Challenging +1.2 This is a standard transformation of random variables question requiring the CDF method and then a straightforward expectation calculation. Part (a) involves routine application of Y = g(X) transformation technique (finding G(y) then differentiating), and part (b) is a direct integration using the derived pdf. While it requires multiple steps and careful algebra, the techniques are well-practiced in FM Statistics with no novel insight needed.
6. The continuous random variable \(X\) has (cumulative) distribution function given by
$$F ( x ) = \left\{ \begin{array} { c c }
0 & x < 1 \\
1 - \frac { 1 } { x ^ { 4 } } & x \geq 1
\end{array} \right.$$
a. Show that the probability density function of \(Y\), where \(Y = \frac { 1 } { X ^ { 2 } }\), is given by
$$g ( y ) = \left\{ \begin{array} { c c }
2 y & 0 < y \leq 1 \\
0 & \text { otherwise }
\end{array} \right.$$
b. Find \(\mathrm { E } ( \sqrt [ 3 ] { Y } )\).
6. The continuous random variable $X$ has (cumulative) distribution function given by
$$F ( x ) = \left\{ \begin{array} { c c }
0 & x < 1 \\
1 - \frac { 1 } { x ^ { 4 } } & x \geq 1
\end{array} \right.$$
a. Show that the probability density function of $Y$, where $Y = \frac { 1 } { X ^ { 2 } }$, is given by
$$g ( y ) = \left\{ \begin{array} { c c }
2 y & 0 < y \leq 1 \\
0 & \text { otherwise }
\end{array} \right.$$
b. Find $\mathrm { E } ( \sqrt [ 3 ] { Y } )$.\\
\hfill \mbox{\textit{SPS SPS FM Statistics 2020 Q6 [9]}}