What is the DC Accuracy When the Waveform Position has Been Changed?

What is the DC Accuracy when waveform position has changed?
I would like to confirm DC Precision when changing the position setting on DL1600. According to the specification of vertical (voltage) axis accuracy, in the case that voltage axis is set to 0.5 V/div: 
  • DC Precision:  ±(1.5% of 8 div + offset voltage precision)
  • Offset Voltage Precision:   ±(1% of set value + 2 mV)  

If DC offset voltage is set to 2.0 V, then the DC precision is ±(1.5% of  8*0.5V + 1% of 2V + 2 mV) = ±82 mV.
However, there is no description about "position" in this equation. What is the equation when I change the position setting?

For example, what is the DC precision when the offset is set to 0 V and the position is set to -4 div?

If DC offset equals 0V and the position is changed, it is necessary to calculate the DC precision supposing the DC offset voltage is set as the same as the voltage corresponding to position changing.
How to calculate DC offset voltage:
Here's how to measure DC offset voltage in the previous example. Since the voltage scale is set to 0.5 V/div, -4 div corresponds to 2V. So the DC precision in this case is the same as the offset voltage is set to 2 V. In conclusion, the DC precision will be:
      ±(1.5% of  8*0.5V + 1% of 4*0.5V + 2 mV)= ±(1.5% of  8*0.5V + 1% of 2V + 2 mV)
    = ±82 mV

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