问题及解答

(a) Show that $b(x;n,1-p)=b(n-x;n,p)$.

(b) Show that $B(x;n,1-p)=1-B(n-x-1;n,p)$. [Hint: At most $x$ $S$'s is equivalent to at least $(n-x)$ $F$'s.]

 

(a) Proof.

\[
\begin{split}
b(x;n,1-p)&=\binom{n}{x}(1-p)^x\cdot[1-(1-p)]^{n-x}\\
&=\binom{n}{x}(1-p)^x\cdot p^{n-x}\\
&=\binom{n}{n-x}p^{n-x}(1-p)^{n-(n-x)}\\
&=b(n-x;n,p)
\end{split}
\]

(b)

\[
\begin{split}
B(x;n,1-p)&=\sum_{y=0}^{x}b(y;n,1-p)\\
&=\sum_{y=0}^{x}b(n-y;n,p)\\
&=\sum_{k=n-x}^{n}b(k;n,p)\\
&=\sum_{k=0}^{n}b(k;n,p)-\sum_{k=0}^{n-x-1}b(k;n,p)\\
&=1-B(n-x-1;n,p)
\end{split}
\]