Question
Consider a sequence of independent trials with probability of success $p$ at each trial. Suppose that $Y$ and $X$ are the times at which the first and second successes happen, respectively.
(a) What are the marginal distributions of $Y$ and $X$?
(b) What is the conditional distribution of $X - y$ given that $Y = y$?
(c) Show that the joint PMF of $X$ and $Y$ is $$p_{X, Y}(x, y) = p^2(1 - p)^{x - 2}\ \forall\ 0 < y < x, x, y \in \mathbb{Z}.$$
My attempt
(a)
$Y$ is intuitively geometric to me, so its marginal PMF should be $$p(1 - p)^{y - 1}\ \forall\ y \in \mathbb{Z}^+.$$
For $X$ to happen, we must have had $1$ success and $x - 2$ failures first and there are $x - 1$ ways for this to happen, so its marginal PMF should be $$p^2(1 - p)^{x - 2}\ \forall\ x > 1, x \in \mathbb{Z}.$$
(b)
By Bayes' theorem, we have \begin{aligned} \mathbb{P}(X - y \mid Y = y) & = \frac {\mathbb{P}(X - y, Y = y)} {\mathbb{P}(Y = y)}\\[2 mm] & = \frac {p(1 - p)^{x - y - 1}} {p(1 - p)^{y - 1}}\\[2 mm] & = (1 - p)^{x - 2y} \end{aligned}
(c)
Here, I am only able to use intuition to derive the answer. Since we want the joint PMF of $X$ and $Y$, we see that for $X$ to happen, $Y$ must obviously happen, and so we simply consider how $X$ can happen, that is, we must have $x - 2$ failures and $2$ successes, so the joint PMF of $X$ and $Y$ is $$p_{X, Y}(x, y) = p^2(1 - p)^{x - 2}\ \forall\ 0 < y < x, x, y \in \mathbb{Z}.$$
Are my answers in parts (a) and (b) correct? In particular, I am skeptical of my answer to part (b) because it does not contain $p$. Where is the probability of my second success then? Moreover, can anyone help with deriving part (c) mathematically and with intuition? Does it involve parts (a) and (b)?