Which formula represents the moment generating function (MGF) of a uniform distribution?

Which formula represents the moment generating function (MGF) of a uniform distribution?

• A. Mx(t) = (e^bt - e^at)/(b-a)
• B. Mx(t) = (e^at - e^bt)/(b-a)
• C. Mx(t) = (b-a)e^t
• D. Mx(t) = 1/(b-a)(e^bt - e^at)

Answer: (A)

The moment generating function (MGF) for a continuous uniform distribution defined on the interval \([a, b]\) is derived from the definition of the MGF. The MGF is calculated using the integral of \(e^{tx}\) times the probability density function (pdf) over the interval of interest. For a uniform distribution, the pdf is constant between \(a\) and \(b\) and equals \(\frac{1}{b-a}\) for \(x \in [a, b]\). The MGF is thus calculated as follows: \[ Mx(t) = \int_a^b e^{tx} \cdot \frac{1}{b-a} \, dx \] Evaluating this integral, the exponential function yields \(e^{tx}\) evaluated from \(a\) to \(b\). Therefore, the result becomes: \[ Mx(t) = \frac{1}{b-a} \left(e^{tb} - e^{ta}\right) \] This can also be rewritten to align with the format given in the choices. Specifically, if we multiply and reorganize slightly, we find: \[ Mx(t) = \frac{e^{tb} - e

The moment generating function (MGF) for a continuous uniform distribution defined on the interval ([a, b]) is derived from the definition of the MGF. The MGF is calculated using the integral of (e^{tx}) times the probability density function (pdf) over the interval of interest.

For a uniform distribution, the pdf is constant between (a) and (b) and equals (\frac{1}{b-a}) for (x \in [a, b]). The MGF is thus calculated as follows:

[

Mx(t) = \int_a^b e^{tx} \cdot \frac{1}{b-a} , dx

]

Evaluating this integral, the exponential function yields (e^{tx}) evaluated from (a) to (b). Therefore, the result becomes:

[

Mx(t) = \frac{1}{b-a} \left(e^{tb} - e^{ta}\right)

]

This can also be rewritten to align with the format given in the choices. Specifically, if we multiply and reorganize slightly, we find:

[

Mx(t) = \frac{e^{tb} - e