Could someone clarify what we meant by the note here: p(x) must be nonzero for all x where f(x) is nonzero? Why does f(x) being nonzero matter?
Resolved. p(x) is in the denominator.
kevintli
Wouldn't the variance of this estimator be very high is your function f is defined over a wide range of inputs, since your uniform probability distribution would have to put very low weight on any particular X value, resulting in a low denominator / high sample values? If you're working with very large environments, would it potentially be better to use a a distribution that is more concentrated around some area of interest?
greeknerd1
What does F_N mean intuitively? Is it the average of all values sampled divided by their corresponding probabilities?
ML2000-LT
This estimator, intuitively, means the average, since we have the 1/N term in the front, of the sum of all values divided by their corresponding probability, but it seemed pretty odd that, it utilized division, since normally isn't multiplying by p(X) makes more sense since, with greater probability the value should have more weight?
Could someone clarify what we meant by the note here: p(x) must be nonzero for all x where f(x) is nonzero? Why does f(x) being nonzero matter?
Resolved. p(x) is in the denominator.
Wouldn't the variance of this estimator be very high is your function f is defined over a wide range of inputs, since your uniform probability distribution would have to put very low weight on any particular X value, resulting in a low denominator / high sample values? If you're working with very large environments, would it potentially be better to use a a distribution that is more concentrated around some area of interest?
What does F_N mean intuitively? Is it the average of all values sampled divided by their corresponding probabilities?
This estimator, intuitively, means the average, since we have the 1/N term in the front, of the sum of all values divided by their corresponding probability, but it seemed pretty odd that, it utilized division, since normally isn't multiplying by p(X) makes more sense since, with greater probability the value should have more weight?