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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..

emonHR
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1 Answers1

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(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.

drhab
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  • 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? –  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)? –  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 ? –  Nov 30 '19 at 15:10
  • if i=1 (i.e. X=1) then what is the value of i-k? –  Nov 30 '19 at 15:12
  • It is important that you work with a fixed $i\in{0,1,2}$ and a fixed $j\in{0,1,2,3}$. Then $i-k$ takes values $i-0=i$ and $i-1$ so you must calculate $P(X-Z=i)$ and $P(X-Z=i-1)$. I repeat that $i$ is fixed here. So you must do this for $i=0$, $i=1$ and $i=2$ separately. Same story for $Y-Z$. – drhab Nov 30 '19 at 15:26