
Distance sampling

Abundance estimation is often accomplished using this equation: \[ \hat{N} = \frac{n}{\hat{p}} \]
Challenges:
Basic idea:
\[ \hat{N} = \frac{n}{\hat{\bar{p}}} \]
Advantages of distance sampling:
Line transects 
Point transects 
In line transect sampling, it is common to record the radial distance and bearing, rather than the perpendicular distance.
However, the analysis must be conducted on the perpendicular distance data.

_Gazella_gutturosa.png)
| Animal | Distance (\(x\)) |
|---|---|
| 1 | 4.4 |
| 2 | 25.3 |
| 3 | 41.8 |
| 4 | 3.1 |
| 5 | 78.5 |
| \(\vdots\) | \(\vdots\) |
| \(n\) | 4.4 |

Fit a detection function, \(\mathrm{g}(x)\), to the data
Assume individuals are “uniformly” distributed with respect to the transect (valid under random sampling): \(p(x) = 1/B\) where \(B\) is the transect half-width.
\(\bar{p}\) is then the average detection function, weighted by \(p(x)\).
\[\bar{p} = \int_0^B g(x)p(x) \mathrm{d} x\].









We still use the same type of detection function (half-normal, hazard, etc…).
We don’t expect a “flat” histogram of distances when \(\bar{p}=1\). Why?
More individuals occur far from the observer than close to the observer because area increases with distance.
When average detection probability is 1, the histogram of detected distances will increase with area because there are more individuals far from the point.
If average detection probability is less than 1, the histogram will increase and then decline.