# Find joint probability mass function of random vector (X,Y).

Flip a fair coin three times. Let X equal the number of heads and Y equal the number of tails.

a) Determine the joint probability mass function of the random vector (X,Y). b) Are X and Y independent? Explain.

I'm not sure what vector has to do with joint probability mass, my teacher never introduce vector into joint probability mass and it confusing me.

For a) i know that the joint probability mass function of X and Y would just X+Y ~ Binomial (n,p). So n would be 3 and i'm guessing p will be 1/3 . Can someone let me know if i'm doing this correctly?

I can understand your confusion; there's a little bit of mixed language happening here. Let me see if I can clear it up.

The joint probability mass for $X$ and $Y$ is the function $$f(i, j):=P(X=i, Y=j).$$ You can think of this as filling out a table, where the column headings are values for $X$, the row headings are values for $Y$, and the cells contain the probability of that combination.

On the other hand, you could think of this as describing the distribution of a single random variable $\vec{V}$ whose outputs are vectors $(i, j)$. In this case, the mass is a function $$f(\vec{v}):=P(\vec{V}=\vec{v}).$$ In this case, you are trying to assign probabilities to each possible output vector $\vec{v}$.

These two approaches are equivalent, which probably explains the way that your instructor interspersed them.

At any rate, in your thought process you've exactly described the distribution of $X+Y$, but that isn't quite the same as finding the join distribution (although it is very close, and gives you lots of information). Try filling out the table as suggested above.

• hmmm if we were to solve this with table the diagonal of the table should all be 0 and the probability of like 1 head n 2 tail and etc correct? – darkflames363 Dec 15 '16 at 2:37
• would there be another way to find it or is the table the only way? – darkflames363 Dec 15 '16 at 2:37
• I'm not sure I follow. The table isn't a way to find it - it is what you are trying to find. – Nick Peterson Dec 15 '16 at 3:18
• In a table with $X$ taking values 0, 1, 2, 3, and $Y$ taking values 0, 1, 2, 3, there are $4 \times 4 = 16$ potential positions to fill with probabilities. But in your problem, only 4 of those 16 probabilities can be positive. – BruceET Dec 15 '16 at 6:36
• @BruceET i mean if were making a 4 by 4 matrix it would be 16 position to fill but almost all of them are 0 beside the positive correlation right? because you flipped it 3times so the sum must be 3 . Example 3 head 0 tail = 1/8 0 head and 3 tail = 1/8 and 1 head 2 tail = 3/8 and 2 head and 1 tail = 3/8 so there actually only 4 probability – darkflames363 Dec 15 '16 at 22:55