0

A fair coin is tossed 4 times. Let X be the number of heads in the first three tosses. Let Y be the number of heads in the last three tosses. Find the joint p.m.f. of X and Y . (Hint: There are only 2^4 = 16 equally likely outcomes when you toss 4 coins. If you are unable to calculate the probabilities using rules we have learned, just list all the possible outcomes!)

enter image description here I'm working on this problem, and getting the answer analytically as the hint suggests is pretty simple, but I can't seem to determine a closed form equation.

I have tried splitting it into some indicator variable type thing, where $f= first$, $m= middle two$, and $l=last$. And then $f+m =x$ and $l+m=y$.

I know that x and y aren't independent, so finding the joint p.m.f should be something like

$P(x \& y) = P(x) * P(y|x)$, and I have some weird equation that almost works, but it's certainly not right and I'd like two know how to properly do it.

Marko
  • 49

2 Answers2

2

It is not clear that there is a neat formula, but we can use indicator functions to get an ugly one.

Let $\mathbb{B} = \{0,1\}$ and $T_k \in \mathbb{B}$ represent the $k$th toss.

Write $X = (T_2+T_3) + T_1, Y=(T_2+T_3)+T_4$. Note that $T_1, T_2+T_3, T_4$ are independent and $P[T_2+T_3 = k] = \begin{cases} {1 \over 4},& k \in \{0,2\} \\ {1 \over 2},& k =1\\ 0,& \text{otherwise} \end{cases}$

We can write \begin{eqnarray} P[X=x,Y=y] &=& P[T_2+T_3 + T_1 = x, T_2+T_3+T_4 = y]\\ &=& \sum_{k=0}^2 P[T_2+T_3 = k, T_1 = x-k, T_4 = y-k]\\ &=& \sum_{k=0}^2 P[T_2+T_3 = k] P[T_1 = x-k] P[T_4 = y-k]\\ &=& {1 \over 4} P[T_1 = x] P[T_4 = y] + {1 \over 2} P[T_1 = x-1] P[T_4 = y-1] + {1 \over 4} P[T_1 = x-2] P[T_4 = y-2] \\ &=& {1 \over 4} ( {1 \over 4} 1_{\mathbb{B}^2}(x,y)+ {1 \over 2} 1_{\mathbb{B}^2}(x-1,y-1)+ {1 \over 4} 1_{\mathbb{B}^2}(x-2,y-2)) \end{eqnarray}

copper.hat
  • 172,524
0

$R_X = $ all possible values of $X$. But when we have $X,Y$, we have a joint PMF. $$P_{XY}(x,y)= P(X=x, Y=y) $$ The range will be $R_{XY} = \big\{(x,y):P_{XY}(x,y)>0\big\}$

The probability of all points must add to $1$: $$\sum_{(x,y)\in R_{XY}}P_{XY}(x,y)=1 $$

Your question is not completely clear to me however. Normally joint pmf of random variable X and Y is usually given through a table or graph, or like you showed, simply solved by hand. The only thing I say you could try is the law of total probability: $$ P_{XY}(x,y)=\sum_{n=0}^{\infty}P(X=x, Y=y|N=n)P_N(n) $$

It also may be helpful to note it's possible to find specific probabilities (for example, $P(X=0,Y\le1)$ like so: $$P_{XY}(0,0)+P_{XY}(0,1)= {1\over16}+{1\over16} $$

kalkuluskueen
  • 19
  • 4