Deriving moment generating function

2 [In this question, you may use the result \(\int _ { 0 } ^ { \infty } u ^ { m } \mathrm { e } ^ { - u } \mathrm {~d} u = m\) ! for any non-negative integer \(m\).]
The random variable \(X\) has probability density function $$\mathrm { f } ( x ) = \begin{cases} \frac { \lambda ^ { k + 1 } x ^ { k } \mathrm { e } ^ { - \lambda x } } { k ! } , & x > 0 \\ 0 , & \text { elsewhere } \end{cases}$$ where \(\lambda > 0\) and \(k\) is a non-negative integer.

1. Show that the moment generating function of \(X\) is \(\left( \frac { \lambda } { \lambda - \theta } \right) ^ { k + 1 }\).
2. The random variable \(Y\) is the sum of \(n\) independent random variables each distributed as \(X\). Find the moment generating function of \(Y\) and hence obtain the mean and variance of \(Y\). [8]
3. State the probability density function of \(Y\).
4. For the case \(\lambda = 1 , k = 2\) and \(n = 5\), it may be shown that the definite integral of the probability density function of \(Y\) between limits 10 and \(\infty\) is 0.9165 . Calculate the corresponding probability that would be given by a Normal approximation and comment briefly.

2

1. 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 \(\theta < \frac { 1 } { \phi }\).
2. 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 \(X\).
3. 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 \(Y\).
4. 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 \(W\).
5. 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.