91Ó°ÊÓ

The uniform distribution on \((0, 1)\) has a probability density function given by: \[f(x) = \begin{cases} 1, & \text{if } 0 \le x \le 1 \\ 0, & \text{otherwise} \end{cases}\]

Now that we have the probability density function, we can use the definition of the Moment Generating Function (MGF): \[M(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx}f(x) \, dx\] Using the PDF of the uniform distribution on \((0,1)\), we have: \[M(t) = \int_{0}^{1} e^{tx} \, dx\] To evaluate this integral, we apply the integration by parts: \[\int e^{tx} \, dx = \frac{1}{t}e^{tx} + C\] So, our MGF becomes: \[M(t) = \left[\frac{1}{t}e^{tx}\right]_0^1 = \frac{e^t - 1}{t}\]

Now that we have the moment generating function, we can differentiate to find the mean and variance. First, we obtain the first moment \(E[X]\) by differentiating the MGF with respect to \(t\): \[E[X] = M'(t) = \frac{d}{dt}\left( \frac{e^t - 1}{t} \right)\] Using the quotient rule for differentiation, we have: \[E[X] = \frac{e^t - te^t}{t^2}\Big|_{t=0}\] \[E[X] = \lim_{t\to 0}\frac{e^t - te^t}{t^2}\] Using L'Hôpital's rule, taking the limit as \(t\) approaches 0: \[E[X] = \lim_{t\to 0}\frac{e^t - e^t - te^t + e^t}{2t} = \frac{1}{2}\] Next, we obtain the second moment \(E[X^2]\) by differentiating the MGF again: \[E[X^2] = M''(t) = \frac{d^2}{dt^2}\left( \frac{e^t - 1}{t} \right)\] Using the quotient rule for differentiation and taking the limit as \(t\) approaches 0: \[E[X^2] = \lim_{t\to 0}\frac{2e^t - te^t - e^t + e^t}{t^3}\] \[E[X^2] = \lim_{t\to 0} \frac{e^t(2-t-e^{t})}{t^3}\] Applying L'Hôpital's rule twice: \[E[X^2] = \lim_{t\to 0} \frac{6 - 6t + t^2}{6} = \frac{1}{3}\] Finally, we can find the variance \(\operatorname{Var}[X]\) using the obtained moments: \[\operatorname{Var}[X] = E[X^2] - \left(E[X]\right)^2 = \frac{1}{3} - \left(\frac{1}{2}\right)^2 = \frac{1}{12}\] So, the mean of the uniform distribution on \((0,1)\) is \(E[X] = \frac{1}{2}\) and its variance is \(\operatorname{Var}[X] = \frac{1}{12}\).