Explore BrainMass

Sampling and gausssian distribution

Assume, I have a Gaussian Distribution given by N~(u0,var) where u0 is mean and var is variance by sampling 999 points.

I used Maximum Likelihood to get the above parameters i.e. u0 and var.

Now from these 999 points, I take one point out (lets call it i) and add a new point (lets call it j). What will the new mean and variance be?

Solution Preview

Definition of the mean (xm) of a number (n) of x(n) sample values is given by

xm = 1/n*∑ {x(n)} (1) summed 1 to n

Now if we take one sample with value (i) out then the new mean of n-1 samples will be

xm = 1/(n-1)*∑ {x(n-1)} (2) summed 1 to n-1

Where, by inspection between (1) & (2) the new sum is

∑{x(n-1)} = ∑{x(n)} - i (3) summed 1 to n-1 & 1 to n

(3) in (2) gives

∑{x(n-1)} = 1/(n-1)*{∑{x(n)} - i} (4)

If we then add in a new sample value (j) the new mean { Xm (new)} of (n) sample values will be

xm (new) = 1/n*{∑{x(n)} - i + j} (5)

xm (new) = (j - i)/n + 1/n*∑{x(n)} (5a)