MGF series expansion

6 The random variable \(X\) has a uniform distribution on the interval \([ - 1,1 ]\), so that its probability density function is given by $$f ( x ) = \begin{cases} \frac { 1 } { 2 } & - 1 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$

1. Show from the definition of the moment generating function that the moment generating function of \(X\) is \(\frac { \sinh t } { t }\).
2. By using the series expansion of \(\sinh t\), find the variance of \(X\) and the value of \(\mathrm { E } \left( X ^ { 4 } \right)\).

The continuous random variable \(X\) has probability density function given by $$f(x) = \begin{cases} 3e^{-x} & 0 \leq x \leq k, \\ 0 & \text{otherwise,} \end{cases}$$ where \(k\) is a constant.

1. Show that \(e^{-k} = \frac{2}{3}\). [2]
2. Show that the moment generating function of \(X\) is given by \(M_X(t) = \frac{3}{1-t}\left(1 - \frac{2}{3}e^{kt}\right)\). [4]
3. By expanding \(M_X(t)\) as a power series in \(t\), up to and including the term in \(t^2\), show that $$M_X(t) = 1 + (1 - 2k)t + (1 - 2k - k^2)t^2 + \ldots.$$ [3] [You may use the standard series for \((1-t)^{-1}\) and \(e^{kt}\) without proof.]
4. Deduce that the exact value of E\((X)\) is \(1 - 2\ln\left(\frac{2}{3}\right)\). [1]