We are given that $X$ is a gamma-distributed random variable with the probability density function (PDF):

$f(x) = \frac{b^a}{\Gamma(a)} x^{a-1} e^{-bx}, \quad 0 < x < \infty,$ where $a > 0$ and $b > 0$, and $\Gamma(a)$ is the Gamma function:

$\Gamma(a) = \int_0^\infty x^{a-1} e^{-x} dx.$

We want to compute the moment-generating function (MGF) or Laplace transform $G_X(t) = E(e^{tX})$ and use it to find the mean $E(X)$ and variance $\text{Var}(X)$.

1. Moment-Generating Function (MGF)

The moment-generating function $G_X(t)$ is defined as the expected value of $e^{tX}$:

$G_X(t) = E(e^{tX}) = \int_0^\infty e^{tx} f(x) dx.$

Substitute the PDF of the gamma distribution:

$G_X(t) = \int_0^\infty e^{tx} \frac{b^a}{\Gamma(a)} x^{a-1} e^{-bx} dx.$

This simplifies to:

$G_X(t) = \frac{b^a}{\Gamma(a)} \int_0^\infty x^{a-1} e^{-(b-t)x} dx.$

The integral $\int_0^\infty x^{a-1} e^{-cx} dx$ is known as the Gamma function $\Gamma(a)$, evaluated at $c$:

$\int_0^\infty x^{a-1} e^{-cx} dx = \frac{\Gamma(a)}{c^a}, \quad \text{for } c > 0.$

Here, $c = b - t$, so we require $b - t > 0$ or $t < b$. Applying this result:

$G_X(t) = \frac{b^a}{\Gamma(a)} \cdot \frac{\Gamma(a)}{(b-t)^a}.$

Simplifying:

$G_X(t) = \left( \frac{b}{b-t} \right)^a, \quad \text{for } t < b.$

Thus, the moment-generating function is:

$G_X(t) = \left( \frac{b}{b-t} \right)^a, \quad t < b.$

2. Finding $E(X)$ and $\text{Var}(X)$

To find the mean $E(X)$ and variance $\text{Var}(X)$, we use the properties of the MGF:

• The mean $E(X)$ is given by:

$E(X) = G_X'(0) = \left. \frac{d}{dt} \left( \frac{b}{b-t} \right)^a \right|_{t=0}.$

First, differentiate $G_X(t) = \left( \frac{b}{b-t} \right)^a$ with respect to $t$:

$G_X'(t) = a \left( \frac{b}{b-t} \right)^{a+1} \cdot \frac{1}{(b-t)^2}.$

At $t = 0$:

$G_X'(0) = a \cdot \frac{1}{b}.$

Thus, the mean is:

$E(X) = \frac{a}{b}.$

• The second moment $E(X^2)$ is found by differentiating the MGF twice:

$E(X^2) = G_X''(0).$

The variance is:

$\text{Var}(X) = E(X^2) - \left( E(X) \right)^2.$

Differentiating the MGF a second time:

$G_X''(t) = a(a+1) \left( \frac{b}{b-t} \right)^{a+2} \cdot \frac{1}{(b-t)^4}.$

At $t = 0$:

$G_X''(0) = a(a+1) \cdot \frac{1}{b^2}.$

Thus:

$E(X^2) = \frac{a(a+1)}{b^2}.$

Finally, the variance is:

$\text{Var}(X) = E(X^2) - \left( E(X) \right)^2 = \frac{a(a+1)}{b^2} - \left( \frac{a}{b} \right)^2 = \frac{a}{b^2}.$

Conclusion:

• The mean $E(X) = \frac{a}{b}$.
• The variance $\text{Var}(X) = \frac{a}{b^2}$.

Do you want any further details or have any questions?

Here are 5 related questions:

1. How would the MGF change if $a$ or $b$ were negative?
2. How can the MGF be used to calculate higher moments (e.g., skewness)?
3. What are the conditions on the parameters $a$ and $b$ to ensure the existence of the MGF?
4. How does the Gamma distribution relate to other distributions, such as the exponential distribution?
5. Can you derive the MGF for other well-known distributions, such as the normal distribution?

Tip: The moment-generating function can be used to derive all moments of a random variable by differentiating it successively at $t = 0$.