Lecture 12: Monte Carlo Integration (51)

Given our observations in slide 40, 41, and 42, it makes me believe that a pp function that resembles the original ff function (off by a scale factor) is probably best for estimating the integral of f.f. Indeed, then, the higher the value of ff (so the more impact it has on the integral value), the more we sample at it. However, actually choosing p(x)=f(x)/cp(x) = f(x) / c, where cc is a constant, feels so unhelpful/pointless to me: in order to sample in accordance with the pp distribution, we'd need to know the CDF of pp, or arp(x)dx\int_a^r p(x)\, dx for different values of rr. Since pp is a scale factor of f,f, this is essentially the same as just finding the integral of arf(x)dx,\int_a^r f(x) \, dx, which is what we're setting out to do in the first place (in fact, the value of cc that is the scale factor for p(x)p(x) is simply abf(x)dx\int_a^b f(x) \, dx!)

So obviously, we don't want p(x)=f(x)abf(x)dx.p(x) = \frac{f(x)}{\int_a^b f(x)\, dx}. However, even estimating p(x)p(x) as approximately f(x)abf(x)dx\frac{f(x)}{\int_a^b f(x)\, dx} seems inefficient, as I believe means that we are essentially estimating f(x)f(x) and therefore abf(x)\int_a^b f(x) in the process anyways. So, I guess my question is: in practice, how do we actually get a good enough function for p(x)p(x)? Clearly, we don't want a very very good function; how do we decide what's good enough with respect to efficiency?

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