This is a continuation of a paper Part I of which was published in the June, 1946 issue of the Annals of Mathematical Statistics. The present paper is divided into two parts, Parts II and III, which are summarized as follows: Part II. The Exact Power Curve and the Distribution of n for Sequential Tests Where z Takes on a Finite Number of Integral Values. Consider a sequential test defined by a decision function $Z_n = \sum^n_{\alpha = 1} z_\alpha$ with boundaries $-b$ and $a$ where $a$ and $b$ are positive integers and $z_\alpha$ is the $\alpha$th observation of a variate $z$ which takes on a finite number of integral values ranging from the negative integer $-r$ to the positive integer $m$ with respective probabilities $p_{-r}, \cdots, p_m$. Let $\xi_{ai} = P\lbrack Z_n = (a + i)\rbrack, (i = 1, 2, \cdots, m - 1)$, and $\xi_{bj} = P\lbrack Z_n = -(b + j)\rbrack, (j = 1, 2, \cdots, r - 1)$. Furthermore, let $A$ be a square matrix of $a + b - 1$ rows and columns with elements defined by: $a_{ii} = 1 - p_0$ for all $i; a_{i,i + k} = -p_k$ for $k = 1, 2, \cdots, m; a_{i,i- j} = -p_{-j}$ for $j = 1, 2, \cdots, r$; and $a_{ij} = 0$ otherwise. It is proved that \begin{equation*}\tag{i} \xi_{bj} = \sum^{r - j - 1}_{i=0} p_{i-r}A_{r-j-i,b},\quad(j = 0, 1, \cdots, r - 1) {(ii)} \xi_{aj} = \sum^{m - j - 1}_{i=0} p_{i+j +1}A_{a+b-i-1,b},\quad(j = 0, 1, \cdots, m - 1),\end{equation*} where $A_{kb}$ is the element of the $k$th row and $b$th column in $A^{-1}$. Let $E_{aj\tau}^n$ and $E_{bj\tau}^n$ be the conditional generating function of $n$ under the restriction that $Z_n = (a + j)$ and $Z_n = -(b + j)$ respectively. Then $\xi_{bj}E_{bj\tau}^n$ is obtained by substituting $\tau p_j$ for each $p_j$ occurring in equation (i) and $\xi_{aj}E_{aj\tau}^n$ is obtained by substituting $\tau p_j$ for each $p_j$ occurring in equation (ii). The probability that $Z_n = a + j$ in exactly $n$ steps is given by the coefficient of $\tau^n$ in the expansion of $\xi_{aj}E_{aj\tau}^n$ in a power series in $\tau$. The probability that $Z_n = -(b + j)$ in exactly $n$ steps is similarly obtained. This method is applied to the derivation of the exact power function and the distribution of $n$ for the sequential binomial probability ratio test. Part III. On Conjugate Distributions. Consider a random variable $X$ with a distribution density $f(x, \theta)$ which satisfies certain specified conditions. Let $\theta_1$ and $\theta_2$ be two values of $\theta$ and let $z = \log (f(x, \theta_2)/f(x, \theta_1))$. For any hypothesis $\theta = \theta'$, let $\varphi(t \mid \theta')$ be the moment generating function of $z$ and $h$ the non-zero value of $t$ for which $\varphi(t \mid \theta') = 1$. We set $F(x) = e^{hz}f(x, \theta')$. Then $f$ and $F$ are conjugate distributions. If $F = f(x, \theta")$, then $\theta'$ and $\theta"$ are defined as conjugate pairs. A method is given for obtaining the totality of conjugate pairs for the general class of distributions which admit a sufficient statistic. It is then shown that the power of the sequential probability ratio test based on such distributions is given explicitly in terms of these pairs. It is proven that within the approximation obtained by neglecting the excess of $\mid Z_n \mid$ over $a$ and $b$ at a decision point the following relationship holds: $P_b(n \mid F) = e^{-hb}P_b(n \mid f) \\ P_a(n \mid F) = e^{ha}P_a(n \mid f)$ where $P_b(n \mid g)$ and $P_a(n \mid g)$ stand for the probability that $Z_n \geq a$ and $Z_n \geq -b$ respectively in exactly $n$ steps under the hypothesis $g$.