Question

Illustrative Example \(8.2 .1\) of this section dealt with a random sample of size \(n=2\) from a gamma distribution with \(\alpha=1, \beta=\theta .\) Thus the mgf of the distribution is \((1-\theta t)^{-1}, t<1 / \theta, \theta \geq 2 .\) Let \(Z=X_{1}+X_{2} .\) Show that \(Z\) has a gamma distribution with \(\alpha=2, \beta=\theta\). Express the power function \(\gamma(\theta)\) of Example 8.2.1 in terms of a single integral. Generalize this for a random sample of size \(n\).

Short answer

Sum of independent gamma-distributed random variables also follows gamma distribution. Specifically, in this case, \(Z = X_{1} + X_{2}\), where both \(X_1\) and \(X_2\) follows gamma distribution with \(\alpha=1, \beta=\theta\), is also gamma distributed with parameters \(\alpha=2, \beta=\theta\). Given this fact, the power function was expressed as a single integral. This was then generalized for a random sample of size \(n\).

Step-by-step solution

Understanding the distribution of Sample Sum

The sum of two independent random variables that are Gamma distributed is also Gamma distributed, with parameters \(\alpha_{1} + \alpha_{2}\) and \(\beta\) if both variables share the same scale parameter \(\beta\). Here, \(Z = X_{1} + X_{2}\) is the sum of two independent samples from a gamma distribution with \(\alpha=1, \beta=\theta\). That gives \(Z\) a gamma distribution with parameters \(\alpha=2, \beta=\theta\)

Express the Power Function in Terms of a Single Integral

The power function \(\gamma(\theta)\) of Example 8.2.1 can be expressed in terms of a single integral as: \(\gamma(\theta) = \int_0^\infty \frac{1}{\Gamma(2)} z e^{-z/\theta} (\frac{z}{\theta})^{2-1} dz\), where \( \Gamma(2) = 1\). This integral represents the probability density function of \( Z \), a gamma distributed variable with parameters (2, \(\theta\)).

Generalizing for Random Sample of Size \(n\)

To generalize this for a sample size \(n\), consider \(Z = X_{1} + X_{2} + ... + X_{n}\), where all \(X_i\) follows same gamma distribution with \(\alpha=1, \beta=\theta\). Then \(Z\) is also gamma-distributed with parameters \(\alpha=n, \beta=\theta\). The power function can thus, be expressed as a single integral: \( \gamma(\theta) = \int_0^\infty \frac{1}{\Gamma(n)} z e^{-z/\theta} (\frac{z}{\theta})^{n-1} dz \)