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QUESTION

 Show that for a Binomial distribution with parameters n and p
   $$ p(x)=\frac{n+1-x}{x}\, \frac{p}{q}\, p(x-1)$$

    \begin{description}

     \item[(i)]
      For $n=5$ and $p=0.4$ use this recurrence relation together with
      the value of $p(0)$ to evaluate all the values of $p(x)$

     \item[(ii)]
      Use the expression to prove that the mode of the
      distribution is the integer value of $(n+1)p$ and the
      distribution has two modes $(n+1)p$ and $(n+1)p-1$ if $(n+1)p$ is
      an integer.
    \end{description}

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ANSWER

 $$p(x)=\left( \begin{array}{c}n\\
   x\end{array}\right)p^xq^{n-x}$$
$$   p(x-1)=\frac{n!}{(x-1)!(n+1-x)!}p^{x-1}q^{n+1-x}$$

   \begin{eqnarray*}
    \frac{p(x)}{p(x-1)}&=&\frac{n!(x-1)!(n+1-x)!}{x!(n-x)!n!}\frac{p^{x-1}q^{n-x}}{p^{x-1}q^{n+1-x}}\\
    &=&\frac{n+1-x}{x}\frac{p}{q}
   \end{eqnarray*}

   \begin{description}

    \item[(i)]

     $p(0)=0.6^5=0.07776$

     $p(1)=\frac{5+1-1}{1}\frac{0.4}{0.6} p(0)=0.2592$

     $p(2)=\frac{5+1-2}{2}\frac{0.4}{0.6} p(1)=0.3456$

     $p(3)=\frac{5+1-3}{3}\frac{0.4}{0.6}p(2)=0.2304$

     $p(4)=\frac{5+1-4}{4}\frac{0.4}{0.6}p(3)=0.0768$

     $p(5)=\frac{5+1-5}{5}\frac{0.4}{0.6}p(4)=0.01024$

    \item[(ii)]
     $$p(x) \geq p(x-1) \textrm{ if } \frac{n+1-x}{x}\frac{p}{1-p}\geq 1$$

     \begin{eqnarray*}
      (n+1-x)p &\geq&x(1-p)\\
      (n+1)p& \geq& x
     \end{eqnarray*}

     Hence $p(x)$ increases if $x \leq (n+1)p$. Otherwise it
     decreases.
     Hence the mode is the integer value of $(n+1)p.$ If this is an
     integer we say $p(k)=p(k-1),$ hence there are two modes
     $(n+1)p$ and $(n+1)p-1.$

   \end{description}

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