When would the variance for a probability distribution give the same result as the standard equation? Variance equation for a probability distribution:
$$
\sigma^2 = \sum_{i=1}^{N}(x_i-\mu)^2P(X=x_i)
$$
Standard variance equation:
$$
\sigma^2 = \frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2
$$
I understand that since the expected value of a probability distribution is the same as the mean, the expected value of the square deviations from the mean (Equation 1) should be the same as the mean of the square deviations from the mean (Equation 2). However, they don't give the same answer when tried with the following test scenario.




Birds spotted
Probability




1
0.1


2
0.1


3
0.3


4
0.3


5
0.1


6
0.1




The table gives the probability of spotting the quantity of birds (from 1 to 6 birds) at any given time.
The mean/expected value is the same and is equal to 3.5. But I get 2 different values for the variance - 1.849 (using equation 1) and 2.916 (using equation 2).
Question: Under what circumstances would the 2 equations give the same result? I'm not sure I fully understand equation 1.
I've written the following python script to compare the results from the 2 equations:
import numpy as np
a = [1,2,3,4,5,6]
prob = [0.1, 0.1, 0.3, 0.3, 0.1, 0.1]

# Variance
variancePD = 0
m = np.mean(a)
for i in range(0,6):
  variancePD += (((a[i] - m)**2) * prob[i])

varianceNormal = 0
for i in range(0,6):
  varianceNormal += ((a[i] - m)**2)
varianceNormal /= len(a)

# Mean
meanPD = 0
for i in range(0,6):
  meanPD += (a[i] * prob[i])

print(f'variancePD: {variancePD}, varianceNormal: {varianceNormal}\
      \nmeanPD: {meanPD}, mean: {m}')


You are actually applying the same formula for the variance (of a discrete random variable) to two different random variables $X$.
Indeed, in the first case, $X$ takes on values $x_{1},\ldots,x_{N}$ with probability $P(X=x_i)$ for $i=1,\ldots,N$. Whereas in the second case, $X$ takes on the same values but $P(X=x_i)=1/N$.
Thus, following your numerical example, in the first case, the variance is
$$\sigma^2 = 0.1(1-3.5)^2+0.1(2-3.5)^2 + \cdots+0.1(6-3.5)^2 = 1.85$$
In the second case, it would be
$$\sigma^2 = (1/6)(1-3.5)^2+(1/6)(2-3.5)^2 + \cdots+(1/6)(6-3.5)^2 = 2.917$$
Agreement between the "two formulas" is obtained only if the two probability distributions are the same.
