Chapter 10.3, Exercise 1, Page 401

Let X be a continuous random variable with values in \([0,2]\) and density \(f_X\) .

Find the moment generating function \(g(t)\) for \(X\) if:

  1. \(f_X(x) = \frac{1}{2}\)
  2. \(f_X(x) = (\frac{1}{2})x\)
  3. \(f_X(x) = 1 - (\frac{1}{2})x\)
  4. \(f_X(x) = |1 - x|\)
  5. \(f_X(x) = (\frac{3}{8})x^2\)

Hint: Use the integral definition, as in Examples 10.15 and 10.16.

The definition of the moment generating function is \(g(t) = \int\limits_{-\infty}^{\infty} e^{tx} \cdot f_X(x) dx\) .

Here, because the support is \([0,2]\) , the MGF is \(g(t) = \int\limits_{0}^{2} e^{tx} \cdot f_X(x) dx\) .

(a) \(f_X(x) = \frac{1}{2}\)

\[\begin{aligned} g(t) &= \int\limits_{0}^{2} e^{tx} \cdot \frac{1}{2} dx \\ \\ &= \left.\frac{e^{tx}}{2t}\right|_{x=0}^{x=2} \\ \\ &= \frac{e^{2t}-e^{0t}}{2t} \\ \\ &= \frac{e^{2t}-1}{2t} \end{aligned}\]

(b) \(f_X(x) = (\frac{1}{2})x\)

\[\begin{aligned} g(t) &= \int\limits_{0}^{2} e^{tx} \cdot (\frac{1}{2})x dx \\ \\ &= \left.\frac{e^{tx}(tx-1)}{2t^2}\right|_{x=0}^{x=2} \\ \\ &= \frac{e^{2t}(2t-1)-e^{0t}(0t-1)}{2t^2} \\ \\ &= \frac{e^{2t}(2t-1)+1}{2t^2} \end{aligned}\]

(c) \(f_X(x) = 1 - (\frac{1}{2})x\)

\[\begin{aligned} g(t) &= \int\limits_{0}^{2} e^{tx} \cdot \left[1 - (\frac{1}{2})x\right] dx \\ \\ &= \int\limits_{0}^{2} e^{tx} dx - \int\limits_{0}^{2} e^{tx} \cdot (\frac{1}{2})x dx \\ \\ &= \left.\frac {e^{tx}}{t} \right|_{x=0}^{x=2} - \frac{e^{2t}(2t-1)+1}{2t^2} \\ \\ &= \frac {e^{2t}-1}{t} - \frac{e^{2t}(2t-1)+1}{2t^2} \\ \\ &= \frac {2t(e^{2t}-1)}{2t^2} - \frac{e^{2t}(2t-1)+1}{2t^2} \\ \\ &= \frac {2te^{2t}-2t - \left[2te^{2t}-e^{2t}+1\right]}{2t^2} \\ \\ &= \frac {e^{2t}-2t-1}{2t^2} \\ \end{aligned}\]

(d) \(f_X(x) = |1 - x|\)

\[\begin{aligned} g(t) &= \int\limits_{0}^{2} e^{tx} \cdot |1 - x| dx \\ \\ &= \int\limits_{0}^{1} e^{tx} \cdot (1 - x) dx + \int\limits_{1}^{2} e^{tx} \cdot (x - 1) dx \\ \\ &= \int\limits_{0}^{1} e^{tx} dx - \int\limits_{0}^{1} xe^{tx} dx + \int\limits_{1}^{2} xe^{tx} dx - \int\limits_{1}^{2} e^{tx} dx \\ \\ &= \left.\frac {e^{tx}}{t} \right|_{x=0}^{x=1} - \left.\frac{e^{tx}(tx-1)}{t^2}\right|_{x=0}^{x=1} + \left.\frac{e^{tx}(tx-1)}{t^2}\right|_{x=1}^{x=2} - \left.\frac {e^{tx}}{t} \right|_{x=1}^{x=2} \\ \\ &= \frac {e^{t}-1}{t} - \frac{e^{t}(t-1) - e^{0}(0-1) }{t^2} + \frac{e^{2t}(2t-1) - e^{t}(t-1) }{t^2} - \frac {e^{2t}-e^{t}}{t} \\ \\ &= \frac {-e^{2t}+2e^{t}-1}{t} + \frac{e^{2t}(2t-1) - 2e^{t}(t-1) -1 }{t^2} \\ \\ &= \frac {-te^{2t}+2te^{t}-t}{t^2} + \frac{2te^{2t}-e^{2t} - 2te^{t}+2e^{t} -1 }{t^2} \\ \\ &= \frac{te^{2t}-e^{2t} +2e^{t} -t-1 }{t^2} \\ \\ &= \frac{e^{2t}(t-1) +2e^{t} -t-1 }{t^2} \\ \end{aligned}\]

(e) \(f_X(x) = (\frac{3}{8})x^2\)

\[\begin{aligned} g(t) &= \int\limits_{0}^{2} e^{tx} \cdot (\frac{3}{8})x^2 dx \\ \\ &= \frac{3}{8}\int\limits_{0}^{2} x^2 e^{tx} dx \\ \\ &= \frac{3}{8} \left[e^{tx} \left( \frac{x^2}{t} - \frac{2x}{t^2} + \frac{2}{t^3} \right) \right]_{x=0}^{x=2} \\ \\ &= \frac{3}{8} \left[e^{2t} \left( \frac{4}{t} - \frac{4}{t^2} + \frac{2}{t^3} \right) - e^{0} \left( \frac{0}{t} - \frac{0}{t^2} + \frac{2}{t^3} \right) \right] \\ \\ &= \frac{3}{8} \left[e^{2t} \left( \frac{4}{t} - \frac{4}{t^2} + \frac{2}{t^3} \right) - \left( \frac{2}{t^3} \right) \right] \\ \\ &= \frac{3}{8} \left[e^{2t} \left( \frac{4t^2-4t+2}{t^3} \right) - \left( \frac{2}{t^3} \right) \right] \\ \\ &= \frac{3}{8t^3} \left[e^{2t} \left( 4t^2-4t+2 \right) - 2 ) \right] \\ \end{aligned}\]