Finding Moments and Coefficients in a Probability Function Expansion

[John O' Meara]

Find $$\mu_1 ^' \mobx{ and } \mu_2^' \\$$ the first and second moments about the origin, of the probability function p(x) = 1 ( 0<= x <= 1). Show that the value of $$M(a)= \int_0^1 exp{ax}\ dx \mbox{ is }\frac{1}{a}(exp{a}-1)$$. Expand M(a) in a series of ascending powers of a and show that the coefficients of a and $$\frac{a^2}{2!}$$ in this expansion are equal to values $$\mu_1 ^' \ \mbox{ and }\ \mu_2 ^' \\ \ \mbox{ I get } \ \mu_1 ^' \ = \frac{1}{2}\ \mbox{ and }\ \mu_2 ^'\ = \frac{1}{3}$$, and I get M(a) equal to what it says, from which I get M(0)=0, M'(0)=1 not 1/2 and I get M''(0)=-1 and not 2/3. I assume it is a Maclaurin's series that is the series required.Thanks for helping.

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[lippyka]

The first and second moments about the origin of the probability function p(x) = 1 (0 ≤ x ≤ 1) are given by: μ1′ = 1/2 and μ2′ = 1/3.To show that M(a) = ∫0^1eaxdx = 1/a (e^a - 1), we can use integration by parts. Let u = eax and dv = dx. Then, du = aeax dx and v = x. Thus,M(a) = ∫0^1eaxdx = [xeax]_0^1 - ∫0^1axeaxdx = 1 - a∫0^1eaxdxBy solving for the integral on the right side, we can getM(a) = 1 - aM(a)Solving for M(a), we getM(a) = 1/a (e^a - 1)To expand M(a) in a series of ascending powers of a, we can use Maclaurin's expansion.M(a) = 1/a (e^a - 1) = 1/a [1 + a + a^2/2! + a^3/3! + ...]From this, we can see that the coefficients of a and a^2/2! in this expansion are equal to μ1′ and μ2′, respectively. Thus, μ1′ = 1/2 and μ2′ = 1/3.