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Consider a fair coin that has 0 on one side and 1 on the other side. We flip this coin, independently, twice. Define the following random variables:

X = the result of the first coin flip

Y = the sum of the results of the two coin flips

Z = X*Y

  • Determine the distribution functions of X,Y and Z.
  • Are X and Y independent random variables?
  • Are X and Z independent random variables?
  • Are Z and Y independent random variables?

I understand that to be independent random variables they must satisfy the equation P(X=x,Y=y) = P(X=x)P(Y=y) but I am not sure how to show what P(X=x,Y=y) is or similarly what P(X=x,Z=z) and P(Y=y,Z=z) is in this case. Any help would be great.

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  • $\begingroup$ First find the distributions. For example $\Pr(Y=0)=\Pr(Y=2)=1/4$ and $\Pr(Y=1)=1/2$. For $Z$, you should find $\Pr(Z=0)=1/2$, $\Pr(Z=1)=1/4$, $\Pr(Z=2)=1/4$. For independence of $X$ and $Y$, note that $\Pr(X=0.Y=0)=1/4$, not equal to $\Pr(X=0)\cdot \Pr(Y=0)$. So not independent. Note that for independence we must have equality in all cases. $\endgroup$ – André Nicolas Apr 6 '15 at 23:13
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The idea here is to make cases.

Your universe is {00,01,10,11}

  • For 00, you have X=0, Y=0, Z=0
  • For 01, you have X=0, Y=1, Z=0
  • For 10, you have X=1, Y=1, Z=1
  • For 11, you have X=1, Y=2, Z=2

Then you can answer to the questions :

First question :

  • X has the distribution : P(X=0)=1/2, P(X=1)=1/2
  • Y has the distribution : P(Y=0)=1/4, P(Y=1)=1/2, P(Y=2)=1/2
  • Z has the distribution : P(Z=0)=1/2, P(Z=1)=1/4, P(Z=2)=1/4

Second question :

  • P(X=0,Y=0)=1/4 (case 00)
  • P(X=0,Y=1)=1/4 (case 01)
  • P(X=1,Y=1)=1/4 (case 10)
  • P(X=1,Y=2)=1/4 (case 11)

Then

  • P(X=0)P(Y=0) = 1/2*1/4 = 1/8

So the variables are not independant

And you continue like this

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  • $\begingroup$ that helps a lot, thanks $\endgroup$ – Kayseta Apr 6 '15 at 23:41

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