tex: NFDBP add expected value T_i

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Paul ALNET 2023-06-04 22:16:30 +02:00
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@ -396,9 +396,32 @@ Finally, plugging this into \eqref{eq:prob} gives us
P(T_i = k) = P(A_{k-1}) - P(A_{k}) = \frac{1}{(k-1)!} - \frac{1}{k!} \qquad \forall k \geq 2 P(T_i = k) = P(A_{k-1}) - P(A_{k}) = \frac{1}{(k-1)!} - \frac{1}{k!} \qquad \forall k \geq 2
\] \]
\subsection{Expected value of $ T_i $}
We now compute the expected value $ \mu $ and variance $ \sigma^2 $ of $ T_i $.
\begin{align*}
\mu = E(T_i) & = \sum_{k=2}^{\infty} k \cdot P(T_i = k) \\
& = \sum_{k=2}^{\infty} (\frac{k}{(k-1)!} - \frac{1}{(k-1)!}) \\
& = \sum_{k=2}^{\infty} \frac{k-1}{(k-1)!} \\
& = \sum_{k=0}^{\infty} \frac{1}{k!} \\
& = e \\
\end{align*}
\begin{align*}
E({T_i}^2) & = \sum_{k=2}^{\infty} k^2 \cdot P(T_i = k) \\
& = \sum_{k=2}^{\infty} (\frac{k^2}{(k-1)!} - \frac{k}{(k-1)!}) \\
& = \sum_{k=2}^{\infty} \frac{(k-1)k}{(k-1)!} \\
& = \sum_{k=2}^{\infty} \frac{k}{(k-2)!} \\
& = \sum_{k=0}^{\infty} \frac{k+2}{k!} \\
& = \sum_{k=0}^{\infty} (\frac{1}{(k-1)!} + \frac{2}{(k)!}) \\
& = \sum_{k=0}^{\infty} \frac{1}{(k)!} - 1 + 2e \\
& = 3e - 1
\end{align*}
\begin{align*}
\sigma^2 = E({T_i}^2) - E(T_i)^2 = 3e - 1 - e^2
\end{align*}
\section{Complexity and implementation optimization} \section{Complexity and implementation optimization}