To solve the given problem, we will first compute the moment generating function (MGF) $G_X(t) = \mathbb{E}(e^{tX})$ for the gamma distribution, and then use this MGF to find the expected value $\mathbb{E}(X)$ and variance $\text{Var}(X)$.

Step 1: Deriving the Moment Generating Function (MGF)

The MGF is defined as:

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

where $f(x)$ is the probability density function (PDF) of the gamma distribution:

$f(x) = \frac{b^a}{\Gamma(a)} x^{a-1} e^{-bx}, \quad 0 < x < \infty$

Substitute $f(x)$ into the expression for $G_X(t)$:

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

Simplify the exponent terms $e^{tx} \cdot e^{-bx} = e^{-(b-t)x}$:

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

This integral is of the standard gamma form:

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

Here, $\lambda = b - t$, so we get:

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

Simplifying further:

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

Step 2: Determining the Values of $t$

The MGF exists only when the integral converges, which means $b - t > 0$. Therefore, the MGF is valid for:

$t < b$

Step 3: Using the MGF to Find $\mathbb{E}(X)$ and $\text{Var}(X)$

Finding $\mathbb{E}(X)$

The expectation $\mathbb{E}(X)$ is the first derivative of $G_X(t)$ evaluated at $t = 0$:

$\mathbb{E}(X) = G_X'(0)$

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

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

At $t = 0$:

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

Thus, the expected value is:

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

Finding $\text{Var}(X)$

The variance $\text{Var}(X)$ is computed using the second derivative of the MGF:

$\text{Var}(X) = G_X''(0) - (G_X'(0))^2$

First, compute the second derivative of $G_X(t)$:

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

At $t = 0$:

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

Now, the variance is:

$\text{Var}(X) = G_X''(0) - (G_X'(0))^2 = \frac{a(a+1)}{b^2} - \left( \frac{a}{b} \right)^2$

Simplify:

$\text{Var}(X) = \frac{a}{b^2}$

Final Results

1. Moment generating function (MGF): $G_X(t) = \left( \frac{b}{b-t} \right)^a, \quad t < b$

2. Expected value: $\mathbb{E}(X) = \frac{a}{b}$

3. Variance: $\text{Var}(X) = \frac{a}{b^2}$

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Would you like further clarification on any of these steps?

Here are 5 related questions:

1. How do you find higher moments from the MGF?
2. What is the relationship between the MGF and cumulants?
3. How is the MGF used in hypothesis testing?
4. How does the gamma distribution generalize to other distributions?
5. How do you compute covariances using joint MGFs?

Tip: The moment generating function is a powerful tool for finding moments (like mean and variance) because its nth derivative at $t = 0$ gives the nth moment of the distribution.