unten eingefügt ist ein Zusatzmaterial aus einem Artikel. Nach dem Finden der Werte für den Heimatbereich sagen die Autoren: "Gamma (13,10) deckt dies gut ab". Wie fanden die Autoren 13 und 10?Ermitteln der Form und des Maßstabs einer Gamma-Verteilung für den angegebenen Bereich
# Moldenhauer and Regelski (1996) state that home range size typically
# ranges from 0.08-0.65 ha, which equates to a radius of
sqrt(0.08/pi*10000) # 15.96 m or
sqrt(0.65/pi*10000) # 45.49 m
# Since parulas were detected by song, let's be safe
# and add a minimum and maximum
# distance at which they could be heard, 50 and 250 m, based upon
# Simons et al. (2009).
# So now we have an area between
(15.96+50)^2*pi/10000 # 1.37 ha, and
(45.49+250)^2*pi/10000 # 27.4 ha
# We note that this is a very large range.
# Following Royle et. al (2011), and assuming a
# chi-squared distribution with 2 degrees of freedom,
# the range of sigma is given by
sqrt(1.37*10000/pi)/sqrt(5.99) # 27 m
sqrt(27.4*10000/pi)/sqrt(5.99) # 120 m
# In our grid spacing, 1 unit = 50m, so our we want a prior with most
# of the density between:
27/50 # 0.54
121/50 # 2.42
# Gamma(13, 10) covers this nicely
qgamma(c(0.001, 0.5, 0.999), 13, 10)
plot(function(x) dgamma(x, 13, 10), 0, 5, xlim=c(0, 3), ylim=c(0, 1.5))**