I don't think I understand what you mean in your assumption, so I can't tell you if it's correct.
The way I think of an "open drain" is: there is a switch inside the IC that either pulls that line to ground, or doesn't pull it to ground. If the signal is to be low, the switch is on, shorting the line to ground, if the signal is to be high, the switch is open, letting it float up to the relevant voltage level. The output is reliant on the voltage rail that the pull up resistor is connected to and the value of the resistor. A very large resistor will mean that the line will take a long time to increase (due to a large resistor providing a low current to charge the parasitic capacitance of the signal line).
Hence, when you ask for "current and voltage" to calculate the pull up value, that isn't quite right. What you're really after is working out the pull up value based on the rise time you want and the capacitance of the signal line (including the pins of connected ICs, the data sheets are usually good at providing this). As a rule of thumb, the charge time is about 5 times the RC constant (5*R*C, where R is pull up resistor and C is the capacitance of all items on the signal line).
In this event, the open drain is on the PGOOD signal, which means that the signal is always going to be low until the rail you're pulling up to is high, and then it will only become high once the power rail you're creating here is within tolerance. As these signals don't need to be particularly quick, the resistors are quite large (often 10K) to reduce power consumption when pulled low. Calculating the resistor value is more critical when dealing with an open drain communication link, such as I2C (or version thereof).
Quick summary: Open drain is either pulled to ground when low, or left floating for high. The value of the pull up resistor is based on the rise time you desire and the capacitance of the signal line and devices on the signal line.