Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Wald distribution ▷ Method of moments

Theorem: Let $y = \left\lbrace y_1, \ldots, y_n \right\rbrace$ be a set of observed data independent and identically distributed according to a Wald distribution with drift rate $\gamma$ and threshold $\alpha$:

$\label{eq:wald} y_i \sim \mathrm{Wald}(\gamma,\alpha), \quad i = 1, \ldots, n \; .$

Then, the method-of-moments estimates for the parameters $\gamma$ and $\alpha$ are given by

$\label{eq:wald-MoM} \begin{split} \hat{\gamma} &= \sqrt{\frac{\bar{y}}{\bar{v}}} \\ \hat{\alpha} &= \sqrt{\frac{\bar{y}^3}{\bar{v}}} \end{split}$

where $\bar{y}$ is the sample mean and $\bar{v}$ is the unbiased sample variance:

$\label{eq:y-mean-var} \begin{split} \bar{y} &= \frac{1}{n} \sum_{i=1}^n y_i \\ \bar{v} &= \frac{1}{n-1} \sum_{i=1}^n (y_i - \bar{y})^2 \; . \end{split}$

Proof: The mean and variance of the Wald distribution in terms of the parameters $\gamma$ and $\alpha$ are given by

$\label{eq:wald-E-Var} \begin{split} \mathrm{E}(X) &= \frac{\alpha}{\gamma} \\ \mathrm{Var}(X) &= \frac{\alpha}{\gamma^3} \; . \end{split}$

Thus, matching the moments requires us to solve the following system of equations for $\gamma$ and $\alpha$:

$\label{eq:wald-mean-var} \begin{split} \bar{y} &= \frac{\alpha}{\gamma} \\ \bar{v} &= \frac{\alpha}{\gamma^3} \; . \end{split}$

To this end, our first step is to express the second equation of \eqref{eq:wald-mean-var} as follows:

$\label{eq:gamma-s1} \begin{split} \bar{v} &= \frac{\alpha}{\gamma^3} \\ & = \frac{\alpha}{\gamma} \cdot \gamma^{-2}\\ & = \bar{y} \cdot \gamma^{-2} \; . \end{split}$

Rearranging \eqref{eq:gamma-s1} gives

$\label{eq:gamma-s2} \gamma^2 = \frac{\bar{y}}{\bar{v}} \; ,$

or equivalently,

$\label{eq:gamma-s3} \gamma = \sqrt{\frac{\bar{y}}{\bar{v}}} \; .$

Our final step is to solve the first equation of \eqref{eq:wald-mean-var} for $\alpha$ and substitute \eqref{eq:gamma-s3} for $\gamma$:

$\label{eq:alpha-s1} \begin{split} \alpha & = \bar{y} \cdot \gamma \\ & = \bar{y} \cdot \sqrt{\frac{\bar{y}}{\bar{v}}}\\ &= \sqrt{\bar{y}^2} \cdot \sqrt{\frac{\bar{y}}{\bar{v}}}\\ &= \sqrt{\frac{\bar{y}^3}{\bar{v}}} \; . \end{split}$

Together, \eqref{eq:gamma-s3} and \eqref{eq:alpha-s1} constitute the method-of-moment estimates of $\gamma$ and $\alpha$.

Sources:

Metadata: ID: P423 | shortcut: wald-mome | author: tomfaulkenberry | date: 2023-10-30, 12:00.