91Ó°ÊÓ

The probability density function (PDF) for the gamma distribution is given by \[f_{X}(x) = \begin{cases}\frac{eta^{ heta} x^{ heta-1} e^{-eta x}}{\Gamma(\theta)}, & x > 0 \0, & \text{otherwise}\end{cases}\]where \( \theta = \alpha \) and \( \beta = \lambda \). This matches the given function for \( x > 0 \), confirming the PDF's form is correctly stated.

The expected value (mean) of the gamma distribution is derived as \[\mu_X = \int_{0}^{\infty} x f_X(x) \, dx \]Substitute the PDF into the integral:\[\mu_X = \int_{0}^{\infty} x \frac{\lambda^{\alpha} x^{\alpha-1} e^{-\lambda x}}{\Gamma(\alpha)} \, dx \]This simplifies to \[\frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_{0}^{\infty} x^{\alpha} e^{-\lambda x} \, dx \]Use the substitution \( y = \lambda x \Rightarrow dy = \lambda dx \). Then, the integral becomes:\[\frac{\lambda^{\alpha}}{\Gamma(\alpha) \lambda} \int_{0}^{\infty} \left(\frac{y}{\lambda}\right)^{\alpha} e^{-y} \left(\frac{dy}{\lambda}\right) = \frac{1}{\lambda} \cdot \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_{0}^{\infty} y^{\alpha} e^{-y} \, dy\]The integral represents \( \Gamma(\alpha + 1) \), thus:\[\mu_X = \frac{1}{\lambda} \cdot \frac{\Gamma(\alpha + 1)}{\Gamma(\alpha)} = \frac{1}{\lambda} \cdot \alpha = \frac{\alpha}{\lambda}\]

Variance of a gamma distribution is found using \[\sigma_X^2 = \mathbb{E}[X^2] - (\mu_X)^2\]First, compute \( \mathbb{E}[X^2] \) as\[\mathbb{E}[X^2] = \int_{0}^{\infty} x^2 f_X(x) \, dx = \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_{0}^{\infty} x^{\alpha+1} e^{-\lambda x} \, dx \]Using the substitution \( y = \lambda x \), the integral can be rewritten as:\[\frac{\lambda^{\alpha}}{\Gamma(\alpha) \lambda^2} \int_{0}^{\infty} y^{\alpha+1} e^{-y} \, dy = \frac{1}{\lambda^2} \cdot \frac{\Gamma(\alpha+2)}{\Gamma(\alpha)}\]Simplifying the ratio using \( \Gamma(\alpha+2) = (\alpha+1)(\alpha)\Gamma(\alpha) \), it gives us:\[\mathbb{E}[X^2] = \frac{\alpha(\alpha+1)}{\lambda^2}\]Now calculate the variance:\[\sigma_X^2 = \frac{\alpha(\alpha+1)}{\lambda^2} - \left(\frac{\alpha}{\lambda}\right)^2 = \frac{\alpha(\alpha+1) - \alpha^2}{\lambda^2} = \frac{\alpha}{\lambda^2}\]

From the above steps, we have shown that for the gamma distribution:- The mean \( \mu_X = \frac{\alpha}{\lambda} \).- The variance \( \sigma_X^2 = \frac{\alpha}{\lambda^2} \).These results are consistent with the properties of the gamma distribution.