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    The question is to find the MGF of X1.X2, where X1 and X2 are i.i.d. random variables following the uniform distribution with parameters a and b, so having MGF:
    \frac{e^{\theta b} - e^{\theta a}}{\theta (b-a)}

    Can you check this working?


\hspace*{40mm}\mathbb{E} \left[ e^{\theta X_1.X_2}\right] = \mathbb{E}_{X_2}\left[ \mathbb{E}_{X_1}\left[ e^{\theta X_1.x_2} \right] \right]

\hspace*{60mm} = \mathbb{E}_{X_2}\left[ \mathbb{E}_{X_1}\left[ e^{(\theta x_2) X_1} \right] \right]


\hspace*{60mm} = \mathbb{E}_{X_2}\left[ \frac{e^{\theta b X_2} - e^{\theta a X_2}}{\theta (b-a) X_2} \right]

\hspace*{60mm} = \left(\frac{1}{b-a}\right) \int_a^b \frac{e^{\theta b x_2} - e^{\theta a x_2}}{\theta (b-a) x_2} \text{  d}x_2

\hspace*{60mm} = \left(\frac{1}{\theta (b-a)^2}\right) \int_a^b \left[ \frac{e^{\theta b x_2}}{x_2} - \frac{e^{\theta a x_2}}{x_2} \right] \text{  d}x_2

    And then I think you can't do the rest because that would involve the exponential integral function (Ei)?
    Does anyone know how to express the answer using that function though?

    Or can you do this (I think line 2 is wrong):

     \mathbb{E} \left[ e^{\theta X_1.X_2}\right] = \mathbb{E}_{X_2}\left[ \mathbb{E}_{X_1}\left[ e^{\theta X_1.x_2} \right] \right]
     \hspace*{20mm} = \mathbb{E}_{X_2}\left[ \left( \mathbb{E}_{X_1}\left[ e^{\theta X_1 \right] }\right)^{x_2} \right]
     \hspace*{20mm} = \mathbb{E}_{X_2}\left[ \left( \frac{e^{\theta b} - e^{\theta a}}{\theta (b-a)} \right)^{X_2} \right]
     \hspace*{20mm} = \left(\frac{1}{b-a}\right) \int_a^b \left( \frac{e^{\theta b} - e^{\theta a}}{\theta (b-a)} \right)^{x_2}\text{ d}x_2
     \hspace*{20mm} = \left(\frac{1}{b-a}\right) \frac{1}{\log\left( \frac{e^{\theta b} - e^{\theta a}}{\theta (b-a)} \right)}\ \left[ \left(\frac{e^{\theta b} - e^{\theta a}}{\theta (b-a)} \right)^{x_2} \right]^b_a  \hspace*{20mm} = \left(\frac{1}{b-a}\right) \frac{1}{\log\left( \frac{e^{\theta b} - e^{\theta a}}{\theta (b-a)} \right)}\ \left[ \left(\frac{e^{\theta b} - e^{\theta a}}{\theta (b-a)} \right)^b - \left(\frac{e^{\theta b} - e^{\theta a}}{\theta (b-a)} \right)^a\right]
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Updated: January 1, 2013
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