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{\bf Question}

A psychologist makes the following assumptions concerning the
behaviour of mice subjected to a particular feeding schedule.  For
any particular trial 80\% of the mice that went right in the
previous experiment will go right in this trial, and 60\% of those
mice that went left in the previous experiment will go right in
this trial.  If 50\% went right in the first trial, what would he
predict for
\begin{description}
\item[(a)] the second trial?
\item[(b)] the third trial?
\item[(c)] the thousandth trial?
\end{description}




\vspace{.25in}

{\bf Answer}

The two states are right($R$) and left($L$)

Transition matrix $P = \bordermatrix{  & R & L \cr  R &  0.8 & 0.2
\cr  L & 0.6 & 0.4 \cr}$

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Initial Distribution ${\bf p}_0 = (0.5,0.5)$

\begin{description}
\item[(i)] $\ds {\bf p}_1 = {\bf p}_0 P= (0.5,0.5)\left(
\begin{array}{cc} 0.8 & 0.2 \\ 0.6 & 0.4 \end{array} \right) =
(0.7 , 0.3)$

We predict that 70 \% go right and 30\% go left.
\item[(ii)]$\ds {\bf p}_2 = {\bf p}_1 P = (0.7 , 0.3) P = (0.74 ,
0.26)$
\item[(iii)] Now provided that $|p+q-1|<1,$
\begin{eqnarray*}
P^n &\rightarrow& \frac{1}{2-p-q} \left(
\begin{array}{cc} 1-q & 1-p \\ 1-q & 1-p
\end{array}\right) \\ &=& \frac{1}{2-.8-.4} \left(
\begin{array}{cc} 0.6 & 0.2 \\ 0.6 & 0.2
\end{array} \right)
\end{eqnarray*}

So $\ds{\bf p}_0 P^n \rightarrow \frac{1}{0.8}(0.5, 0.5) \left(
\begin{array}{cc} 0.6 & 0.2 \\ 0.6 & 0.2
\end{array} \right) = (0.75, 0.25)$

So 75\% go right for $n$ large.
\end{description}





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