(Solution Library) Let y_X(t)=ln [M_X(t)], where M_X(t) is a mfg. The function y_X(t) is called the cumulant generating function of X, and the value of the
Question: Let \(y_{X}(t)=\ln \left[M_{X}(t)\right]\), where \(M_{X}(t)\) is a mfg. The function \(y_{X}(t)\) is called the cumulant generating function of \(X\), and the value of the \(r\) th derivative evaluated at \(t=0\), \(\mathrm{K}_{r}=\psi_{X}^{(y)}(0)\), is called the \(\mathrm{r}\) th cumulant of \(\mathrm{X}\)
- Show that \(\mu=\psi_{X}^{\prime}(0)\).
- Show that \(\sigma^{2}=\psi_{X}^{\prime \prime}(0)\) "
-
Use \(y_{X}(t)\) to find \(\mathrm{m}\) and \(\mathrm{s}^{2}\) for the random variable with \(\mathrm{pdf}_{,} f_{X}(x)=e^{-(x+2)}\) if \(-2
Price: $2.99
Solution: The downloadable solution consists of 2 pages
Deliverable: Word Document 