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# faq2 - it’s a good choice a lot of the time What is the...

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FREQUENTLY ASKED QUESTIONS October 7, 2011 Content Questions Why might I not want to use the Inverse Transform method? It seems like the obvious choice for most, but maybe not all, circum- stances. It’s a matter of both computing eﬃciency, and accuracy. If the inverse of the integral of the function you’re selecting from can be computed analytically, the inverse transform method is almost certainly the best choice. However if the inverse of the integral must be computed numerically, the cost of that computation could be such that you’re better oﬀ doing acceptance-rejection or something else. Sometimes there’s a fast inversion method for a speciﬁc function, sometimes not. You can always compute a table to do a numerical inverse– however you may need a very ﬁnely-grained table for suﬃcient ac- curacy, and that costs speed and memory. So it may or may not be worth it, depending on the form of the function you are dealing with. But in fact, the inverse transform method is quite commonly used and
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Unformatted text preview: it’s a good choice a lot of the time. What is the distinction between the inverse transform method and saying Q ( v ) = P ( u ) | du dv | , then computing the derivative and plugging in, where P ( u ) is uniform? Is this the same thing? No, this expression can’t be used directly for the problem we’re trying to solve. We know Q ( v ): it’s the function we are trying to pick values from. If u is a uniform variable, P ( u ) is just a constant, 1 / ( b-a ), where [ a,b ] is the interval we are selecting over, and it’s just 1 for [0 , 1]. Selecting u and plugging in to Q ( v ) = | du dv | is just going to give values of Q ( v ), not variables v distributed according to Q ( v ). But if you integrate Q ( v ) = | dF dv | (writing u ( v ) = F ( v )), you get the expression relating the value of uniformly-selected u to Q ( v ): this is the inverse transform expression u = R v F ( v ) dv . Then v = F-1 ( u ) will have the distribution you want....
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