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EnumaElish

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What is the goal that you are trying to accomplish?

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EnumaElish

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You could divide each distribution's domain into "bins" (frequency ranges) and use the test that way. A number of non-parametric tests can also be used, e.g. the runs test. See this thread.

P.S. From that earlier thread:

*Originally posted by ***EnumaElish**

There are several non-parametric tests for assessing whether 2 samples are from the same distribution. For example, the "runs" test. Suppose the two samples are [itex]u_1<...<u_n[/itex] and [itex]v_1<...<v_n[/itex]. Suppose you "mix" the samples. If the resulting mix looks something like [itex]u_1< v_1 < u_2 < u_3 < u_4 < v_2 < v_3 <[/itex] ... [itex] < u_{n-1} < v_{n-1} < v_n < u_n[/itex] then the chances that they are from the same distribution is greater than if they looked like [itex]u_1<...<u_n<v_1<...<v_n[/itex]. The latter example has a smaller number of runs (only two: first all u's then all v's) than the former (at least seven runs: one u, one v, u's, v's, ..., u's, v's, one u). This and similar tests are usually described in standard probability textbooks like Mood, Graybill and Boes.

P.S. From that earlier thread:

There are several non-parametric tests for assessing whether 2 samples are from the same distribution. For example, the "runs" test. Suppose the two samples are [itex]u_1<...<u_n[/itex] and [itex]v_1<...<v_n[/itex]. Suppose you "mix" the samples. If the resulting mix looks something like [itex]u_1< v_1 < u_2 < u_3 < u_4 < v_2 < v_3 <[/itex] ... [itex] < u_{n-1} < v_{n-1} < v_n < u_n[/itex] then the chances that they are from the same distribution is greater than if they looked like [itex]u_1<...<u_n<v_1<...<v_n[/itex]. The latter example has a smaller number of runs (only two: first all u's then all v's) than the former (at least seven runs: one u, one v, u's, v's, ..., u's, v's, one u). This and similar tests are usually described in standard probability textbooks like Mood, Graybill and Boes.

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EnumaElish

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yes, or more specifically I would like to show how close to identical they are

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EnumaElish

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1. You could make two variables X(t) = value of the "true" disrtibution (expensive simulation) at point t and Y(t) = value of the alternative dist. (practical simulation) at point t. Then run the regression Y(t) = a + b X(t) for as many t's as you can (or like), then show that the joint hypothesis "(a = 0) AND (b = 1)" is highly statistically significant.

2. Plot X(t) and Y(t) on the same graph. Select a lower bound T0 and an upper bound T1. Let's assume X(T0) = Y(T0) and X(T1) = Y(T1), i.e. both T0 and T1 are crossing points. Divide the interval [T0,T1] into arbitrary subintervals {s(1),...,s(N)}. Define string variable z(i) = "x" if the integral of X(t) - Y(t) > 0 over subinterval s(i); z(i) = "y" otherwise. You'll end up with a string like xxxyyyxyxyx... whose length = N. Now apply the RUNS TEST that I described above.

I may post again if I can think of anything else.

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thank you

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EnumaElish

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N.B. The "runs test" addresses the directionality of the error e(t) = X(t) - Y(t); the regression addresses the magnitude of the errors. Technically, the regression minimizes the sum of e(t)^{2} = sum of [X(t) - Y(t)]^{2} over all t in the sample. Ideally one should apply both techniques to cover the directionality as well as the magnitude of the errors.

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