13.4 - Parametric Methods

13.4 - Parametric Methods

We will introduce a new concept here: Effective Half-Width of the Transect: $$\textbf{w}$$

Imagine an equivalent strip plot, with perfect detectability out to some distance $$\textbf{w}$$, in which the same number of animals would be seen, on average, as are seen from the transect with decreasing detectability. We have:

$$f(0)=\dfrac{1}{w}$$

In terms of effective half-width, the density estimate vased on an estimate of $$\textbf{w}$$ is:

$$\hat{D}=\dfrac{y}{2L\hat{w}}$$

Given the above formula, one may proceed either to estimate $$f(0)$$ or to estimate $$\textbf{w}$$.

In parametric method, we assumed that the distribution of the detection distance $$x$$ will have the same shape as the detectability function.

There are numerous density functions $$g(x)$$, we introduce two of them:

1. The exponential detectability function:

$$g(x)=\text{exp}(-x/w)$$

The corresponding maximum likelihood estimator for $$\textbf{w}$$:

$$\hat{w}=\bar{x}$$

2. The half-normal detectability function:

$$g(x)=\text{exp}\left( \dfrac{-\pi x^2}{4w^2} \right)$$

The corresponding maximum likelihood estimate for w:

$$\hat{w}=\sqrt{\dfrac{\pi}{2y}\sum_{i=1}^{y}x_{i}^{2}}$$

Example for Exponential Detectability Function

(with the data from the bird example)

$$\bar{x}=(0+0+1+ \cdots + 44)/18=16.39$$

$$\hat{D}=\frac{y}{2L\hat{w}}=\frac{18}{2 \times 100 \times 16.39}=0.055$$

Example for Half-Normal Detectability

(with the data from the bird example)

$$(1/n)\Sigma x_{i}^{2}=(1/18)(0^2 + \cdots + 44^2)=417.5$$

$$\hat{w}=\sqrt{\frac{\pi}{2y}\sum_{i=1}^{y}x_{i}^{2}}=\sqrt{\left(\frac{3.1417}{2}\right)(417.5)}=25.61$$

$$\hat{D}=\frac{y}{2L\hat{w}}=\frac{18}{2 \times 100 \times 25.61}=0.0035$$

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