# Dependent Coin Toss 100% rigged

I toss two coins which can both land Heads or Tails with probability 0.5 Define random variables: X = 1 if first coin lands Heads and 0 otherwise Y = 1 if second coin lands Heads and 0 otherwise

Now consider three cases: a) The two tosses are independent b) The coins are rigged so the second coin lands the same way as the first c) The coins are rigged so that the second coin lands the opposite way to the first.

For all three cases: • Calculate E(X), E(Y), σ(X), σ(Y), COV( X, Y) and ρ(X, Y) • Write down the probability distribution of X + Y • Using the probability distribution of X + Y, calculate E(X + Y), σ( X + Y)

I can calculate the E(x) etc. for the two tosses when they are independent, i just cant get my head around the E(x) and E(y) when the coins are rigged. As the formula i believe is the sum of Xi * Pi . I am unsure to what the probability of the second rigged toss is as it relies on the outcome of the first toss

would like if possible an explanation of how to get the E(x) and E(y) for the case b. as it is the base for calculating all of the other figures.

• Welcome at SE. What results have you got? And if you have no results, then what did you try and where did you get stuck? Add that to your question. Mar 2 '15 at 9:33
• OK, done considering, calculating, writing, using and calculating again. Now what? Mar 2 '15 at 9:33

The fact that the coins are rigged in cases b) and c) does not change the distribution of $X$ and does not change the distribution of $Y$. However it does change the distribution of $(X,Y)$.
In case b) $Y=X$ so that $X+Y=2X$ and e.g. $\rm{Cov}(X,Y)=\rm{Cov}(X,X)=\rm{Var}(X)$.
In case c) $Y=1-X$ so that $X+Y=1$ et cetera.