\textbf{Exercise 3 — Candidates who have followed the specialization}

Each young parent uses only one brand of baby food jars each month. Three brands $\mathrm { X } , \mathrm { Y }$ and Z share the market. Let $n$ be a natural integer.\\
We denote: $\quad X _ { n }$ the event ``brand X is used in month $n$ '',\\
$Y _ { n }$ the event ``brand Y is used in month $n$ '',\\
$Z _ { n }$ the event ``brand Z is used in month $n$ ''.\\
The probabilities of events $X _ { n } , Y _ { n } , Z _ { n }$ are denoted respectively $x _ { n } , y _ { n } , z _ { n }$.\\
The advertising campaign of each brand causes the distribution to change.\\
A buyer of brand X in month $n$ has the following month:\\
$50 \%$ chance of remaining loyal to this brand,\\
$40 \%$ chance of buying brand Y,\\
$10 \%$ chance of buying brand $Z$.\\
A buyer of brand Y in month $n$ has the following month:\\
$30 \%$ chance of remaining loyal to this brand,\\
$50 \%$ chance of buying brand X,\\
$20 \%$ chance of buying brand $Z$.\\
A buyer of brand Z in month $n$ has the following month:\\
$70 \%$ chance of remaining loyal to this brand,\\
$10 \%$ chance of buying brand X,\\
$20 \%$ chance of buying brand Y.

\begin{enumerate}
  \item a. Express $x _ { n + 1 }$ as a function of $x _ { n } , y _ { n }$ and $z _ { n }$.
\end{enumerate}

We admit that:\\
$y _ { n + 1 } = 0.4 x _ { n } + 0.3 y _ { n } + 0.2 z _ { n }$ and that $z _ { n + 1 } = 0.1 x _ { n } + 0.2 y _ { n } + 0.7 z _ { n }$.\\
b. Express $z _ { n }$ as a function of $x _ { n }$ and $y _ { n }$. Deduce the expression of $x _ { n + 1 }$ and $y _ { n + 1 }$ as functions of $x _ { n }$ and $y _ { n }$.\\
2. We define the sequence $\left( U _ { n } \right)$ by $U _ { n } = \binom { x _ { n } } { y _ { n } }$ for every natural integer $n$.

We admit that, for every natural integer $n$, $U _ { n + 1 } = A \times U _ { n } + B$ where $A = \left( \begin{array} { l l } 0.4 & 0.4 \\ 0.2 & 0.1 \end{array} \right)$ and $B = \binom { 0.1 } { 0.2 }$.

At the beginning of the statistical study (January 2014: $n = 0$), we estimate that $U _ { 0 } = \binom { 0.5 } { 0.3 }$.\\
Consider the following algorithm:

\begin{center}
\begin{tabular}{|l|l|}
\hline
Variables & \begin{tabular}{l}
$n$ and $i$ natural integers. \\
$A$, $B$ and $U$ matrices \\
\end{tabular} \\
\hline
Input and initialization & \begin{tabular}{l}
Request the value of $n$ $i$ takes the value 0 \\
$A$ takes the value $\left( \begin{array} { l l } 0.4 & 0.4 \\ 0.2 & 0.1 \end{array} \right)$ \\
$B$ takes the value $\binom { 0.1 } { 0.2 }$ \\
$U$ takes the value $\binom { 0.5 } { 0.3 }$ \\
\end{tabular} \\
\hline
Processing & \begin{tabular}{l}
While $i < n$ \\
$U$ takes the value $A \times U + B$ \\
$i$ takes the value $i + 1$ \\
End while \\
\end{tabular} \\
\hline
Output & Display $U$ \\
\hline
\end{tabular}
\end{center}

a. Give the results displayed by this algorithm for $n = 1$ then for $n = 3$.\\
b. What is the probability of using brand X in April?

In the rest of the exercise, we seek to determine an expression of $U _ { n }$ as a function of $n$.\\
We denote by $I$ the matrix $\left( \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right)$ and $N$ the matrix $I - A$.\\
3. We denote by $C$ a column matrix with two rows.\\
a. Prove that $C = A \times C + B$ is equivalent to $N \times C = B$.\\
b. We admit that $N$ is an invertible matrix and that $N ^ { - 1 } = \left( \begin{array} { l l } \frac { 45 } { 23 } & \frac { 20 } { 23 } \\ \frac { 10 } { 23 } & \frac { 30 } { 23 } \end{array} \right)$.

Deduce that $C = \binom { \frac { 17 } { 46 } } { \frac { 7 } { 23 } }$.\\
4. We denote by $V _ { n }$ the matrix such that $V _ { n } = U _ { n } - C$ for every natural integer $n$.\\
a. Show that, for every natural integer $n$, $V _ { n + 1 } = A \times V _ { n }$.\\
b. We admit that $U _ { n } = A ^ { n } \times \left( U _ { 0 } - C \right) + C$.

What are the probabilities of using brands $\mathrm { X } , \mathrm { Y }$ and Z in May?