📂Probability Distribution

Derivation of the Beta Distribution from Two Independent Gamma Distributions

Theorem

If two random variables $X_{1},X_{2}$ are independent and $X_{1} \sim \Gamma ( \alpha_{1} , 1)$, $X_{2} \sim \Gamma ( \alpha_{2} , 1)$, then $${{ X_{1} } \over { X_{1} + X_{2} }} \sim \text{beta} \left( \alpha_{1} , \alpha_{2} \right)$$

Explanation

If two data points follow the gamma distribution and are independent, it could be possible to explain the ratio of their sum using probability distribution theory. Specifically, the gamma distribution allows for relatively free movement among various probability distributions, so it’s good to know as a fact.

Derivation¹

Strategy: Direct deduction from the joint density function of gamma distributions.

Definition of gamma distribution: For $k, \theta > 0$, the following probability density function for a continuous probability distribution $\Gamma ( k , \theta )$ is called the gamma distribution. $$f(x) = {{ 1 } \over { \Gamma ( k ) \theta^{k} }} x^{k - 1} e^{ - x / \theta} \qquad , x > 0$$

Definition of beta distribution: For $\alpha , \beta > 0$, the following probability density function for a continuous probability distribution $\text{Beta}(\alpha,\beta)$ is called the beta distribution. $$f(x) = {{ 1 } \over { B(\alpha,\beta) }} x^{\alpha - 1} (1-x)^{\beta - 1} \qquad , x \in [0,1]$$

Since $X_{1}, X_{2}$ is independent, the joint density function $h$ for $x_{1} , x_{2} \in (0, \infty)$ is as follows. $$h(x_{1} , x_{2}) = {{ 1 } \over { \Gamma (\alpha_{1}) \Gamma (\alpha_{2}) }} x_{1}^{\alpha_{1} -1 } x_{2}^{\alpha_{2} -1 } e^{-x_{1} - x_{2}}$$ Now, if we let $Y_{1} := X_{1} + X_{2}$ and $Y_{2} := X_{1} / (X_{1} + X_{2})$, then $x_{1} = y_{1} y_{2}$ and $x_{2} = y_{1} ( 1 - y_{2} )$, so $$J = \begin{vmatrix} y_{2} & y_{1} \\ 1 - y_{2} & -y_{1} \end{vmatrix} = - y_{1} \ne 0$$ Therefore, the joint density function of $Y_{1}, Y_{2}$ for $y_{1} \in (0,\infty) , y_{2} \in (0,1)$ is $$\begin{align*} g(y_{1},y_{2}) =& |y_{1}| {{ 1 } \over { \Gamma (\alpha_{1}) \Gamma (\alpha_{2}) }} (y_{1} y_{2})^{\alpha_{1} -1 } \left[ y_{1} ( 1 - y_{2} ) \right]^{\alpha_{2} -1 } e^{-y_{1}} \\ =& {{ 1 } \over { \Gamma (\alpha_{1}) \Gamma (\alpha_{2}) }} y_{1}^{\alpha_{1} + \alpha_{2} - 1} e^{-y_{1}} \cdot y_{2}^{\alpha_{1} - 1} (1-y_{2})^{\alpha_{2} - 1} \\ =& {{ 1 } \over { \Gamma ( \alpha_{1} + \alpha_{2}) }} y_{1}^{\alpha_{1} + \alpha_{2} - 1} e^{-y_{1}} \cdot {{ \Gamma ( \alpha_{1} + \alpha_{2}) } \over { \Gamma (\alpha_{1}) \Gamma (\alpha_{2}) }} y_{2}^{\alpha_{1} - 1} (1-y_{2})^{\alpha_{2} - 1} \end{align*}$$ The marginal density function of $Y_{1},Y_{2}$ $g_{1}, g_{2}$ is $$g_{1}(y_{1}) = {{ 1 } \over { \Gamma ( \alpha_{1} + \alpha_{2}) }} y_{1}^{\alpha_{1} + \alpha_{2} - 1} e^{-y_{1}} \\ g_{2}(y_{2}) = {{ \Gamma ( \alpha_{1} + \alpha_{2}) } \over { \Gamma (\alpha_{1}) \Gamma (\alpha_{2}) }} y_{2}^{\alpha_{1} - 1} (1-y_{2})^{\alpha_{2} - 1}$$ Thus, $$Y_{1} \sim \Gamma ( \alpha_{1} + \alpha_{2} ,1 ) \\ Y_{2} \sim \text{beta} (\alpha_{1} , \alpha_{2})$$

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