Using Continuous Uniform MGF to find E(X)

[Darth Frodo]

Continuous Uniform MGF is $M_{x}(z) = E(e^zx) = \frac{e^{zb} - e^{za}}{zb - za}$

$\frac{d}{dz}M_{x}(z) = E(X)$

Using the Product Rule

$\ U = e^{bz} - e^{az}$

$\ V = (zb - za)^{-1}$

$\ U' = be^{bz} - ae^{az}$

$\ V' = -1(zb - za)^{-2}(b - a)$

$\frac{dM}{dz} = UV' + VU'$

$\frac{dM}{dz} = (e^{bz} - e^{az})(-1(zb - za)^{-2}(b - a)) + ((zb - za)^{-1})(be^{bz} - ae^{az})$ evaluated at z = 1


$\ (e^{b}-e^{a})(-1)(b - a)^{-2}(b - a) + (b - a)^{-1}(be^{b} - ae^{a})$


$\frac{e^{a} - e^{b} + be^{b} - ae^{a} }{b - a}$


The answer is $\frac{b + a}{2}$


I'd really appreciate it if someone could tell me where I'm going wrong. Thanks.

 

[brmath]

You didn't define your terms. What is MGF and what is E(X). These may be standard, but I don't know them.

 

[Darth Frodo]

Moment Generating Function and Expected Value of the Continuous random variable X

 

[Ray Vickson]

Continuous Uniform MGF is $M_{x}(z) = E(e^zx) = \frac{e^{zb} - e^{za}}{zb - za}$

$\frac{d}{dz}M_{x}(z) = E(X)$

Using the Product Rule

$\ U = e^{bz} - e^{az}$

$\ V = (zb - za)^{-1}$

$\ U' = be^{bz} - ae^{az}$

$\ V' = -1(zb - za)^{-2}(b - a)$

$\frac{dM}{dz} = UV' + VU'$

$\frac{dM}{dz} = (e^{bz} - e^{az})(-1(zb - za)^{-2}(b - a)) + ((zb - za)^{-1})(be^{bz} - ae^{az})$ evaluated at z = 1


$\ (e^{b}-e^{a})(-1)(b - a)^{-2}(b - a) + (b - a)^{-1}(be^{b} - ae^{a})$


$\frac{e^{a} - e^{b} + be^{b} - ae^{a} }{b - a}$


The answer is $\frac{b + a}{2}$


I'd really appreciate it if someone could tell me where I'm going wrong. Thanks.

You need to evaluate M'(z) at z = 0, not at z = 1! Of course, you need to worry about the fact that your formula for M(z) does not apply right at z = 0, so you need to look at limits as z → 0, and use l'Hospital's rule, for example. BTW: the formula you wrote for EX is wrong; it should be
$$EX = \left. \frac{d M(z)}{dz} \right|_{z = 0} = M'(0).$$ For $z \neq 0$ the derivative of M does not have any particular relation to EX.

However, that would be doing it the hard way. From
$$M_X(z) = E e^{zX} = 1 + z \,EX + \frac{z^2}{2} E X^2 + \cdots,$$
we see that we can get the moments of X from the moment-generating-function---that's why it is so named---so all you need is the series expansion of M(z) around z = 0 (that is, the Maclaurin series). That is most easily obtained by just getting the series expansion of the numerator, then dividing by z.