From Wikipedia's depth of field article,
$$\mathrm{DOF} \approx {2u^2NC \over f^2} $$ for a given circle of confusion (\$C\$), focal length (\$f\$), F-number (\$N\$), and distance to subject (\$u\$).
According to the depth of field (DOF) formula, the DOF is linearly proportional to the circle of confusion (\$C\$), in which \$C\$ is also linearly proportional to the pixel size. Thus DOF is linearly proportional to the pixel size.
If that assumption is correct, then does that means some of the DSLR with APS-C sensor size and low pixel number would have a shallower DOF than some of the full frame DSLR with a high pixel number, assuming using full frame equivalent lens by considering the crop factor and same distance to the capturing subject?
Now calculating the ratio between DOF of Full frame (FF) and that of crop body (c), also applying the crop factor to the focal length and f-number, we get:
$$\begin{align*} {\mathrm{DOF_{FF}} \over \mathrm{DOF}_c} &= \frac{{P_\mathrm{FF}\cdot cN / (cf)^2}}{P_c N / f^2} \\ &= {P_\mathrm{FF} \over c P_c}\,, \end{align*}$$
for: \$P\$ = pixel number; \$f\$ = focal length of the lens on the crop body; \$N\$ = f-number of the lens on the crop body; and \$c\$ = crop factor.
Thus, C would have a shallower DOF than that of a FF when using FF equivalent lens, when:
$$ {P_\mathrm{FF} \over P_c} > c\,. $$
For example the Pentax K100D has 6 megapixels with an APS-C sized sensor, comparing to Pentax K-1 with a full frame sensor and 36 megapixels.
As such, [![enter image description here][4]][4]
$$ {\mathrm{DOF_{K-1}}\over\mathrm{DOF_{K100D}}} = {36\over6}\cdot{1\over1.5} = 3.733. $$
The DOF of K-1 would be about 4 times deeper than that of K100D.
Please correct me if I my logic is not sound. Thank you for your time and effort!