The probability density function of the random variable \(X\) is
$$\mathrm { f } ( x ) = \frac { x ^ { k - 1 } \mathrm { e } ^ { - x / \phi } } { \phi ^ { k } ( k - 1 ) ! } , x > 0$$
where \(k\) is a known positive integer and \(\phi\) is an unknown parameter ( \(\phi > 0\) ). Show that the moment generating function (mgf) of \(X\) is
$$\mathrm { M } _ { X } ( \theta ) = ( 1 - \phi \theta ) ^ { - k }$$
for \(\theta < \frac { 1 } { \phi }\).
Write down the mgf of the random variable \(W = \sum _ { i = 1 } ^ { n } X _ { i }\) where \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\) are independent random variables each with the same distribution as \(X\).
Write down the mgf of the random variable \(Y = \frac { 2 W } { \phi }\). Given that the mgf of the random variable \(V\) having the \(\chi _ { m } ^ { 2 }\) distribution is \(\mathrm { M } _ { V } ( \theta ) = ( 1 - 2 \theta ) ^ { - m / 2 }\) (for \(\theta < \frac { 1 } { 2 }\) ), deduce the distribution of \(Y\).
Deduce that \(\mathrm { P } \left( l < \frac { 2 W } { \phi } < u \right) = 0.95\) where \(l\) and \(u\) are the lower and upper \(2 \frac { 1 } { 2 } \%\) points of the \(\chi _ { 2 n k } ^ { 2 }\) distribution. Hence deduce that a \(95 \%\) confidence interval for \(\phi\) is given by \(\left( \frac { 2 w } { u } , \frac { 2 w } { l } \right)\) where \(w\) is an observation on the random variable \(W\).
For the case \(k = 2\) and \(n = 10\), use percentage points of the \(\chi ^ { 2 }\) distribution to write down, in terms of \(w\), an expression for a \(95 \%\) confidence interval for \(\phi\). By considering the \(\operatorname { mgf }\) of \(W\), find in terms of \(\phi\) the expected length of this interval.