# Joint distribution table

A fair coin is tossed four times. Let X and Y be the numbers of tails obtained in the first two tosses and the last three tosses, respectively.
$$(a)$$ State the distributions of X and Y .
$$(b)$$ Describe the joint distribution of X and Y by a clearly labelled table and use this to find the marginal distributions of X and Y .

for a) I believe it is $$X\sim B(2,0.5)$$ and $$Y\sim B(3,0.5)$$ but I am having trouble constructing the table again. I would appreciate a lot if someone could construct it for me as I could really use an example..

(a)

If $$B$$ in your question denotes "binomial" then your answer is correct.

(b)

For convenience let $$Z$$ denote the number of tails obtained in the second toss.

Then $$X-Z$$ is the number of tails in the first toss, $$Z$$ is the number of tails obtained in the second toss and $$Y-Z$$ is the number of tails obtained in the last $$2$$ tosses.

This indicates that they are independent and also makes clear how they are distributed.

Now for every pair $$\left(i,j\right)$$ with $$i\in\left\{ 0,1,2\right\}$$ and $$j\in\left\{ 0,1,2,3\right\}$$ find $$P\left(X=i,Y=j\right)$$ on base of:

\begin{aligned}P\left(X=i,Y=j\right) & =\sum_{k=0}^{1}P\left(X=i,Z=k,Y=j\right)\\ & =\sum_{k=0}^{1}P\left(X-Z=i-k,Z=k,Y-Z=j-k\right)\\ & =\sum_{k=0}^{1}P\left(X-Z=i-k\right)P\left(Z=k\right)P\left(Y-Z=j-k\right)\\ & =\frac{1}{2}\sum_{k=0}^{1}P\left(X-Z=i-k\right)P\left(Y-Z=j-k\right) \end{aligned} The marginals can be calculated by means of:

• $$P(X=i)=\sum_{j=0}^3P(X=i,Y=j)$$ for $$i=0,1,2$$
• $$P(Y=j)=\sum_{i=0}^2P(X=i,Y=j)$$ for $$j=0,1,2,3$$

Actually the marginals are calculated in (a) already, so check them.

• Hi, how did you get 1/2 for P(Z=k) as I get 1? Also how can this help me to construct the distribution table for X and Y? – user720013 Nov 30 '19 at 14:18
• Do you agree that the probability on obtaining $0$ tails in the second toss equals $0.5$ and that the probability on obtaining $1$ tail in the second toss also equals $0.5$? That tells us that $P(Z=k)=0.5$ for $k=0$ and also for $k=1$, right? What I give you is a way to construct the distribution table for $(X,Y)$ (On LHS you find $P(X=i,Y=j)$ and on RHS the outcomes) so I do not really understand your last question. Of course you must work out the RHS but that can be done because the distributions of $X-Z$ and $Y-Z$ are easily found. – drhab Nov 30 '19 at 14:24
• yes so then the sum of those (the sum from k=0 to k=1 ) is 0.5+0.5=1. is this correct: X-Z~B(1,0.5) and Y-Z~B(2,0.5)? – user720013 Nov 30 '19 at 14:30
• Yes, that is correct. Btw, on RHS we only meet $P(Z=k)$ and not the sum $P(Z=0)+P(Z=1)$. – drhab Nov 30 '19 at 14:35
• Ok I understand thank you! one more thing : does i-k take values 0 and 1 and j-k takes values 0,1,2 ? – user720013 Nov 30 '19 at 15:10