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Five observations from an underlying loss distribution are: $0.1,\ 0.2,\ 0.5,\ 0.7,\ 1.3$. Find the value of the Kolmogorov-Smirnov test statistic for test that the underlying distribution has p.d.f. $f(x) = 4/(1+x)^5$.

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    The full question is "Five observations from an underlying loss distribution are: 0.1, 0.2, 0.5, 0.7, 1.3. Find the value of the Kolmogorov-Smirnov test statistic for test that the underlying distribution has p.d.f. f(x) = 4/(1+x)^5,02012-04-02

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$ F(x) = \int_0^x f(u)\;dx = \int_0^x \frac{4 \; du}{(1+u)^5} = \left.\frac{-1}{(1+u)^4}\right|_{u=0}^{u=x} = 1-\frac{1}{(1+x)^4}. $ Now notice that at $0.1$, the empirical c.d.f. leaps from $0$ up to $1/5 = 0.2$. The value of the hypothetical c.d.f. at that point is $F(0.1) = 1-\dfrac{1}{(1+0.1)^4}$. Find the amount by which that differs from $0$ and from $0.2$. Find the larger of the two. Do the same at the other four data points. Find the largest of the numbers that you get. That's the maximum discrepancy statistic. In some table (or maybe you get it from software), you should be able to find where that is in the distribution of the Kolmogorov-Smirnov distribution for samples of size $5$, and thus whether it's significant.