$U$ stands for the number of trials to get the first head, $V$ stands for the number of trials to get two heads.

I used hand-waving proof, saying that you could not have the two heads trials without having the first head occur, but I don't know how to mathematically prove that the two random variables must be dependent.

  • $\begingroup$ Find functions $f$ and $g$ so that $Ef(U)G(V)\ne Ef(U) Eg(V)$. $\endgroup$ – kimchi lover Oct 15 '18 at 12:08
  • 2
    $\begingroup$ What's the probability that $U=3$ and $V=2$? What's the probability of those events separately? $\endgroup$ – lulu Oct 15 '18 at 12:10

The variables are independent if $P(U=u, V=v)=P(U=u)\cdot P(V=v)$ for all values $u,v$.

Therefore, to prove that the variables are not independent, all you need to do is find one pair $u,v$ such that

$P(U=u, V=v)\neq P(U=u)\cdot P(V=v)$

To do that, you can go one of two ways:

Option $1$: Try a little brute force search. Plug in a couple values $u,v$ and see if the equation holds.

Option $2$: Since the right side of the equation will always be nonzero, think about how you could make the left side zero. Can you think of some pair of numbers $u,v$ such that it's impossible to get two heads in $v$ trials, but only one head in $u$ trials?

  • $\begingroup$ I attempted to do option 1 but don't know how to derive the joint distribution of u,v...which is why I posted this question to see if someone could give me some suggestion on the most generic way of solving this question... $\endgroup$ – Chloe Zhou Oct 15 '18 at 14:16
  • $\begingroup$ @ChloeZhou Maybe try option 2? $\endgroup$ – 5xum Oct 16 '18 at 7:40

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