# Kolmogorov-Smirnov test

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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So what is your question? You need the maximum discrepancy between the empirical c.d.f. and the hypothetical c.d.f. (i.e. the c.d.f. given by the null hypothesis). The hypothetical c.d.f. is an anti-derivative of the p.d.f. you gave. You neglected to specify that the p.d.f. is equal to the function you gave on a certain interval and $0$ elsewhere. Are you doing stenography and passing on to us a question written by someone else rather than asking a question of your own? – Michael Hardy Apr 1 '12 at 21:36
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)^5,0<x<∞.", and that's it. – nong Apr 2 '12 at 3:24

$$F(x) = \int_0^x f(u)\;dx = \int_0^x \frac{4 \; du}{(1+u)^5} = \left.\frac{-1}{(1+u)^4}\right|_{u=0}^{u=x} = 1-\frac{1}{(1+x)^4}.$$ 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.