Question

(a) Show that the moment generating function of a uniform random variable \newline X~Uniform(a, b), with probability density function \newline $p_X(x) = \begin{cases} \frac{1}{b-a}, & a \le x \le b \\ 0, & \text{otherwise} \end{cases}$ \newline is given by \newline $M_X(t) = \frac{e^{tb} - e^{ta}}{t(b-a)}$. \newline (b) Show, that the first moment (i.e. the expected value) of a uniform random variable \newline X~ Uniform(a, b) is given by \newline $E[X] = \frac{a+b}{2}$, \newline and that the variance of a uniform random variable X ~ Uniform(a, b), denoted \newline $\sigma_X^2$, is given by \newline $\sigma_X^2 = \frac{(b-a)^2}{12}$.

          (a) Show that the moment generating function of a uniform random variable \newline X~Uniform(a, b), with probability density function \newline $p_X(x) = \begin{cases} \frac{1}{b-a}, & a \le x \le b \\ 0, & \text{otherwise} \end{cases}$ \newline is given by \newline $M_X(t) = \frac{e^{tb} - e^{ta}}{t(b-a)}$. \newline (b) Show, that the first moment (i.e. the expected value) of a uniform random variable \newline X~ Uniform(a, b) is given by \newline $E[X] = \frac{a+b}{2}$, \newline and that the variance of a uniform random variable X ~ Uniform(a, b), denoted \newline $\sigma_X^2$, is given by \newline $\sigma_X^2 = \frac{(b-a)^2}{12}$.

(a) Show that the moment generating function of a uniform random variable X Uniform(a, b), with probability density function pX(x) = (1)/(b-a), a ≤ x ≤ b
0, otherwise is given by MX(t) = (e^tb - e^ta)/(t(b-a)). (b) Show, that the first moment (i.e. the expected value) of a uniform random variable X Uniform(a, b) is given by E[X] = (a+b)/(2), and that the variance of a uniform random variable X Uniform(a, b), denoted ^2, is given by ^2 = ((b-a)^2)/(12).

Added by Craig J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ameer Said

03/01/2022

Step 1

First, we need to find the moment generating function (MGF) of a uniform random variable X with probability density function (PDF) given by: $$ P_X(x) = \begin{cases} \frac{1}{b-a} & \text{if } a \le x \le b \\ 0 & \text{otherwise} \end{cases} $$ Show more…

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Show that the moment generating function of a uniform random variable Uniform(a,b) with probability density function Px (x) = 1/(b-a) for a â‰¤ x â‰¤ b, otherwise 0, is given by Mx (t) = (e^(bt) - e^(at))/(t(b-a)). [6] Show that the first moment (i.e. the expected value) of a uniform random variable Uniform(a,b) is given by E[X] = (a + b)/2, and that the variance of uniform random variable X is given by (b-a)^2/12. Uniform(a, b), denoted as U(a, b). [10]