The moment generating function (MGF) of the chi-square distribution can be derived based on the definition of the MGF and the properties of the chi-square distribution.

Definition of Moment Generating Function (MGF)

The MGF of a random variable $X$ is given by:

$M_X(t) = \mathbb{E}[e^{tX}]$

For the chi-square distribution with $k$ degrees of freedom, denoted as $X \sim \chi^2(k)$, the probability density function (PDF) is:

$f_X(x) = \frac{x^{(k/2) - 1} e^{-x/2}}{2^{k/2} \Gamma(k/2)}, \quad x \geq 0$

Deriving the MGF

Now, to find the MGF, we compute $\mathbb{E}[e^{tX}]$:

$M_X(t) = \int_0^\infty e^{tx} \frac{x^{(k/2) - 1} e^{-x/2}}{2^{k/2} \Gamma(k/2)} \, dx$

Simplifying the exponential terms:

$M_X(t) = \frac{1}{2^{k/2} \Gamma(k/2)} \int_0^\infty x^{(k/2) - 1} e^{-(1/2 - t)x} \, dx$

This integral is of the form of the Gamma distribution, where the MGF of a Gamma random variable with shape parameter $\alpha$ and rate parameter $\beta$ is:

$M_X(t) = (1 - \frac{t}{\beta})^{-\alpha} \quad \text{for} \quad t < \beta$

In the case of the chi-square distribution, the shape parameter $\alpha = k/2$ and the rate parameter $\beta = 1/2$. Therefore, the MGF of the chi-square distribution is:

$M_X(t) = \left( 1 - 2t \right)^{-k/2} \quad \text{for} \quad t < \frac{1}{2}$

Final Answer:

Thus, the moment generating function of a chi-square distribution with $k$ degrees of freedom is:

$M_X(t) = \left( 1 - 2t \right)^{-k/2}, \quad t < \frac{1}{2}$

Would you like a more detailed explanation of any part of this derivation? Or do you have any questions on the concept?

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Here are some related questions you might find interesting:

1. What are the moments of the chi-square distribution?
2. How does the moment generating function help in finding the moments of a distribution?
3. What is the relationship between the chi-square distribution and the Gamma distribution?
4. How can the MGF be used to find the distribution of the sum of independent chi-square random variables?
5. How does the MGF of the chi-square distribution relate to its characteristic function?

Tip: Always check the range of $t$ when working with MGFs, as it determines where the function converges!