# Fair coin flipped twice independently

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.

• 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. – André Nicolas Apr 6 '15 at 23:13

The idea here is to make cases.

• 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

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