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There are two random variables $X$ and $Y$, each of which can take on the values $0$ or $1$. Furthermore:

  • $P(X=0,Y=0)=p$,
  • $P(X=0)=1/2$
  • and $P(Y=0)=1/2$.

So these two shouldn't be independent in general, since $\frac{1}{2}^2$ is not necessarily $p$. The distributions of the 2 variables look like this

$$ \begin{array}{c|cc} x & P(X=x) \\ \hline 0 & 1/2 \\ 1 & 1/2 \\ \end{array} $$ $$ \begin{array}{c|cc} y & P(Y=y) \\ \hline 0 & 1/2 \\ 1 & 1/2 \\ \end{array} $$

Now what does the distribution of $X+Y$ look like? If I just add them, the sum surpasses $1$.
I found a formula for independent variables, but the ones in this example aren't.

Edit:

$$ \begin{array}{c|cc} z & P(Z=z) \\ \hline 0 & p \\ 1 & 1-2p \\ 2 & p \\ \end{array} $$

355durch113
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  • Hint: Compute the distribution of (X,Y). – Did Apr 10 '16 at 07:47
  • @Did I'm just guessing now, since I'm new to probability, but something like this? – 355durch113 Apr 10 '16 at 07:50
  • Quote: ...the distribution of (X,Y), not of X+Y. – Did Apr 10 '16 at 08:00
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    @Did Alright, following this, I get something similar, namely $P(0,0)=p$, $P(1,0)=1/2-p$, $P(0,1)=1/2-p$, $P(1,1)=p$. Correct? – 355durch113 Apr 10 '16 at 08:11
  • And now you can deduce from this (which you do not explain how you proved it...) the distribution of X+Y. – Did Apr 10 '16 at 08:20
  • @Did For instance, if I want to find out $P(X=1,Y=0)$, I need to subtract from the cases where $Y=0$ the the cases where $X$ is $0$, too (probability-wise). I really can't think of another way, except maybe multiplying the respective probabilities for the variables to be 0 or 1. Or are you saying that what I edited in is the solution? – 355durch113 Apr 10 '16 at 08:30
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    What I am saying is that if P(X=0,Y=0)=p, P(X=1,Y=0)=1/2−p, P(X=0,Y=1)=1/2−p and P(X=1,Y=1)=p, then P(X+Y=0)=p, P(X+Y=1)=1-2p and P(X+Y=2)=p. But that, for your solution to be complete, you should show how P(X=0,Y=0)=p, P(X=0)=1/2 and P(Y=0)=1/2, implies that P(X=0,Y=0)=p, P(X=1,Y=0)=1/2−p, P(X=0,Y=1)=1/2−p and P(X=1,Y=1)=p. – Did Apr 10 '16 at 08:37
  • @Did I see, I think I can show that more rigorously than in my previous comment, thanks a lot! – 355durch113 Apr 10 '16 at 08:39

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