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I have problem stated as : Coin is tossed 2 times and it is a fair coin. We are letting X - # of heads, Y indicator function of the event {X=2}. Need to find the join mass function of (X,Y).

I was given a hint: P(X=0,Y=0)=P(X=0,{X=0 U X=1}) = P(X=0) = 1/4

I read various materials on join p.m.f and found that P(X=0,Y=0) means P(X=) and P(Y=0) - does this mean we sum the 2 probabilities? I am not sure what the comma means in that formula. Some documentation stated it was an intersection.

Could you please explain why 1/4 above and if possible illustrate with P(X=0,Y=1)?

Thank you.

1 Answers 1

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Yes, the comma is an intersection, or, put in words, $P(X=x,Y=y)$ is the probability that both $X=x$ and $Y=y.$

So we must compute $P(X=x,Y=y).$ There are four outcomes for the coinflips, $HH,$ $HT,$ $TH$ or $TT$, each with the same probability $1/4$ of occurring. Let's compute the values of $(X,Y)$ for each of those.

  1. For $HH,$ $X=2$ since there are two heads. $Y=1$ because $Y$ is the indicator that there are two heads. So $(X,Y) = (2,1)$

  2. For $TH$ or $HT$, $X=1$ since there is one head. $Y=0$ because we don't have $X=2.$

  3. For $TT$, $X=0$ since there are no heads and $Y=0$ because $X\ne 2.$

Thus there are three possible values for $(X,Y).$ We have $$ P(X=2,Y=1) = 1/4\\P(X=1,Y=0)=1/2\\P(X=0,Y=0)=1/4.$$ (The middle option has probability $1/2$ because it represents two of the four outcomes: $HT$ and $TH.$)

$P(X=x,Y=y) = 0$ if $x$ and $y$ are not any of the combinations listed above, i.e $(x,y)\ne$ $(2,1)$, $(1,0),$ or $(0,0)$.