Potential bug in remove_var? 
#### Code Sample, a copy-pastable example if possible

```python
# Your code here
cdef inline void remove_var(double val, double *nobs, double *mean_x,
                            double *ssqdm_x) nogil:
    """ remove a value from the var calc """
    cdef double delta

    # Not NaN
    if val == val:
        nobs[0] = nobs[0] - 1
        if nobs[0]:
            # a part of Welford's method for the online variance-calculation
            # https://en.wikipedia.org/wiki/Algorithms_for_calculating_variance
            delta = val - mean_x[0]
            mean_x[0] = mean_x[0] - delta / nobs[0]
            ssqdm_x[0] = ssqdm_x[0] - ((nobs[0] + 1) * delta ** 2) / nobs[0]
        else:
            mean_x[0] = 0
ssqdm_x[0] = 0


nobs is a pointer to a double and is incremented and decrement accordingly as add_var and remove_var are called. if nobs == 0, then you want to reset mean_x and ssqdm_x to 0 as you do in the else condition. my question is this -- isn't it possible that nobs could be some epsilon value and not exactly equal to 0.0 ? I dont know floating point math that well, and it could be fine since all we do is incrememt by1 or decrement by 1 so it could be fine. just wondering in case this should be:
if abs(nbos[0]) < 1e-11 or something

thanks


```
#### Problem description

[this should explain **why** the current behaviour is a problem and why the expected output is a better solution.]

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#### Expected Output

#### Output of ``pd.show_versions()``

<details>

[paste the output of ``pd.show_versions()`` here below this line]

</details>
