The random variable X has moment generating function
$$m_X(t) = \frac{6}{\sqrt{(1-t)(4-t)(9-t)}}$$
for $t < 1$
Compute $E(X)$
Solution:
This can be rewritten to:
$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} $
Having the distribution of $Gamma(\alpha = 1, \lambda = 1/2), Gamma(\alpha = 4, \lambda = 1/2), Gamma(\alpha = 9, \lambda = 1/2)$
Since Gamma has $E(X) = \frac{\alpha}{\lambda} = 0.5/1 + 0.5/4 + 0.5/9 = 49/72$
I have a few questions about this solution.
Isn't $E(X) = m'_X(0)$?
And I thought the gamma distribution was
$Gamma(a, b) = \frac{b^a x^{a-1} e^{-bx}}{\Gamma{(a)}}$ for $x > 0$
not $Gamma(\alpha = 1, \lambda = 1/2) = \left(\frac{1}{1-t}\right)^{1/2}$
Could someone help me understand this please.
The moment generating function of $Gamma(\alpha = 1, \lambda = \frac12) = \left( \frac1{1-t}\right)^\frac12$, what you have written is the pdf.
There is more than one method to solve the problem.
Their method has identified $X=X_1+X_2+X_3$ where each $X_i$ is a gamma distribution of which we know their mean and to find the expectation of $X$, we just have to sum of expectation of each $X_i$.