found mgf, but $E(x^3)$ is not defined

Question

A question asks me to find the MGF for the continuous random variable $f(x) = 3x^2$ on $[0,1]$. Using the MGF, it asks me to find $E[X^3]$. I'm having trouble evaluating $E[X^3]$

I found the MGF using integration by parts twice, and got:

$$(e^t(3t^2-6t+6)-2)/t^3$$

Now to find $E[X^3]$, I would take the 3rd derivative of the MGF and evaluate it at $t = 0$. The problem is there is a $t$ in the denominator, which would give an invalid result. How do I calculate $E[X^3]$? Is there something I did wrong in deriving the MGF that results in this problem?

Answer 1

$$ M_X (t )=\Bbb E (e^{tX})=\int_0^1 3x^2 e^{t x} dx= \frac{3e^t (t^2-2t+2)-6}{t^3}=1+\frac{3t}{4}+\frac{3t^2}{10}+\frac{t^3}{12}+o (t^4) $$ where we used the Taylor expansion $$ e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\frac{t^5}{5!}+\frac{t^6}{6!}+o (t^7) $$

Thus $$\Bbb E (X^3)= M^{(3)}(0)=3!\frac{1}{12}=\frac{1}{2}$$

Answer 2

Assuming I haven't make a mistake (big assumption), I get $E[tf(X)] = {2 \over 3} { e^{3t}(3t-1)+1 \over t^2}$, this has a removable discontinuity at $t=0$, and expanding the first few terms gives $E[tf(X)] = 1 + 2t + {9 \over 4} t^2 + {9 \over 5} t^3 + \cdots$.

Alternative:

I suspect that $f$ is the cdf. of the distribution in which case we have $E[e^{tX}] = \int_0^1 e^{tx} 3 x^2 dx$. Then we have $E[e^{tX}] = \sum_{k=0}^\infty t^k \int_0^1 {x^k \over k!} 2 x^2 dx = \sum_{k=0}^\infty t^k {1\over k!} E X^k = \sum_{k=0}^\infty {3 \over k! (2+k+1)}t^k $.

Hence $E X^k = {3 \over (2+k+1)}$, in particular, $EX^3 = {1 \over 2}$.