0
$\begingroup$

I understand if a fair coin is tossed repeatedly, the probability of getting a head during each toss does not change and is independent from that of other tosses.

What about the number of tosses required to get heads - are they dependent?

For example let X be the random variable representing number of tosses required to get the first head, and Y the random variable representing the number of tosses to get the first two heads. Are the two variables dependent?

Intuitively it seems that Y depends on X. What is a more 'formal' reasoning than mere intuition?

$\endgroup$
  • $\begingroup$ Not sure I follow. As it stands , of course they are dependent. If it takes you $20$ tosses to get the first head, then it is impossible that you got two heads in the first $5$ tosses. Is that what you meant? $\endgroup$ – lulu Jul 13 at 13:52
  • $\begingroup$ The number of tosses has a negative binomial distribution. The questions is whether X (number of tosses getting the 1st head) and Y (number of tosses getting the first two heads) are independent. $\endgroup$ – Cee Cee Jul 13 at 13:56
  • 1
    $\begingroup$ And as I remarked, they are obviously dependent. Do you think $P(X=3, Y=2)\,=\,P(X=3)\times P(Y=2)$? Or am I misunderstanding the question? $\endgroup$ – lulu Jul 13 at 13:58
  • $\begingroup$ I do not think P(X=x, Y= y) = P((X = x) * P(Y = y). What I am asking is a more formal "proof". I have the distribution function for P(X) and P(Y) respectively, both follow the negative binomial distribution. But I cannot figure out the joint distribution P(X, Y).... $\endgroup$ – Cee Cee Jul 13 at 14:03
  • $\begingroup$ What I wrote is a formal proof. Just compute both sides and verify that they are not equal. $\endgroup$ – lulu Jul 13 at 14:03
0
$\begingroup$

We have $Pr(X=3) >0$ and $Pr(Y=2)>0$

But $$Pr(X=3, Y=2) =0 \ne Pr(X=3)Pr(Y=2)$$

Hence, they are dependent.

Note that we do not need the whole distribution explicitly.

We know the joint distribution requireds $X < Y$ to be positive. But $X$ follows a geometric distribution while $Y$ follows the sum of two geometric distributions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.