long length of yarn at the tex7le mill. The yarn is then cut into pieces. The lengths of the different pieces are independent, and the length of each piece is distributed according to the same PDF $f_X(x)$. Let $R$ be the length of the piece that includes the dot. Determine the expected value of $R$ in each of the following cases. In each part below, express your answer in terms of $\lambda$ using \edXxml{<a href="/courses/MITx/6.041x/1T2014/courseware/Unit_0_Overview/ Homework_mechanics_and_standard_nota7on6/2" target="_blank">standard nota7on</a>}. Enter 'lambda' for $\lambda$. \begin{enumerate} \item Suppose that $f_X(x) = \begin{cases}\lambda e^{-‐\lambda x}, & x\geq 0,\\ 0, & x<0.\end{cases}$ \edXinline{$\E[R]=\,$} \edXabox{type="formula" size="20" expect="2/lambda" samples="lambda@0:10#20" tolerance="3\%" math="1" inline="1"} \item Suppose that $f_X(x) = \begin{cases}\frac{\lambda^3 x^2 e^{-‐\lambda x}}{2}, & x\geq 0,\\ 0, & x<0.\end{cases}$ \edXinline{$\E[R]=\,$} \edXabox{type="formula" size="20" expect="4/lambda" samples="lambda@0:10#20" tolerance="3\%" math="1" inline="1"} \end{enumerate} \begin{edXsolu.on} \begin{enumerate} \item Here, the lengths of the pieces of yarn are independent and exponen7ally distributed with parameter $\lambda$. As explained on pages 322-‐324 of the text, due to the memorylessness of the exponen7al, the distribu7on of the length of the piece of yarn containing the dot is a second order Erlang. Thus, $\E[R]=2\E[X]=2/\lambda$. \item Here, $X$ is an Erlang of order 3. Think of sec7ons on the yarn, each exponen7ally distributed with parameter $\lambda$. We can then interpret each piece of yarn as {\em three} consecu7ve sec7ons of exponen7ally distributed lengths. The piece of yarn with the dot will have the dot in one of these three sec7ons. By the standard random incidence analysis, the expected length of that sec7on will be $2/\lambda$. However, the piece of yarn containing the dot also consists of two other sec7ons, each with an expected length of $1/\lambda$. Thus, the total expected length of the piece of yarn containing the dot is $4/\lambda$. %In general, for processes in which the interarrival interval lengths are i.i.d.\ with common distribu7on $f_X(x)$, the expected length of the interval containing an arbitrary point is $\frac{\E[X^2]} {\E[X]}$. For each part of the present problem, this formula is certainly valid. \end{enumerate} \end{edXsolu.on}
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