Marginal probability mass function

[magnifik]

A biased coin lands heads with probability p and tails with probability q. An experiment
consists of tossing the coin until a second head appears. Let T1 denote the number of
tosses until the first head appears and let T2 denote the number of tosses (counted from
the start) until the second head appears. For example, in the outcome TTHTTTH we
would have T1 = 3 and T2 = 7. Find the marginal PMF for each of T1 and T2.

I know the joint mass function is
P(T1 = j, T2 = n) = p²q^n-2 for 1 < j < n -1

I need assistance in finding the marginal PMF from there...

 

[KataKoniK]

To find the marginal PMF for T1, we need to sum over all possible values of T2. This means that we need to sum over all values of n such that n > j. So, the marginal PMF for T1 is:

P(T1 = j) = ∑ P(T1 = j, T2 = n) = ∑ p2qn-2 = p2qj-1 ∑ qn-1

= p2qj-1 (1/(1-q)) = p2qj-1/(p+q-1)

Similarly, to find the marginal PMF for T2, we need to sum over all possible values of T1. This means that we need to sum over all values of j such that j < n. So, the marginal PMF for T2 is:

P(T2 = n) = ∑ P(T1 = j, T2 = n) = ∑ p2qn-2 = p2qn-2 ∑ pj-1

= p2qn-2 (1/(1-p)) = p2qn-2/(p+q-1)

Both of these marginal PMFs follow a geometric distribution, with parameter p and q respectively. This means that the expected value for T1 is 1/p and the expected value for T2 is 1/q.

In summary, the marginal PMF for T1 is p2qj-1/(p+q-1) and the marginal PMF for T2 is p2qn-2/(p+q-1), both following a geometric distribution. These distributions can be used to calculate probabilities or expected values for T1 and T2 in any given experiment.