Finding $E(X)$ using moment generating function

Question

Say:

$m_X(t) = \left(\frac{1}{1-t}\right)^{1/2} \cdot \left(\frac{4}{4-t}\right)^{1/2} \cdot \left(\frac{9}{9-t}\right)^{1/2} $

We want. $E(X)$.

The only theorem in my textbook is that

$m_X^k(0) = E(X^k)$

As in, the derivative is proportional to the power k to $E(X)$.

Gamma is: $X$ ~ $Gamma(\alpha, \lambda)$ with mgt $$\left(\frac{\lambda}{\lambda-t}\right)^\alpha$$

so $m_X(t) = Gamma(\alpha=1/2, \lambda = 1) \cdot Gamma(\alpha=1/2, \lambda = 4) \cdot Gamma(\alpha=1/2, \lambda = 9)$.

$E(X) = \frac{\alpha}{\lambda}$

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Solution:

$E(X) = 1/2/1 + 1/2/4 + 1/2/9 = 49/72$

What formula does this use to get above?

Answer 1

Hint: Using the theorem, with $k=1$ you get that $m_X'(0) = E(X)$, so find the first derivative of $m_X(t)$ at $0$ to get the answer.

Answer 2

The case $k=1$ gives $E(X)=m_X'(0)$. Since $m_X(0)$, $E(X)$ is also the value at $t=0$ of $$\dfrac{d}{dt}\ln m_X=\dfrac{-1}{2}\dfrac{d}{dt}(\ln (1-t)+\ln (4-t)+\ln (9-t))=\frac{1}{2}(\dfrac{1}{1-t}+\dfrac{1}{4-t}+\dfrac{1}{9-t}).$$For $t=0$, this gives $\tfrac{49}{72}$.