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Chapter 2: Problem 60

Calculate the moment generating function of the uniform distribution on \((0,1)\). Obtain \(E[X]\) and \(\operatorname{Var}[X]\) by differentiating.

Short Answer

Expert verified
The moment generating function (MGF) of the uniform distribution on \((0,1)\) is \(M_X(t) = \frac{1}{t}(e^t - 1)\). By differentiating, we obtain \(E[X] = \frac{1}{2}\) and \(Var[X] = \frac{1}{12}\).

Step by step solution

01

Find the Probability Density Function (PDF) of X

For the uniform distribution on the interval \((0,1)\), the PDF is as follows: \[f_X(x) = \begin{cases} 1 & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases} \]
02

Define and Calculate the Moment Generating Function (MGF) of X

The MGF of a random variable \(X\) with PDF \(f_X(x)\) is denoted as \(M_X(t)\) and defined as: \[M_X(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx} f_X(x) dx\] Now, we will calculate the MGF of X using the PDF we derived in the previous step: \[M_X(t) = \int_{0}^{1} e^{tx} dx\] To integrate, we will use the formula for the integral of an exponential function: \(\int e^{ax} dx = \frac{1}{a}e^{ax} + C\). Thus, we have: \[M_X(t) = \frac{1}{t} (e^{t} - 1)\]
03

Find E[X] by Differentiating MGF of X

Now, differentiate \(M_X(t)\) with respect to \(t\) and evaluate it at \(t = 0\) to find the expected value of \(X\). \[E[X] = \left.\frac{dM_X(t)}{dt}\right|_{t=0}\] Differentiating, we have: \[\frac{dM_X(t)}{dt} = \frac{e^{t}(t - 1) + 1}{t^2}\] Now, evaluate at \(t = 0\): \[E[X] = \left. \frac{e^{t}(t - 1) + 1}{t^2}\right|_{t=0} = \frac{1}{2}\] Thus, the expected value of \(X\) is \(\frac{1}{2}\).
04

Find Var[X] by Differentiating MGF of X

Now, differentiate the MGF of X once more w.r.t. \(t\) to find \(E[X^2]\) and then use the formula \(Var[X] = E[X^2] - (E[X])^2\) to calculate the variance of \(X\). First, find \(E[X^2]\) by differentiating the MGF of X: \[E[X^2] = \left.\frac{d^2M_X(t)}{dt^2}\right|_{t=0}\] Differentiating, we have: \[\frac{d^2M_X(t)}{dt^2} = -\frac{e^{t}(t^2-2t+2) - 2}{t^3}\] Now, evaluate at \(t = 0\): \[E[X^2] = \left. -\frac{e^{t}(t^2-2t+2) - 2}{t^3}\right|_{t=0} = \frac{1}{3}\] Now, calculate the variance of \(X\) using the formula: \[Var[X] = E[X^2] - (E[X])^2 = \frac{1}{3} - \left(\frac{1}{2}\right)^2 = \frac{1}{12}\] Thus, the variance of \(X\) is \(\frac{1}{12}\).

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