\begin{propriete}
\[
E\left( {aX + b} \right) = aE(X) + b\ \text{\quad et\quad } \ V\left( {aX} \right) =
{a^2}V\left( X \right).
\]
\end{propriete}

\begin{preuve}~\\
$
\begin{aligned}[t]
  E\left( {aX + b} \right) &= {p_1}\left( {a{x_1} + b} \right) +
  {p_2}\left( {a{x_2} + b} \right) + ... + {p_n}\left( {a{x_n} + b} \right) \\
  &= a{p_1}{x_1} + b{p_1} + a{p_2}{x_2} + b{p_2} + ... +  a{p_n}{x_n} + b{p_n}\\
  &= a{p_1}{x_1} + a{p_2}{x_2} + ... + a{p_n}{x_n} +  b{p_1} + b{p_2} + ... + b{p_n}\\
  &= a\left( {{p_1}{x_1} + {p_2}{x_2} + ... + {p_n}{x_n}} \right) +  b\left( {{p_1} + {p_2} + ... + {p_n}}  \right)\\
  &= aE\left( X \right) + b \times 1 \\
  E\left( {aX + b} \right)&= aE\left( X \right) + b
\end{aligned}
$

\bigskip

D'apr\`{e}s la seconde formule de la variance :

$V\left( {aX} \right) = {p_1}{\left( {a{x_1}} \right)^2} + {p_2}{\left( {a{x_2}}
\right)^2} + ... + {p_n}{\left( {a{x_n}} \right)^2} - {\left[ {E\left( {aX}
\right)} \right]^2}$

\bigskip

D'apr\`es la formule pr\'ec\'edente: $E\left( {aX} \right) =
aE\left( X \right)$, donc:

$
\begin{aligned}
  V\left( {aX} \right) &= {p_1}{a^2}{x^2}_1 + {p_2}{a^2}{x^2}_2 +
  ... +
  {p_n}{a^2}{x^2}_n - {\left[ {aE\left( X \right)} \right]^2}{\rm{ }}\\
  &= {a^2}\left( {{p_1}{x^2}_1 + {p_2}{x^2}_2 + ... +
      {p_n}{x^2}_n} \right) - {a^2}{\left[ {E\left( X \right)} \right]^2}\\
  &= {a^2}\left( {{p_1}{x^2}_1 + {p_2}{x^2}_2 + ... + {p_n}{x^2}_n -
      {{\left[ {E\left( X \right)} \right]}^2}} \right)\\
  V\left( {aX} \right) &= {a^2}V\left(
    X \right)
\end{aligned}
$
\end{preuve}
\begin{remarque}
  $V\left( {aX + b} \right) = {a^2}V\left( X \right)\text{ et }\sigma
  \left( {aX + b} \right) = \left| a \right|\sigma \left( X \right)$
  
  En effet :
  \begin{itemize}
  \item $
    \begin{aligned}[t]
      V\left( {aX + b} \right) &= \sum\limits_{i = 1}^{i = n}
      {{p_i}{{\left( {a{x_i} + b - E\left( {aX + b} \right)}
            \right)}^2}}\\
      &= \sum\limits_{i = 1}^{i = n} {{p_i}{{\left(
              {a{x_i} + b - aE\left( X \right) - b} \right)}^2}}\\
      &= \sum\limits_{i = 1}^{i = n}
      {{p_i}{{\left( {a{x_i} - aE\left( X \right)} \right)}^2}}\\
      &= \sum\limits_{i = 1}^{i = n} {{p_i}{a^2}{{\left( {{x_i} -
                E\left( X \right)} \right)}^2}}\\
      V\left( {aX + b} \right) &= {a^2}V\left( X \right)
    \end{aligned}
    $
    
  \item $
    \begin{aligned}[t]
      \sigma \left( {aX + b} \right) &= \sqrt {V\left( {aX + b}
        \right)}\\
      &= \sqrt {{a^2}V\left( X \right)} = \left| a \right|\sqrt
      {V\left( X \right)}\\
      \sigma \left( {aX + b} \right)&= \left| a \right|\sigma \left( X \right)
    \end{aligned}
    $
  \end{itemize}
\end{remarque}
\begin{exemple}
  On donne $E(X) = 3$ et $V(X) =16$.  

  Calculer: 
  \begin{enumerate}
  \item $E(-2X + 5)$
  \item $V(-2X + 5)$
  \item $\sigma(-2X + 5)$
  \end{enumerate}

  \correction 

  \begin{enumerate}
  \item $E(-2X + 5) = -2E(X) + 5 = -2 \times 3 + 5 = -1$
  \item $V(-2X + 5) = (-2)^2 \times V(X) = 4 \times 16 = 64$
  \item $\sigma(-2X + 5) = |-2| \sigma (X) = 2\sqrt{V(X)} = 2 \times \sqrt{16} = 8$
  \end{enumerate}
\end{exemple}
