# In the Dropout paper, why would increasing the dropout increase the error rate if the capacity is constant?

In the original paper on dropout, in section 7.3.2, we see that while keeping $$pn$$ constant, we get a (test) error increase by decreasing retainment below 0.6. Why would that happen? If $$pn$$ is constant, the capacity of the network should be constant, correct?

In the images below (figure 9 from section 7.3) $$n$$ refers to a unit in the network and $$p$$ refers to the probability of retaining a unit in the network. The combination of both, $$pn$$ thus refers to the "effective" number of units. The left image thus shows how training and test error change when varying the probability of retaining a unit ($$p$$) whereby the number of units ($$n$$) is fixed. The right image shows how the errors change when the "effective" number of units ($$pn$$) is fixed, and the amount of units ($$n$$) is adopted with an increasing $$p$$.
The test error is U-shaped with a sweet spot around 0.5 or 0.6 in both cases. In the left image where the number of units ($$n$$) is fixed, this likely occurs since the ensemble-like properties of dropout networks do not emerge with too little dropout ($$p$$ below 0.3) or too much dropout ($$p$$ above 0.8). Those results could however be purely due to a change in capacity. The right image keeps the "capacity" constant and only looks at the "pure dropout effect" since the capacity (namely the number of units $$pn$$) is fixed. This rules out a decrease in capacity as the driving factor. The interaction of the decrease in capacity due to an increase in dropout is hinted at with the following sentence: