaxPolaris
  • PaxPolaris
for E(XY) when Y is 0, XY is just 0,so E[XY] is 0 when Y is 1, XY=X...\[ X= \sim \mathcal{N}(1,1)\]so E[XY]=1 so I think E[XY] is also 0.5, although not sure.
mysesshou
  • mysesshou
Ok. all help is appreciated, esp to help me see it different ways.
mysesshou
  • mysesshou
tricky problem, eh?
PaxPolaris
  • PaxPolaris
:) so my guess for cov(X,Y) = 0.5-0.25= 0.25
PaxPolaris
  • PaxPolaris
I think I was wrong about E[XY] = 0.5 ...you can't just take the probabilities of the expected values and add them up here... When Y=0, XY= 0 When Y=1, XY= ∼N(1,1) graphing, |dw:1332346799591:dw|
PaxPolaris
  • PaxPolaris
then taking average,|dw:1332347194314:dw| as you can see E[XY] is still 1, only its std. deviation changed. E[XY]=1
mysesshou
  • mysesshou
hmm. hadn't thought about plotting it. :)
mysesshou
  • mysesshou
Thanks !

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