Hi Kuwala

Firstly we need to find \(P[K=k, S\le s]\).

We know that \(S=T-K\) and the expression \(\{K=k\}\) is equivalent to \(\{k\le T \lt k+1\}\), hence:

\(P[K=k,S\le s] = P[k\le T \lt k+1,T-K\le s]\)

\(= P[k\le T \lt k+1, T \le k+s]\)

\(= P[k\le T\le k+s]\) since we know \(s \varepsilon [0,1)\).

Now we know that T is exponentially distributed so we can easily find the \(P[K=k, S \le s]\) using integration. I will leave that for you to do.

We end up finding \(P[K=k, S\le s]\) = \(e^{-\lambda k}\) \((1-e^{-\lambda s})\) that is for any nonnegative integer k and \(s \varepsilon [0,1)\).

The joint cumulative distribution of K and S can now be found.

\(F(k,s)\) = \(P[K\le k, S\le s]\)

= \(\sum_{r=0}^k\)\(e^{-\lambda r}\) \((1-e^{-\lambda s})\)

I will leave the rest for you to do.

Thank you so much Joshua