Theorem 17

$$G_{m,n} (x) = F^n (x) \sum^{m-1}_{k=0} \begin{pmatrix} n \\ k\end{pmatrix} \left[ \frac{1-F(x)}{F(x)} \right]^k.$$

Proof: The following table of recursion relationships can be prepared:

$$\begin{array} {r@{~}c@{~}l} \text{Pr} (t_1 \leq x) & = & \te... ...m-1} \leq x) + \text{Pr} (t_m \leq x; t_{m-1} \gt x)\end{array}$$

or in explicit terms

 

$$\text{Pr} (t_1 \leq x; t_0 \gt x) \equiv \text{Pr} (t_1 \leq x).$$ (101)

where the convention is adopted that

$$\text{Pr} (t_1 \leq x; t_0 \gt x) \equiv \text{Pr} (t_1 \leq x).$$

Hence, the problem of finding $\text{Pr} (t_m \leq x)$ is reduced to that of finding $\text{Pr} (t_j \leq x ; t_{j-1} \gt x)$.

Now, for one observation, by Definition 8

$$\begin{array} {r@{~}c@{~}l} \text{Pr} (t \gt x) & = & 1 - F(... ... \text{(call this the non-occurrence of event $A$)}.\end{array}$$

But, $\text{Pr} (t_j \leq x ; t_{j-1} \gt x)$ is the probability that in n observations event A occurs j-1 times and fails to occur n - j + 1 times. Hence, by Theorem 1

 

$$\text{Pr}(t_j \leq x ; t_{j-1} \gt x) = \begin{pmatrix} n \\ j-1\end{pmatrix} F^{n-j+1} (x) [ 1 - F(x)]^{j-1}.$$ (102)

If j = 1, the convention adopted above that

$$\text{Pr} (t_1 \leq x ) \equiv \text{Pr} (t_1 \leq x; t_0 \g... ...{pmatrix} n \\ 0 \end{pmatrix} F^n (x) [1 - F(x)]^0 = F^n (x)$$

is satisfied by the binomial frequency distribution, Eq. (102). Substituting Eq. (102) into Eq. (101),

$$\begin{array} {r@{~}c@{~}l} G_{m,n} (x) = \text{Pr} (t_m \le... ...{pmatrix} \left[ \frac{1-F(x)}{F(x)} \right]^{j-1}. \end{array}$$

Finally, set k = j - 1 to obtain the required results as

$$G_{m,n} (x) = F^n (x) \sum^{m-1}_{k=0} \begin{pmatrix} n \\ k\end{pmatrix} \left[ \frac{1-F(x)}{F(x)} \right]^k.$$