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 based 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 the 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:
- The exponential detectability function:
\(g(x)=\text{exp}(-x/w)\)
The corresponding maximum likelihood estimator for \(\textbf{w}\):
\(\hat{w}=\bar{x}\)
- 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 of 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 of Half-Normal Detectability
(with the data from the bird example)
\((1/n)\Sigma x_{i}^{2}=(1/18)(0^2 + \cdots + 44^2)=417.5\)
\(\hat{w}=\sqrt{\frac{\pi}{2y}\sum_{i=1}^{y}x_{i}^{2}}=\sqrt{\left(\frac{3.1417}{2}\right)(417.5)}=25.61\)
\(\hat{D}=\frac{y}{2L\hat{w}}=\frac{18}{2 \times 100 \times 25.61}=0.0035\)