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true |
Thomas J. Faulkenberry |
Tarleton State University |
faulkenberry@tarleton.edu |
2023-10-30 05:00:00 -0700 |
Method of moments for Wald-distributed data |
Probability Distributions |
Univariate continuous distributions |
Wald distribution |
Method of moments |
|
P423 |
wald-mome |
tomfaulkenberry |
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 $\alpha$.