# How to show $U$ and $V$ are not independent random variable?

$$U$$ 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.

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

• 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... – Chloe Zhou Oct 15 '18 at 14:16
• @ChloeZhou Maybe try option 2? – 5xum Oct 16 '18 at 7:40